Solved Math Exams

A1

Find

a

and

b

such that the following function is differentiable everywhere.

f(x) = \begin{cases} x^2 + 6x + 12 & \text{if } x < -2 \\ ax + b & \text{if } x \geq -2 \end{cases}

Exercise Tags

continuity

Piecewise functions

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Step 1: Ensure Continuity at

x = -2

For

f(x)

to be differentiable everywhere, it must first be continuous everywhere. This means the two pieces of the function must meet at the point where their definitions change, which is at

x = -2

. The value of the top function at

x = -2

must equal the value of the bottom function at

x = -2

.

Evaluate the top part of the function at

x = -2

:

\begin{aligned}& f(-2) = (-2)^2 + 6(-2) + 12 \\& \phantom{f(-2)} = 4 - 12 + 12 \\& \phantom{f(-2)} = 4\end{aligned}

Now, set the bottom part of the function equal to this value at

x = -2

:

\begin{aligned}& a(-2) + b = 4 \\& -2a + b = 4 \quad (Equation \ 1)\end{aligned}

Step 2: Ensure Differentiability (Smoothness) at

x = -2

For

f(x)

to be differentiable at

x = -2

, the slopes of the two pieces must be equal at that point. We find the slope by taking the derivative of each part of the function.

Take the derivative of the top part,

x^2 + 6x + 12

:

\begin{aligned}& \frac{d}{dx}(x^2 + 6x + 12) = 2x + 6\end{aligned}

Evaluate this derivative at

x = -2

:

\begin{aligned}& f'(-2) = 2(-2) + 6 \\& \phantom{f'(-2)} = -4 + 6 \\& \phantom{f'(-2)} = 2\end{aligned}

Take the derivative of the bottom part,

ax + b

:

\begin{aligned}& \frac{d}{dx}(ax + b) = a\end{aligned}

For differentiability, the derivatives must be equal at

x = -2

:

\begin{aligned}& a = 2 \quad (Equation \ 2)\end{aligned}

For a piecewise function to be differentiable at the point where its definition changes, both the function values and their derivatives must be equal at that point.

Step 3: Solve for

a

and

b

Now we have a system of two equations with two unknowns:
1.

-2a + b = 4

2.

a = 2

Substitute

a = 2

from Equation 2 into Equation 1:

\begin{aligned}& -2(2) + b = 4 \\& -4 + b = 4 \\& b = 4 + 4 \\& b = 8\end{aligned}

Thus, the values for

a

and

b

that make the function differentiable everywhere are

a = 2

and

b = 8

.

The final solution is:

\boxed{a = 2, b = 8}

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The final solution is:

\boxed{a = 2, \quad b = 8}

A2

Consider the function

f(x)

whose graph is given below. Find the set of all numbers

x

satisfying the following properties. No justifications required.

Exercise Tags

Curve Sketching

concavity

critical points

find critical numbers

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Sub-questions:

A2 Part a)

f

is not differentiable at

x

.

Exercise Tags

graphing

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Let's start from left to right.

The first place to look is where

x=-4

. But this is just a derivative of 0, so that's totally fine.

Next we look at

x=-3

. This is a 'cusp' since it's a 'sharp turn' or 'corner'.

Any 'sharp turn' in a function are called a cusp and the derivative does not exist there.

So at

x=-3

we know the derivative does not exist.

Next we look at

x=-1

. Since the function is not even continuous there, we therefore also know that the function is not differentiable there.

So we add

x=-1

to our list.

The next point in question is when

x=1

. Since it looks like the slope is completely vertical at

x=1

, that makes the slope essentially infinite. That means the derivative does not exist there!

The derivative does not exist for any

x

value where the function is totally vertical.

Moving on to the next point in question would be where

x=4

. The function is not continuous there so therefore it is also not differentiable.

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\{ x \mid x \in \{-3, -1, 1\} \}

which is the same as:

x=-3, x=-1, x=1

A2 Part b)

f'(x) < 0

.

Exercise Tags

graphing

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This is asking for all the

x

values where the slope is strictly less than 0.

From the graph we can see the slope is negative starting when

x = -1

and continues to be negative until

x=2

.

There are two important exceptions when

x=-1

and when

x=1

. The derivative does not exist at either of those points, so we can't include them in our final answer. Let's explain this:

When

x=-1

we have no left side limit, so the derivative does not exist.

When

x=1

the derivative does not exist because the left hand limit and right hand limit both approach infinity.

So for our final answer we have only:

\{ x \mid -1 < x < 2 \text{ and } x \neq 1 \}

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The set of

x

such that

x

is in the interval

[-1, 2]

but not equal to

-1

or

1

is given by:

\{ x \mid -1 < x < 2 \text{ and } x \neq 1 \}

Another way to say it is this:

(-1, 2) \setminus \{-1, 1\}

A2 Part c)

f''(x) =0

.

Exercise Tags

graphing

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Here we are looking for

x

values where the second derivative is 0.

NOTE: These points do not necessarily indicate an inflection point! For instance take the parabola

y=x^4

when

x=0

.

From the graph we can see an inflection point at

x=-4

together with a tangent line of finite slope (in this case the slope is 0), so the second derivative must be 0 there.

We also definitively have an inflection point when

x=5

since the graph changes from concave up to concave down together with a tangent line. So for sure we have

f''(x)=0

here at

x=5

too.

We can't really tell what's happening when

x=2

, but for this case the answer key will almost certainly not include this point. It's a local min which can indeed sometimes have

f''(x)=0

(an example was given earlier with

f(x)=x^4

at

x=0

).

Careful with

x=1

. Since the first derivative doesn't exist there, the second derivative can't exist either. So we have

f''(x) =0

only at

x = -4

and

x=5

.

Final solution is:

\{ x \mid x = -4, x=5 \}

or just :

\{-4, 5 \}

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Final solution is:

\{ x \mid x = -4, x=5 \}

or just :

\{-4, 5 \}

A2 Part d)

f''(x) >0

.

Exercise Tags

concavity

graphing

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Here we are looking for when the graph

f(x)

is concave up.

We can see a small concave up segment from when

x

is between

-4

and

-3

.

Then it looks like another small segment when

x

is between 1 and 4.

Then finally there is a small interval where

f(x)

is concave up right near then end when

x

is between 4 and 5.

Note that x=4 is not included.

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(-4, -3) \cup (1, 4) \cup (4, 5)

A2 Part e)

f

has an absolute minimum at

x

.

Exercise Tags

graphing

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The absolute minimum can clearly be seen when

x=2

.

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The absolute minimum can clearly be seen when

x=2

.

A3

Find the derivative of

\sqrt{\cosh(2x)}

Exercise Tags

hyperbolic trig functions

taking derivatives

Differentiation: general

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Step 1: Recall the derivative of

\cosh(u)

The derivative of

\cosh(u)

with respect to

u

is:

\frac{d}{du} \cosh(u) = \sinh(u)

Step 2: Use the chain rule

We are differentiating

\cosh(2x)

, so we need to apply the chain rule because we have a function inside another function (i.e.,

2x

inside

\cosh

).

According to the chain rule, if you have a composite function

\cosh(g(x))

, its derivative is:

\frac{d}{dx} \cosh(g(x)) = \sinh(g(x)) \cdot g'(x)

Step 3: Apply the chain rule to

\cosh(2x)

Here,

g(x) = 2x

, so:

\frac{d}{dx} \cosh(2x) = \sinh(2x) \cdot \frac{d}{dx}(2x)

Step 4: Differentiate

2x

The derivative of

2x

with respect to

x

is:

\frac{d}{dx}(2x) = 2

Step 5: Combine the results

Now, substitute the derivative of

2x

into the expression:

\frac{d}{dx} \cosh(2x) = \sinh(2x) \cdot 2

Thus, the derivative of

\cosh(2x)

is:

\boxed{\frac{d}{dx} \cosh(2x) = 2\sinh(2x)}

Summary

\boxed{\frac{d}{dx} \cosh(2x) = 2\sinh(2x)}

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\boxed{\frac{d}{dx} \cosh(2x) = 2\sinh(2x)}

A4

Find the derivative of

\frac{\arcsin(x)}{\arccos(x)}

Exercise Tags

quotient rule

taking derivatives

inverse trig functions

Differentiation: general

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Step 1: Recall the quotient rule

To find the derivative of:

\frac{\arcsin(x)}{\arccos(x)},

we will apply the quotient rule.

The quotient rule states that for a function of the form

\frac{f(x)}{g(x)}

, the derivative is:

\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{g(x) f'(x) - f(x) g'(x)}{(g(x))^2}.

Here,

f(x) = \arcsin(x)

and

g(x) = \arccos(x)

.

Step 2: Derivatives of the functions

We need the derivatives of both

f(x)

and

g(x)

:

- The derivative of

f(x) = \arcsin(x)

is:

f'(x) = \frac{1}{\sqrt{1 - x^2}}.

- The derivative of

g(x) = \arccos(x)

is:

g'(x) = \frac{-1}{\sqrt{1 - x^2}}.

Step 3: Apply the quotient rule

Substitute the functions and their derivatives into the quotient rule:

\begin{aligned}& \frac{d}{dx} \left( \frac{\arcsin(x)}{\arccos(x)} \right) \\& = \frac{\arccos(x) \cdot \frac{1}{\sqrt{1 - x^2}} - \arcsin(x) \cdot \frac{-1}{\sqrt{1 - x^2}}}{(\arccos(x))^2}.\end{aligned}

Step 4: Simplify the expression

Simplify the numerator:

\frac{\arccos(x)}{\sqrt{1 - x^2}} + \frac{\arcsin(x)}{\sqrt{1 - x^2}}.

Using the identity

\arcsin(x) + \arccos(x) = \frac{\pi}{2}

, the expression becomes:

\frac{\frac{\pi}{2}}{\sqrt{1 - x^2}}.

Using the identity

\arcsin(x) + \arccos(x) = \frac{\pi}{2}

is a key trick that simplifies many trigonometric expressions involving inverse trigonometric functions.

Step 5: Final simplified result

Thus, the derivative of

\frac{\arcsin(x)}{\arccos(x)}

is:

\boxed{\frac{\pi}{2\sqrt{1 - x^2}}}.

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\boxed{\frac{\pi}{2\sqrt{1 - x^2}}}.

A5

Find the derivative of

e^{(x^e)} + x^{(e^x)}

Exercise Tags

Differentiation: general

differentiation: logarithmic

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Step 1: Understand the expression

The expression is

e^{(x^e)} + x^{(e^x)}

, which contains two terms:

1. The first term is

e^{(x^e)}

. This is an exponential function where the base is

e

and the exponent is

x^e

.
2. The second term is

x^{(e^x)}

. Here, the base is

x

, and the exponent is

e^x

.

For differentiation, we need to treat each term carefully using rules such as the chain rule and logarithmic differentiation.

Step 2: Differentiate the first term

The first term is

e^{(x^e)}

. Applying the chain rule:

\frac{d}{dx} e^{(x^e)} = e^{(x^e)} \cdot \frac{d}{dx}(x^e).

Now, differentiate

x^e

. Since

e

is a constant, the derivative of

x^e

is:

\frac{d}{dx}(x^e) = e \cdot x^{e - 1}.

So, the derivative of the first term becomes:

\frac{d}{dx} e^{(x^e)} = e^{(x^e)} \cdot e \cdot x^{e - 1}.

Step 3: Differentiate the second term

The second term is

x^{(e^x)}

. To differentiate this, we use logarithmic differentiation. Let:

y = x^{(e^x)} \implies \ln y = e^x \ln x.

Now differentiate both sides implicitly:

\frac{1}{y} \cdot \frac{dy}{dx} = e^x \ln x + \frac{e^x}{x}.

Solving for

\frac{dy}{dx}

:

\frac{dy}{dx} = x^{(e^x)} \cdot \left( e^x \ln x + \frac{e^x}{x} \right).

Step 4: Combine the results

Now, combine the derivatives of both terms:

\begin{aligned}&\frac{d}{dx} \left( e^{(x^e)} + x^{(e^x)} \right) \\&= e^{(x^e)} \cdot e \cdot x^{e - 1} \\&\quad + x^{(e^x)} \cdot \left( e^x \ln x + \frac{e^x}{x} \right)\end{aligned}

Thus, the derivative of the expression is:

\boxed{\begin{aligned}e^{(x^e)} \cdot e \cdot x^{e - 1} &+ x^{(e^x)} \cdot \left( e^x \ln x + \frac{e^x}{x} \right)\end{aligned}}.

Prof's perspective

This question tests your understanding of **differentiation rules for composite and complex functions**. The goal is to assess whether you can correctly apply the chain rule for exponential functions and **logarithmic differentiation** for functions where both the base and exponent are dependent on

x

.

By mastering this problem, you are developing the ability to handle advanced differentiation tasks involving multiple rules, which is essential for success in calculus.

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\boxed{e^{(x^e)} \cdot e \cdot x^{e - 1} + x^{(e^x)} \cdot \left( e^x \ln x + \frac{e^x}{x} \right)}.

A6

Compute the following limit:

\lim_{x \to 0} \frac{5^{(1/x)}}{2^{(1/x^2)}}

Exercise Tags

limits: with logarithms

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Step 1: Understand the expression

We are tasked with finding the limit:

\lim_{x \to 0} \frac{5^{(1/x)}}{2^{(1/x^2)}}

This expression involves two exponential terms:

5^{(1/x)}

and

2^{(1/x^2)}

.

Step 2: Investigate the behavior as

x \to 0

Let’s analyze the behavior of both terms:

1. The term

5^{(1/x)}

: As

x \to 0

, the exponent

1/x

grows large, making

5^{(1/x)}

very large.

2. The term

2^{(1/x^2)}

: As

x \to 0

, the exponent

1/x^2

grows even faster, making

2^{(1/x^2)}

grow much larger than

5^{(1/x)}

.

This results in an indeterminate form:

\frac{\infty}{\infty}

.

Step 3: Apply logarithms

To simplify the expression, we take the natural logarithm. Define:

L = \lim_{x \to 0} \frac{5^{(1/x)}}{2^{(1/x^2)}}

Taking the natural logarithm of both sides:

\ln L = \lim_{x \to 0} \left( \ln 5^{(1/x)} - \ln 2^{(1/x^2)} \right)

Simplifying further:

\ln L = \lim_{x \to 0} \left( \frac{\ln 5}{x} - \frac{\ln 2}{x^2} \right)

This limit involves terms that grow at different rates as

x \to 0

, making it useful to find a common denominator to better analyze their behavior.

Step 4: Get a common denominator

The least common denominator between

x

and

x^2

is

x^2

. Rewrite each term with this common denominator:

\frac{\ln 5}{x} = \frac{\ln 5 \cdot x}{x^2}, \quad \frac{\ln 2}{x^2} = \frac{\ln 2}{x^2}.

Now, combine the terms:

\frac{\ln 5 \cdot x - \ln 2}{x^2}.

Step 5: Evaluate the limit

As

x \to 0

, the term

\ln 5 \cdot x

vanishes, leaving:

\lim_{x \to 0} \left( \frac{\ln 5 \cdot x - \ln 2}{x^2} \right) = \lim_{x \to 0} \frac{-\ln 2}{x^2}.

Since the numerator is a negative constant and the denominator grows large, the limit is:

\lim_{x \to 0} \frac{-\ln 2}{x^2} = -\infty.

Step 6: Evaluate the original limit

Since:

\ln L \to -\infty \implies L \to 0

Step 5: Conclusion

The limit is:

\boxed{\lim_{x \to 0} \frac{5^{(1/x)}}{2^{(1/x^2)}} = 0}

Prof's perspective

The professor likely asked this question to assess your understanding of indeterminate forms and your ability to apply logarithmic transformations effectively. This problem reinforces the idea that in calculus, growth rates of different functions (like exponential terms) are crucial for evaluating limits.

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\boxed{\lim_{x \to 0} \frac{5^{(1/x)}}{2^{(1/x^2)}} = 0}

A7

Compute the following limit:

\lim_{x \to 0} \frac{\sinh(x) - x}{\sin(x) - x}

Exercise Tags

L'hopitals rule

limits: general

hyperbolic trig functions

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Step 1: Identify the form of the limit

We are tasked with solving:

\lim_{x \to 0} \frac{\sinh(x) - x}{\sin(x) - x}

As

x \to 0

, both

\sinh(x) - x

and

\sin(x) - x

approach 0. This results in the indeterminate form

\frac{0}{0}

, making this a case for applying L'Hopital's Rule.

Remember, L'Hopital's Rule applies when the limit results in either

\frac{0}{0}

or

\frac{\infty}{\infty}

.

Step 2: Differentiate the numerator and denominator

We apply L'Hopital's Rule by differentiating the numerator and denominator separately.

1. **Differentiate the numerator**

\sinh(x) - x

:
- The derivative of

\sinh(x)

is

\cosh(x)

, and the derivative of

x

is 1. Thus:

\frac{d}{dx} \left( \sinh(x) - x \right) = \cosh(x) - 1

2. **Differentiate the denominator**

\sin(x) - x

:
- The derivative of

\sin(x)

is

\cos(x)

, and the derivative of

x

is 1. Thus:

\frac{d}{dx} \left( \sin(x) - x \right) = \cos(x) - 1

Step 3: Apply the limit again

Now, the limit becomes:

\lim_{x \to 0} \frac{\cosh(x) - 1}{\cos(x) - 1}

Evaluating both the numerator and the denominator at

x = 0

:

-

\cosh(0) = 1

, so

\cosh(x) - 1 \to 0

.
-

\cos(0) = 1

, so

\cos(x) - 1 \to 0

.

We encounter the indeterminate form

\frac{0}{0}

again, so we apply L'Hopital's Rule a second time.

Applying L'Hopital's Rule multiple times is perfectly valid when necessary.

Step 4: Differentiate again

1. **Differentiate the numerator**

\cosh(x) - 1

:
- The derivative of

\cosh(x)

is

\sinh(x)

.

\frac{d}{dx} \left( \cosh(x) - 1 \right) = \sinh(x)

2. **Differentiate the denominator**

\cos(x) - 1

:
- The derivative of

\cos(x)

is

-\sin(x)

.

\frac{d}{dx} \left( \cos(x) - 1 \right) = -\sin(x)

Step 5: Evaluate the limit

Now, the limit becomes:

\lim_{x \to 0} \frac{\sinh(x)}{-\sin(x)}

For small

x

, we use the small-angle approximations

\sinh(x) \approx x

and

\sin(x) \approx x

. Thus:

\lim_{x \to 0} \frac{x}{-x} = -1

For very small values of

x

,

\sin(x)

and

\sinh(x)

behave similarly, leading to:

\lim_{x \to 0} \frac{\sinh(x)}{\sin(x)} = 1

Step 6: Final Answer

Thus, the value of the limit is:

\boxed{-1}

Prof's perspective

This problem tests your ability to use L'Hopital's Rule effectively and repeatedly. It emphasizes the importance of knowing small-angle approximations for functions like

\sin(x)

and

\sinh(x)

. Additionally, the exercise aims to develop an intuition for recognizing when two functions behave similarly for small inputs.

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The value of the limit is:

\boxed{-1}

A8

Compute the following limit:

\lim_{x \to \frac{\pi}{2}} \sec(x) - \tan(x)

Exercise Tags

L'hopitals rule

limits: general

trig identities

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Step 1: Identify the form of the limit

We are tasked with solving:

\lim_{x \to \frac{\pi}{2}} \sec(x) - \tan(x)

As

x

approaches

\frac{\pi}{2}

, both

\sec(x)

and

\tan(x)

tend to infinity, creating the indeterminate form

\infty - \infty

. To properly evaluate this limit, we need to manipulate the expression into a form suitable for limit evaluation.

When dealing with indeterminate forms like

\infty - \infty

, it’s helpful to simplify the terms into a single fraction.

Step 2: Rewrite the expression in terms of

\sin(x)

and

\cos(x)

We express

\sec(x)

and

\tan(x)

using sine and cosine functions:

\sec(x) = \frac{1}{\cos(x)}, \quad \tan(x) = \frac{\sin(x)}{\cos(x)}.

Now, the limit becomes:

\lim_{x \to \frac{\pi}{2}} \left( \frac{1}{\cos(x)} - \frac{\sin(x)}{\cos(x)} \right).

We can factor out

\frac{1}{\cos(x)}

from both terms:

\lim_{x \to \frac{\pi}{2}} \frac{1 - \sin(x)}{\cos(x)}.

Step 3: Analyze the behavior as

x \to \frac{\pi}{2}

As

x

approaches

\frac{\pi}{2}

:

-

\sin\left( \frac{\pi}{2} \right) = 1

.
-

\cos\left( \frac{\pi}{2} \right) = 0

.

Thus, the expression approaches the indeterminate form

\frac{0}{0}

, making it suitable for applying L'Hopital's Rule.

Step 4: Apply L'Hopital's Rule

We differentiate the numerator and the denominator separately:

1. **Differentiate the numerator**

1 - \sin(x)

:
- The derivative of

1

is 0, and the derivative of

\sin(x)

is

\cos(x)

. Thus:

\frac{d}{dx} \left( 1 - \sin(x) \right) = -\cos(x).

2. **Differentiate the denominator**

\cos(x)

:
- The derivative of

\cos(x)

is

-\sin(x)

. Thus:

\frac{d}{dx} \left( \cos(x) \right) = -\sin(x).

Step 5: Simplify and evaluate the limit

Now, the limit becomes:

\lim_{x \to \frac{\pi}{2}} \frac{-\cos(x)}{-\sin(x)} = \lim_{x \to \frac{\pi}{2}} \frac{\cos(x)}{\sin(x)}.

As

x \to \frac{\pi}{2}

:

-

\cos\left( \frac{\pi}{2} \right) = 0

.
-

\sin\left( \frac{\pi}{2} \right) = 1

.

Thus, the limit simplifies to:

\frac{0}{1} = 0.

Step 6: Final Answer

The value of the limit is:

\boxed{0}.

Prof's perspective

This problem is designed to test your understanding of **trigonometric identities** and how to handle **indeterminate forms** using L'Hopital's Rule.

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The value of the limit is:

\boxed{0}

A9

Find the equation of the line tangent to

f(x) = \tan(x)

at

x = \frac{\pi}{4}

.

Exercise Tags

point slope form

equation of tangent

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Step 1: Recall the formula for the tangent line

To find the equation of the tangent line to a function

f(x)

at a specific point

x = a

, we use the **point-slope form** of a line:

y - f(a) = f'(a)(x - a)

Here,

f(x) = \tan(x)

, and we need to find the tangent line at

x = \frac{\pi}{4}

.

Step 2: Find

f(a)

First, evaluate the function at

x = \frac{\pi}{4}

:

f\left( \frac{\pi}{4} \right) = \tan\left( \frac{\pi}{4} \right) = 1.

Thus, the point on the curve is

\left( \frac{\pi}{4}, 1 \right)

.

Step 3: Find

f'(x)

The derivative of

f(x) = \tan(x)

is:

f'(x) = \sec^2(x).

Now, evaluate the derivative at

x = \frac{\pi}{4}

:

f'\left( \frac{\pi}{4} \right) = \sec^2\left( \frac{\pi}{4} \right)

= \left( \frac{1}{\cos\left( \frac{\pi}{4} \right)} \right)^2

= \left( \frac{1}{\frac{\sqrt{2}}{2}} \right)^2

= 2.

Thus, the slope of the tangent line at

x = \frac{\pi}{4}

is

2

.

Remember, the slope of the tangent line is the derivative evaluated at the given point.

Step 4: Write the equation of the tangent line

We have the slope

m = 2

and the point

\left( \frac{\pi}{4}, 1 \right)

. Using the point-slope form:

y - 1 = 2\left( x - \frac{\pi}{4} \right).

Simplify the equation:

y - 1 = 2x - \frac{\pi}{2}

y = 2x - \frac{\pi}{2} + 1.

Thus, the equation of the tangent line is:

y = 2x - \frac{\pi}{2} + 1.

Final Answer

The equation of the line tangent to

f(x) = \tan(x)

at

x = \frac{\pi}{4}

is:

\boxed{y = 2x - \frac{\pi}{2} + 1}.

Prof's perspective

This problem helps you practice finding the **equation of a tangent line** using the **point-slope formula**.

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The equation of the line tangent to

f(x) = \tan(x)

at

x = \frac{\pi}{4}

is:

\boxed{y = 2x - \frac{\pi}{2} + 1}

Sub-questions:

A9 Part a)

Deduce a linear approximation of

\tan\left(\frac{\pi}{5}\right)

.

Exercise Tags

linear approximation

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Step 1: Use the equation of the tangent line from part (a)

From part (a), the equation of the tangent line to

f(x) = \tan(x)

at

x = \frac{\pi}{4}

was:

y = 2x - \frac{\pi}{2} + 1.

This equation gives us a linear approximation for

\tan(x)

around

x = \frac{\pi}{4}

.

Step 2: Apply the linear approximation

Linear approximations are most accurate near the point where the tangent line is calculated. Here, we use the tangent line to estimate

\tan\left( \frac{\pi}{5} \right)

.

1. **Substitute

x = \frac{\pi}{5}

into the tangent line equation:**

y = 2\left( \frac{\pi}{5} \right) - \frac{\pi}{2} + 1.

2. **Simplify the expression:**

We start by rewriting the fractions with a common denominator. The least common denominator of

\frac{2\pi}{5}

and

\frac{\pi}{2}

is 10:

y = \frac{4\pi}{10} - \frac{5\pi}{10} + 1.

Now simplify:

y = -\frac{\pi}{10} + 1.

Thus, the linear approximation for

\tan\left( \frac{\pi}{5} \right)

is:

y = 1 - \frac{\pi}{10}.

Step 3: Final Answer

The linear approximation of

\tan\left( \frac{\pi}{5} \right)

based on the tangent line at

x = \frac{\pi}{4}

is:

\boxed{1 - \frac{\pi}{10}}.

Linear approximations work best for values close to the point of tangency. Here, the approximation is most accurate for

x

near

\frac{\pi}{4}

.

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The linear approximation of

\tan\left( \frac{\pi}{5} \right)

based on the tangent line at

x = \frac{\pi}{4}

is:

\boxed{1 - \frac{\pi}{10}}

A10

The volume of a right circular cylinder of height

h

and radius

r

is

V = \pi r^2 h

. At the instant when

h = 6 \, \text{cm}

and increasing at the rate of

2 \, \text{cm}

per second, we observe that the radius

r

is

10 \, \text{cm}

and is decreasing at the rate of

1 \, \text{cm}

per second. How fast is the volume changing at this time?

Exercise Tags

related rates

Volume

Shapes

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Step 1: Write down the formula for the volume of the cylinder

The volume

V

of a right circular cylinder with height

h

and radius

r

is given by the formula:

V = \pi r^2 h

We are asked to find how fast the volume is changing with respect to time, i.e.,

\frac{dV}{dt}

, at a specific moment when:
-

h = 6 \, \text{cm}

and increasing at a rate of

\frac{dh}{dt} = 2 \, \text{cm/s}

,
-

r = 10 \, \text{cm}

and decreasing at a rate of

\frac{dr}{dt} = -1 \, \text{cm/s}

.

Step 2: Differentiate the volume equation with respect to time

To find how fast the volume is changing, we differentiate

V = \pi r^2 h

with respect to time

t

using the product rule. The radius

r

and height

h

are both functions of time, so we apply the chain rule:

\frac{dV}{dt} = \pi \left( 2r \frac{dr}{dt} h + r^2 \frac{dh}{dt} \right)

This equation accounts for the rates of change of both the radius and the height.

Step 3: Plug in the given values

We are given the following information at the instant of interest:
-

h = 6 \, \text{cm}

,
-

\frac{dh}{dt} = 2 \, \text{cm/s}

,
-

r = 10 \, \text{cm}

,
-

\frac{dr}{dt} = -1 \, \text{cm/s}

.

Substitute these values into the differentiated volume equation:

\frac{dV}{dt} = \pi \left( 2(10)(-1)(6) + (10)^2(2) \right)

Step 4: Simplify the expression

Now, simplify each term in the expression:

\frac{dV}{dt} = \pi \left( 2 \times 10 \times -1 \times 6 + 100 \times 2 \right)

\frac{dV}{dt} = \pi \left( -120 + 200 \right)

\frac{dV}{dt} = \pi (80)

Thus, the rate of change of the volume is:

\frac{dV}{dt} = 80\pi \, \text{cm}^3/\text{s}

Step 5: Final Answer

The volume of the cylinder is increasing at a rate of:

\boxed{80\pi \, \text{cm}^3/\text{s}}

Use the product rule when differentiating expressions with multiple variables that change with time.

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The volume of the cylinder is increasing at a rate of:

\boxed{80\pi \, \text{cm}^3/\text{s}}

A11

Given:

f(x) = \left( \frac{x + 3}{x + 1} \right)^2

, find the following.

Exercise Tags

Curve Sketching

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Sub-questions:

A11 Part a)

The domain of

f

.

Exercise Tags

find the domain

graphing

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Step 1: Analyze the function

The function given is:

f(x) = \left( \frac{x + 3}{x + 1} \right)^2

This is a rational function, meaning it is a fraction with a polynomial in the numerator and denominator. The square does not affect the domain restrictions, so we can focus on finding when the denominator equals zero.

Step 2: Find when the denominator equals zero

The denominator is

x + 1

, and the function is undefined when the denominator is zero because division by zero is not allowed.

So, we set the denominator equal to zero:

x + 1 = 0

Solving for

x

:

x = -1

This means the function is undefined at

x = -1

.

Step 3: State the domain

The domain of

f(x)

is all real numbers except

x = -1

. In interval notation, the domain is:

\boxed{(-\infty, -1) \cup (-1, \infty)}

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The domain of

f(x)

is all real numbers except

x = -1

. In interval notation, the domain is:

\boxed{(-\infty, -1) \cup (-1, \infty)}

A11 Part b)

Critical numbers.

Exercise Tags

find critical numbers

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Step 1: Recall the definition of critical numbers

Critical numbers occur where the derivative of a function

f(x)

is either zero or undefined. However, these points must be within the domain of the original function to be considered valid critical numbers.

Step 2: Find the derivative of

f(x)

Given:

f(x) = \left( \frac{x + 3}{x + 1} \right)^2

To differentiate, we apply the chain rule and quotient rule. Let:

f(x) = g(x)^2 \quad \text{where} \quad g(x) = \frac{x + 3}{x + 1}

Using the chain rule:

f'(x) = 2g(x) \cdot g'(x)

Now, we apply the quotient rule to find

g'(x)

. For

g(x) = \frac{u(x)}{v(x)}

, the quotient rule states:

g'(x) = \frac{u'(x) v(x) - u(x) v'(x)}{v(x)^2}

For

u(x) = x + 3

and

v(x) = x + 1

:

u'(x) = 1, \quad v'(x) = 1

Substitute these into the quotient rule:

g'(x) = \frac{1 \cdot (x + 1) - (x + 3) \cdot 1}{(x + 1)^2}

Simplify:

g'(x) = \frac{x + 1 - x - 3}{(x + 1)^2} = \frac{-2}{(x + 1)^2}

Now, substitute

g'(x)

into the derivative of

f(x)

:

f'(x) = 2 \cdot \left( \frac{x + 3}{x + 1} \right) \cdot \frac{-2}{(x + 1)^2}

Simplify:

f'(x) = \frac{-4(x + 3)}{(x + 1)^3}

Step 3: Find where

f'(x) = 0

or undefined

To find the critical numbers, we need to determine where the derivative is zero or undefined.

- **When

f'(x) = 0

:**

\frac{-4(x + 3)}{(x + 1)^3} = 0

The fraction equals zero when the numerator is zero:

x + 3 = 0 \quad \Rightarrow \quad x = -3

- **When

f'(x)

is undefined:**

The derivative is undefined when the denominator is zero:

(x + 1)^3 = 0 \quad \Rightarrow \quad x = -1

However,

f(x)

itself is undefined at

x = -1

, so this value is not a valid critical number.

Step 4: State the critical numbers

Since

x = -1

is outside the domain of the function, the only valid critical number is:

\boxed{x = -3}

Prof's perspective

This problem tests your ability to find critical numbers by correctly applying the chain rule and quotient rule. A key takeaway is that critical numbers must be within the function’s domain—points where the derivative is undefined but are outside the domain are not considered valid. The professor wants you to understand how to handle both cases where the derivative is zero and where it is undefined while also considering the original function’s domain.

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The critical number is the value where the derivative equals zero. So, the critical number of the function is:

\boxed{x = -3}

A11 Part c)

Possible points of inflection.

Exercise Tags

inflection points

graphing

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Step 1: Recall the definition of points of inflection

A point of inflection occurs where the concavity of the function changes, which happens when the second derivative,

f''(x)

, changes sign. To find potential points of inflection, we need to:

1. Find the second derivative

f''(x)

.
2. Determine where

f''(x) = 0

or undefined.
3. Check if the concavity changes at those points.

A point of inflection occurs only if the concavity changes from concave up to concave down (or vice versa).

Step 2: Differentiate

f'(x)

to find

f''(x)

We already know that:

f'(x) = \frac{-4(x + 3)}{(x + 1)^3}

Now, let's differentiate

f'(x)

to find

f''(x)

using the quotient rule. The quotient rule for

\frac{u(x)}{v(x)}

is:

f''(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}

For

f'(x) = \frac{-4(x + 3)}{(x + 1)^3}

, we have:
-

u(x) = -4(x + 3) \quad \Rightarrow \quad u'(x) = -4

-

v(x) = (x + 1)^3 \quad \Rightarrow \quad v'(x) = 3(x + 1)^2

Substitute these into the quotient rule:

f''(x) = \frac{-4(x + 1)^3 - (-4)(x + 3) \cdot 3(x + 1)^2}{(x + 1)^6}

Simplify the numerator:

f''(x) = \frac{-4(x + 1)^3 + 12(x + 3)(x + 1)^2}{(x + 1)^6}

Factor out

(x + 1)^2

from the numerator:

f''(x) = \frac{(x + 1)^2 \left( -4(x + 1) + 12(x + 3) \right)}{(x + 1)^6}

Simplify further:

f''(x) = \frac{-4(x + 1) + 12(x + 3)}{(x + 1)^4}

f''(x) = \frac{-4x - 4 + 12x + 36}{(x + 1)^4}

f''(x) = \frac{8x + 32}{(x + 1)^4}

f''(x) = \frac{8(x + 4)}{(x + 1)^4}

Step 3: Find where

f''(x) = 0

or undefined

Now, we set

f''(x) = 0

and solve for

x

:

\frac{8(x + 4)}{(x + 1)^4} = 0

The fraction is zero when the numerator is zero, so:

x + 4 = 0 \quad \Rightarrow \quad x = -4

Also,

f''(x)

is undefined when the denominator is zero, which happens when:

x + 1 = 0 \quad \Rightarrow \quad x = -1

However,

x = -1

is not in the domain of the function (as found in part (a)).

Points where

f''(x)

is undefined may still be points of inflection if the concavity changes across those points, but only if they lie in the domain.

Step 4: Check for concavity changes

To confirm that

x = -4

is a point of inflection, we check the sign of

f''(x)

on either side of

x = -4

.

- For

x < -4

(e.g.,

x = -5

),

f''(x) = \frac{8(-5 + 4)}{(-5 + 1)^4} = \frac{8(-1)}{(-4)^4} = \frac{-8}{256} < 0

, so the function is concave down.
- For

x > -4

(e.g.,

x = -3

),

f''(x) = \frac{8(-3 + 4)}{(-3 + 1)^4} = \frac{8(1)}{(-2)^4} = \frac{8}{16} > 0

, so the function is concave up.

Since the concavity changes from concave down to concave up,

x = -4

is a point of inflection.

Step 5: State the possible points of inflection

The possible point of inflection is:

\boxed{x = -4}

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The possible point of inflection is:

\boxed{x = -4}

A11 Part d)

Horizontal asymptotes.

Exercise Tags

asymptotes

graphing

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Step 1: Recall the definition of horizontal asymptotes

Horizontal asymptotes occur when the function approaches a constant value as

x

tends to infinity or negative infinity. To find horizontal asymptotes, we calculate the limits of the function

f(x)

as

x \to \infty

and

x \to -\infty

.

The given function is:

f(x) = \left( \frac{x + 3}{x + 1} \right)^2

We'll examine the limits as

x \to \infty

and

x \to -\infty

.

Remember, horizontal asymptotes describe the behavior of the function as

x

tends to very large positive or negative values.

Step 2: Find the limit as

x \to \infty

First, we calculate the limit of

f(x)

as

x \to \infty

:

\lim_{x \to \infty} \left( \frac{x + 3}{x + 1} \right)^2

Divide both the numerator and the denominator by

x

to simplify the expression:

\lim_{x \to \infty} \left( \frac{\frac{x + 3}{x}}{\frac{x + 1}{x}} \right)^2 = \lim_{x \to \infty} \left( \frac{1 + \frac{3}{x}}{1 + \frac{1}{x}} \right)^2

As

x \to \infty

, both

\frac{3}{x}

and

\frac{1}{x}

approach zero, so the expression simplifies to:

\lim_{x \to \infty} \left( \frac{1 + 0}{1 + 0} \right)^2 = 1^2 = 1

Thus, the horizontal asymptote as

x \to \infty

is

y = 1

.

Step 3: Find the limit as

x \to -\infty

Next, we calculate the limit of

f(x)

as

x \to -\infty

:

\lim_{x \to -\infty} \left( \frac{x + 3}{x + 1} \right)^2

Again, divide the numerator and the denominator by

x

:

\lim_{x \to -\infty} \left( \frac{\frac{x + 3}{x}}{\frac{x + 1}{x}} \right)^2 = \lim_{x \to -\infty} \left( \frac{1 + \frac{3}{x}}{1 + \frac{1}{x}} \right)^2

As

x \to -\infty

, both

\frac{3}{x}

and

\frac{1}{x}

approach zero, so the expression simplifies to:

\lim_{x \to -\infty} \left( \frac{1 + 0}{1 + 0} \right)^2 = 1^2 = 1

Thus, the horizontal asymptote as

x \to -\infty

is also

y = 1

.

Step 4: State the horizontal asymptotes

Since both limits as

x \to \infty

and

x \to -\infty

yield the same value, the function has a horizontal asymptote at:

\boxed{y = 1}

Horizontal asymptotes tell us the behavior of the function as

x

approaches extreme values.

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The function has a horizontal asymptote at:

\boxed{y = 1}

A11 Part e)

Vertical asymptotes.

Exercise Tags

asymptotes

graphing

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Step 1: Recall the definition of vertical asymptotes

A vertical asymptote occurs where the function approaches infinity (or negative infinity) as

x

approaches a certain value. Vertical asymptotes are typically found by identifying values of

x

that make the denominator of a rational function equal to zero, causing the function to be undefined.

The given function is:

f(x) = \left( \frac{x + 3}{x + 1} \right)^2

For this function, we look at when the denominator

x + 1

is zero because division by zero creates vertical asymptotes.

Vertical asymptotes occur where the denominator of a rational function is zero and the function becomes undefined.

Step 2: Find where the denominator equals zero

Set the denominator equal to zero:

x + 1 = 0

Solving for

x

:

x = -1

Thus, the function is undefined at

x = -1

. This means there is a potential vertical asymptote at

x = -1

.

Step 3: Confirm the vertical asymptote behavior

To confirm that

x = -1

is a vertical asymptote, we analyze the behavior of the function as

x

approaches

-1

from both the left and the right:

1. **As

x \to -1^+

(approaching from the right):**

When

x

approaches

-1

from values greater than

-1

,

x + 1

is positive but very small. Thus,

\frac{x + 3}{x + 1}

becomes very large, and squaring it makes

f(x)

approach

\infty

.

2. **As

x \to -1^-

(approaching from the left):**

When

x

approaches

-1

from values smaller than

-1

,

x + 1

is negative but very small. In this case,

\frac{x + 3}{x + 1}

becomes very large but negative, and squaring it still makes

f(x)

approach

\infty

.

In both cases, as

x

approaches

-1

, the function tends to infinity, confirming that

x = -1

is a vertical asymptote.

Step 4: State the vertical asymptote

The vertical asymptote occurs at:

\boxed{x = -1}

Remember, vertical asymptotes occur where the function becomes undefined and approaches

\infty

or

-\infty

.

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The vertical asymptote occurs at:

\boxed{x = -1}

A11 Part f)

Fill a table of signs with all the information above in order to find the intervals where

f

is increasing/decreasing and the intervals of concavity.

Exercise Tags

concavity tables

graphing

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Step 1: Review the critical points and inflection points

From the previous parts, we know the following:

- The **critical number** of

f(x)

is

x = -3

(from part (b)).
- The **point of inflection** is at

x = -4

(from part (c)).
- There is a **vertical asymptote** at

x = -1

(from part (e)).
- The **domain** of the function is

(-\infty, -1) \cup (-1, \infty)

(from part (a)).

We will use these points to create a table of signs for

f'(x)

and

f''(x)

.

Step 2: Create the table of signs for

f'(x)

We already found

f'(x) = \frac{-4(x + 3)}{(x + 1)^3}

. Now we check the sign of

f'(x)

in each interval determined by the critical point

x = -3

and the vertical asymptote

x = -1

.

Interval	Test Point	Sign of $f'(x)$	Behavior of $f(x)$
$(-\infty, -3)$	$x = -4$	Positive	Increasing
$(-3, -1)$	$x = -2$	Negative	Decreasing
$(-1, \infty)$	$x = 0$	Negative	Decreasing

So, the function is increasing on

(-\infty, -3)

and decreasing on

(-3, \infty)

.

Step 3: Create the table of signs for

f''(x)

We already found

f''(x) = \frac{8(x + 4)}{(x + 1)^4}

. Now we check the sign of

f''(x)

in each interval determined by the point of inflection

x = -4

and the vertical asymptote

x = -1

.

Interval	Test Point	Sign of $f''(x)$	Concavity of $f(x)$
$(-\infty, -4)$	$x = -5$	$f''(-5) = \frac{8(-5 + 4)}{(-5 + 1)^4} < 0$	Concave Down
$(-4, -1)$	$x = -3$	$f''(-3) = \frac{8(-3 + 4)}{(-3 + 1)^4} > 0$	Concave Up
$(-1, \infty)$	$x = 0$	$f''(0) = \frac{8(0 + 4)}{(0 + 1)^4} > 0$	Concave Up

So, the function is concave down on

(-\infty, -4)

and concave up on

(-4, \infty)

.

Step 4: Summarize the intervals of increase/decrease and concavity

- **Increasing**:

(-\infty, -3)

- **Decreasing**:

(-3, -1) \cup (-1, \infty)

- **Concave Down**:

(-\infty, -4)

- **Concave Up**:

(-4, -1) \cup (-1, \infty)

Remember, increasing/decreasing behavior is determined by the sign of

f'(x)

, and concavity is determined by the sign of

f''(x)

.

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Interval	Test Point	Sign of $f'(x)$	Behavior of $f(x)$
$(-\infty, -3)$	$x = -4$	Positive	Increasing
$(-3, -1)$	$x = -2$	Negative	Decreasing
$(-1, \infty)$	$x = 0$	Negative	Decreasing

Interval	Test Point	Sign of $f''(x)$	Concavity of $f(x)$
$(-\infty, -4)$	$x = -5$	$f''(-5) = \frac{8(-5 + 4)}{(-5 + 1)^4} < 0$	Concave Down
$(-4, -1)$	$x = -3$	$f''(-3) = \frac{8(-3 + 4)}{(-3 + 1)^4} > 0$	Concave Up
$(-1, \infty)$	$x = 0$	$f''(0) = \frac{8(0 + 4)}{(0 + 1)^4} > 0$	Concave Up

A11 Part g)

Sketch the graph of

f

, labelling all asymptotes, extrema and points of inflection on your picture.

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graphing

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Sketch the graph of

f

, labelling all asymptotes, extrema and points of inflection on your picture.

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A12

Your neighbourhood pizza place sells pizza slices "by the perimeter". You can only afford a pizza slice with a perimeter of 60 cm. What radius should your pizza have in order to maximize its area? Don't forget to show that your answer is the maximum.

Hint: The area of a sector of angle

\theta

radians in a circle of radius

r

is

A = \frac{1}{2} \theta r^2

, and the arc length of this sector is

r\theta

.

Exercise Tags

maximizing and minimizing

optimization

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Step 1: Understanding the problem

You want to maximize the area of a pizza slice, given that the perimeter of the slice is 60 cm. The perimeter consists of two straight sides (the radii) and the curved side (the arc of the circle).

Let:
-

r

be the radius of the pizza slice.
-

\theta

be the central angle of the pizza slice in radians.

The perimeter

P

of the pizza slice is given by the sum of the two straight sides (each of length

r

) and the length of the arc:

P = 2r + r\theta

You are told that the perimeter is 60 cm, so:

60 = 2r + r\theta

The area

A

of the pizza slice is given by the formula for the area of a sector:

A = \frac{1}{2} r^2 \theta

Step 2: Express

\theta

in terms of

r

From the perimeter equation

60 = 2r + r\theta

, solve for

\theta

:

r\theta = 60 - 2r

\theta = \frac{60 - 2r}{r} = \frac{60}{r} - 2

We now have

\theta

in terms of

r

, which we will substitute into the area formula to express the area as a function of

r

.

Step 3: Substitute

\theta

into the area formula

Substitute

\theta = \frac{60}{r} - 2

into the area equation:

A = \frac{1}{2} r^2 \left( \frac{60}{r} - 2 \right)

Simplify the expression:

A = \frac{1}{2} r^2 \left( \frac{60}{r} - 2 \right) = \frac{1}{2} \left( 60r - 2r^2 \right) = 30r - r^2

So, the area function in terms of

r

is:

A(r) = 30r - r^2

Step 4: Take the derivative

To maximize the area, we need to take the derivative of

A(r)

with respect to

r

and set it equal to zero. The derivative is:

A'(r) = 30 - 2r

Set

A'(r) = 0

to find the critical points:

30 - 2r = 0

Solving for

r

:

r = 15

Step 5: Verify that it is a maximum

To confirm that

r = 15

gives a maximum, take the second derivative of

A(r)

:

A''(r) = -2

Since

A''(r) = -2

, which is negative, the function is concave down at

r = 15

, confirming that this is a maximum.

Because the second derivative is negative, this shows that the area is maximized at

r = 15

.

Step 6: Final Answer

The radius that maximizes the area of the pizza slice is:

\boxed{r = 15 \, \text{cm}}

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The radius that maximizes the area of the pizza slice is:

\boxed{r = 15 \, \text{cm}}

A13

Consider the following equation

3-2x=e^x-cos(x)

.

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Intermediate Value Theorem

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Sub-questions:

A13 Part a)

Use the Intermediate Value Theorem to show that the equation has at least one
solution.

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Step 1: Recall the Intermediate Value Theorem

The Intermediate Value Theorem (IVT) states that if a continuous function

f(x)

takes opposite signs at two points

a

and

b

(i.e.,

f(a) \cdot f(b) < 0

), then there exists at least one value

c

in the interval

(a, b)

such that

f(c) = 0

.

We will use this theorem to show that the equation

3 - 2x = e^x - \cos(x)

has at least one solution.

For the IVT to apply, the function must be continuous and change signs between two points.

Step 2: Define the function

Rearrange the equation

3 - 2x = e^x - \cos(x)

to one side:

f(x) = 3 - 2x - e^x + \cos(x)

We are looking for values

a

and

b

such that

f(a) \cdot f(b) < 0

, indicating the function changes sign over the interval.

Step 3: Evaluate

f(x)

at two points

We will evaluate

f(x)

at

x = 0

and

x = 1

exactly, without approximations.

1. **For

x = 0

**:

f(0) = 3 - 2(0) - e^0 + \cos(0) = 3 - 1 + 1 = 3

Thus,

f(0) = 3

.

2. **For

x = 1

**:

f(1) = 3 - 2(1) - e^1 + \cos(1)

Using exact values:

f(1) = 3 - 2 - e + \cos(1)

So:

f(1) = 1 - e + \cos(1)

Step 4: Apply the Intermediate Value Theorem

Now we compare the signs of

f(0)

and

f(1)

:

-

f(0) = 3

is positive.
-

f(1) = 1 - e + \cos(1)

.

Since

e > 2.7

and

\cos(1) < 1

, the expression

1 - e + \cos(1)

is negative.

Therefore,

f(0) \cdot f(1) < 0

, meaning the function changes sign over the interval

(0, 1)

. By the Intermediate Value Theorem, there is at least one value

c

in the interval

(0, 1)

such that:

f(c) = 0.

Step 5: State the solution

Thus, the equation

3 - 2x = e^x - \cos(x)

has at least one solution in the interval:

\boxed{c \in (0, 1)}.

The IVT guarantees a root in the interval where the function changes signs, even if the exact root is not known.

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\boxed{c \in (0, 1)}

A13 Part b)

9. (b) Use the Mean Value Theorem to show that the equation has at most one solution.

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Mean Value Theorem

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Step 1: Recall the Mean Value Theorem (MVT)

The Mean Value Theorem (MVT) states that if a function

f(x)

is continuous on a closed interval

[a, b]

and differentiable on the open interval

(a, b)

, then there exists at least one point

c

in the interval

(a, b)

such that:

f'(c) = \frac{f(b) - f(a)}{b - a}.

To show that the given equation

3 - 2x = e^x - \cos(x)

has at most one solution, we need to prove that the function derived from this equation is either strictly increasing or strictly decreasing. This ensures the function cannot cross the x-axis more than once.

The MVT helps confirm that if the derivative does not equal zero, the function is strictly increasing or decreasing.

Step 2: Define the function and find its derivative

We rearrange the given equation into the form:

f(x) = 3 - 2x - e^x + \cos(x).

Now, differentiate

f(x)

to determine its behavior:

f'(x) = \frac{d}{dx} \left( 3 - 2x - e^x + \cos(x) \right).

Applying standard derivative rules:

f'(x) = -2 - e^x - \sin(x).

Step 3: Analyze the derivative

We now analyze the derivative

f'(x) = -2 - e^x - \sin(x)

to determine whether it is always positive, negative, or zero:

1. **The term

e^x

** is always positive for all

x

, so

-e^x

is always negative.
2. **The term

\sin(x)

** oscillates between

-1

and

1

. Thus,

-\sin(x)

oscillates between

1

and

-1

.
3. **The constant term

-2

** is always negative.

Together:

f'(x) = -2 - e^x - \sin(x).

Since

-2 - e^x < -1

for all

x

, the oscillations of

-\sin(x)

cannot make the derivative positive. Thus:

f'(x) < 0 \quad \text{for all} \quad x.

Step 4: Apply the Mean Value Theorem

Since

f'(x) < 0

for all

x

, the function is strictly decreasing. A strictly decreasing function can intersect the x-axis at most once, meaning the equation has at most one solution.

A strictly decreasing function implies that the function values are unique for each

x

, ensuring no more than one solution exists.

Step 5: Final Answer

Using the MVT, we conclude that the equation:

3 - 2x = e^x - \cos(x)

can have at most one solution.

\boxed{\text{At most one solution exists.}}

Prof's perspective

This problem tests your ability to apply the Mean Value Theorem in an abstract setting. The goal is to help students recognize how derivative behavior determines whether a function is increasing or decreasing. The professor wants to ensure you understand that the uniqueness of solutions relies on whether the function’s derivative maintains a consistent sign.

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Step 1: Define the function and find its derivative

Given:

f(x) = 3 - 2x - e^x + \cos(x)

The derivative is:

f'(x) = -2 - e^x - \sin(x)

Step 2: Analyze the derivative

Since

-e^x

is always negative and

-2 - \sin(x)

is also always negative, we conclude that:

f'(x) < 0 \quad \text{for all } x

Step 3: Conclusion

Because

f'(x) < 0

for all

x

, the function is strictly decreasing, so the equation has at most one solution:

\boxed{\text{At most one solution.}}

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