Final, Fall 2011

$$\lim_{t \to -3} \frac{t^2 - 9}{3t^2 + 11t + 6}$$

For the numerator:
$$t^2 - 9 = (t-3)(t+3)$$

For the denominator:




$$\lim_{t \to -3} \frac{t^2 - 9}{3t^2 + 11t + 6}$$
$$= \lim_{t \to -3} \frac{(t-3)(t+3)}{(3t+2)(t+3)}$$
$$= \lim_{t \to -3} \frac{(t-3)\cancel{(t+3)}}{(3t+2)\cancel{(t+3)}}$$
$$= \lim_{t \to -3} \frac{t-3}{3t+2}$$



$$\lim_{t \to -3} \frac{t-3}{3t+2}$$
$$= \frac{-3-3}{3(-3)+2}$$
$$= \frac{-6}{-9+2}$$
$$= \frac{-6}{-7}$$
$$= \frac{6}{7}$$

Therefore:
$$\boxed{\lim_{t \to -3} \frac{t^2 - 9}{3t^2 + 11t + 6} = \frac{6}{7}}$$

I understand now. Let me correct the formatting with proper math brackets and vertical alignment:


Evaluate the following limit:
$$\lim_{h \to 0} \frac{(-4 + h)^2 - 16}{h}$$








$$\begin{align}$$
\lim_{h \to 0} (-8 + h) &= -8 + 0 \\
&= -8
\end{align}

Therefore:
$$\boxed{\lim_{h \to 0} \frac{(-4 + h)^2 - 16}{h} = -8}$$

As $x \to \infty$:
- $\sin(x)/e^x \to 0$ since $e^x$ grows much faster than $\sin(x)$
- $\cos(x)/e^x \to 0$ for the same reason

When dealing with limits involving exponentials and trigonometric functions, remember that exponential functions like $e^x$ grow much faster than any polynomial or trigonometric function. Since sine and cosine are always bounded between -1 and 1, the ratio $\sin(x)/e^x$ or $\cos(x)/e^x$ will always approach 0 as $x$ approaches infinity.





Therefore:
$$\boxed{\lim_{x \to \infty} \frac{3e^x + 5\sin(x)}{e^x + \cos(x)} = 3}$$

The vertical asymptote(s) of $f(x) = \frac{\sqrt{3x^6 + 7}}{x^3 - 5x^2 + 6x}$ is (are) best described by:

(a) $x = 0$, (b) $x = 2$, (c) $x = 3$, (d) two asymptotes, (e) three asymptotes.



Vertical asymptotes occur at values where the denominator equals zero.

$$\begin{align}$$
x^3 - 5x^2 + 6x &= 0 \\
x(x^2 - 5x + 6) &= 0 \\
x(x - 2)(x - 3) &= 0
\end{align}

This gives us $x = 0$, $x = 2$, or $x = 3$ as potential vertical asymptotes.



For these points to be vertical asymptotes, we need to verify that the numerator is non-zero at these points.

The numerator is $\sqrt{3x^6 + 7}$. Since $3x^6$ is always non-negative (as it's raised to an even power) and we're adding the positive constant 7, the expression under the square root is always positive. Therefore, the numerator is always positive and never zero.

Since the numerator is never zero at any of these points, all three values $x = 0$, $x = 2$, and $x = 3$ are vertical asymptotes.



Since we've confirmed that the function has vertical asymptotes at $x = 0$, $x = 2$, and $x = 3$, there are three vertical asymptotes in total.

Therefore:
$$\boxed{\text{The answer is (e) three asymptotes}}$$

When analyzing rational functions with radicals, remember that expressions like $x^{2n} + k$ (where k > 0) are always positive, ensuring that vertical asymptotes occur at all zeros of the denominator.

Let $f(x) = 3x^{\frac{2}{3}} - 5x^{-\frac{3}{4}}$. Find $f'(1)$.



We need to use the power rule for differentiation: $\frac{d}{dx}(x^n) = nx^{n-1}$

For the first term: $3x^{\frac{2}{3}}$
$$\begin{align}$$
\frac{d}{dx}(3x^{\frac{2}{3}}) &= 3 \cdot \frac{2}{3} \cdot x^{\frac{2}{3}-1} \\
&= 2 \cdot x^{-\frac{1}{3}} \\
&= \frac{2}{x^{\frac{1}{3}}}
\end{align}

For the second term: $-5x^{-\frac{3}{4}}$
$$\begin{align}$$
\frac{d}{dx}(-5x^{-\frac{3}{4}}) &= -5 \cdot (-\frac{3}{4}) \cdot x^{-\frac{3}{4}-1} \\
&= -5 \cdot (-\frac{3}{4}) \cdot x^{-\frac{7}{4}} \\
&= \frac{15}{4} \cdot x^{-\frac{7}{4}} \\
&= \frac{15}{4 \cdot x^{\frac{7}{4}}}
\end{align}

Now we combine these results to find $f'(x)$:
$$\begin{align}$$
f'(x) &= \frac{2}{x^{\frac{1}{3}}} + \frac{15}{4 \cdot x^{\frac{7}{4}}}
\end{align}



When we substitute $x = 1$:
$$\begin{align}$$
f'(1) &= \frac{2}{1^{\frac{1}{3}}} + \frac{15}{4 \cdot 1^{\frac{7}{4}}} \\
&= \frac{2}{1} + \frac{15}{4 \cdot 1} \\
&= 2 + \frac{15}{4} \\
&= 2 + \frac{15}{4} \\
&= \frac{8}{4} + \frac{15}{4} \\
&= \frac{23}{4}
\end{align}

Therefore:
$$\boxed{f'(1) = \frac{23}{4}}$$

We need to differentiate both sides of the equation with respect to $x$:
$$\begin{align}$$
\tanh(x) &= xy
\end{align}

Differentiating the left side:
$$\begin{align}$$
\frac{d}{dx}[\tanh(x)] &= \text{sech}^2(x)
\end{align}

Differentiating the right side:
$$\begin{align}$$
\frac{d}{dx}[xy] &= y + x\frac{dy}{dx}
\end{align}



$$\begin{align}$$
\text{sech}^2(x) &= y + x\frac{dy}{dx} \\
\text{sech}^2(x) - y &= x\frac{dy}{dx} \\
\frac{dy}{dx} &= \frac{\text{sech}^2(x) - y}{x}
\end{align}



At the point $(1, \tanh(1))$, we have $x = 1$ and $y = \tanh(1)$.

Let's substitute these values:
$$\begin{align}$$
\frac{dy}{dx}\bigg|_{(1,\tanh(1))} &= \frac{\text{sech}^2(1) - \tanh(1)}{1} \\
&= \text{sech}^2(1) - \tanh(1)
\end{align}

Now we need to calculate these values:
$$\begin{align}$$
\text{sech}^2(1) &= \frac{1}{\cosh^2(1)} \\
&= \frac{1}{(\frac{e + e^{-1}}{2})^2} \\
&= \frac{4}{(e + e^{-1})^2}
\end{align}

And:
$$\begin{align}$$
\tanh(1) &= \frac{\sinh(1)}{\cosh(1)} \\
&= \frac{\frac{e - e^{-1}}{2}}{\frac{e + e^{-1}}{2}} \\
&= \frac{e - e^{-1}}{e + e^{-1}}
\end{align}

Therefore:

$$\begin{align}$$
\frac{dy}{dx} &= \frac{4}{(e + e^{-1})^2} - \frac{e - e^{-1}}{e + e^{-1}} \\
&= \frac{4}{(e + e^{-1})^2} - \frac{(e - e^{-1})(e + e^{-1})}{(e + e^{-1})^2} \\
&= \frac{4 - (e^2 - e^{-2})}{(e + e^{-1})^2} \\
&= \frac{4 - (e^2 - e^{-2})}{(e + e^{-1})^2}
\end{align}


We can simplify further:
$$\begin{align}$$
\frac{dy}{dx} &= \frac{4 - e^2 + e^{-2}}{(e + e^{-1})^2}
\end{align}

Therefore:
$$\boxed{\frac{dy}{dx} = \frac{4 - e^2 + e^{-2}}{(e + e^{-1})^2}}$$

When differentiating hyperbolic functions, remember that $\frac{d}{dx}[\tanh(x)] = \text{sech}^2(x) = \frac{1}{\cosh^2(x)}$, similar to how the derivative of $\tan(x)$ is $\sec^2(x)$ for trigonometric functions.

For a piecewise function to be continuous at the transition point (in this case $x = 0$), the limits from both sides must be equal:

$$\begin{align}$$
& \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)
\end{align}



The left-hand limit uses the first piece of the function:

$$\begin{align}$$
& \lim_{x \to 0^-} f(x) \\
&= \lim_{x \to 0^-} \sinh(x) \\
&= \sinh(0) \\
&= 0
\end{align}



The right-hand limit uses the second piece:

$$\begin{align}$$
& \lim_{x \to 0^+} f(x) \\
&= \lim_{x \to 0^+} (b - 3\cos(x)) \\
&= b - 3\cos(0) \\
&= b - 3(1) \\
&= b - 3
\end{align}



For continuity, these limits must be equal:

$$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)$$
$$\sinh(0) = b - 3\cos(0)$$
$$0 = b - 3(1)$$
$$0 = b - 3$$
$$b = 3$$

When working with piecewise functions, always check the value of each piece at the transition point to ensure continuity.

Therefore, the value of $b$ that makes $f(x)$ continuous everywhere is:

$$\boxed{b = 3}$$

We have a function with two terms:
1. $x^3e^{2x}$
2. $3(x - 2)^2e^{-x}$

We'll need to find the derivative of each term using the product rule, then evaluate at $x = 0$.



For the first term, $x^3e^{2x}$, we use the product rule:

$$\begin{align}$$
& \frac{d}{dx}[x^3e^{2x}] \\
&= \frac{d}{dx}[x^3] \cdot e^{2x} + x^3 \cdot \frac{d}{dx}[e^{2x}] \\
&= 3x^2 \cdot e^{2x} + x^3 \cdot 2e^{2x} \\
&= 3x^2e^{2x} + 2x^3e^{2x} \\
&= e^{2x}(3x^2 + 2x^3)
\end{align}



For the second term, $3(x - 2)^2e^{-x}$, we use the product rule again:


$$\begin{align}$$
& \frac{d}{dx}[3(x - 2)^2e^{-x}] \\
&= \frac{d}{dx}[3(x - 2)^2] \cdot e^{-x} + 3(x - 2)^2 \cdot \frac{d}{dx}[e^{-x}] \\
&= 3 \cdot 2(x - 2) \cdot e^{-x} + 3(x - 2)^2 \cdot (-1)e^{-x} \\
&= 6(x - 2)e^{-x} - 3(x - 2)^2e^{-x} \\
&= e^{-x}[6(x - 2) - 3(x - 2)^2]
\end{align}


When differentiating terms with $e^{ax}$, remember that $\frac{d}{dx}[e^{ax}] = a \cdot e^{ax}$. The coefficient comes out front!



The complete derivative is:


$$\begin{align}$$
& f'(x) \\
&= e^{2x}(3x^2 + 2x^3) \\
&+ e^{-x}[6(x - 2) - 3(x - 2)^2]
\end{align}

Now we evaluate at $x = 0$:

$$\begin{align}$$
& f'(0) \\
&= e^{2(0)}(3(0)^2 + 2(0)^3) \\
&+ e^{-(0)}[6(0 - 2) - 3(0 - 2)^2] \\
&= e^0(0) \\
&+ e^0[6(-2) - 3(-2)^2] \\
&= 1 \cdot 0 \\
&+ 1 \cdot [(-12) - 3 \cdot 4] \\
&= 0 \\
&+ [-12 - 12] \\
&= -24
\end{align}
Therefore:

$$\boxed{f'(0) = -24}$$

Let $f(x) = \frac{\tan(x)}{x^2 - 2x + 5}$. Find $f'(0)$.



Since we have a fraction, we'll use the quotient rule.

If $f(x) = \frac{g(x)}{h(x)}$, then:

$$\begin{align}$$
f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}
\end{align}



Let's set:
- $g(x) = \tan(x)$
- $h(x) = x^2 - 2x + 5$



$$\begin{align}$$
& g'(x) = \frac{d}{dx}[\tan(x)] \\
&= \sec^2(x)
\end{align}

$$\begin{align}$$
& h'(x) = \frac{d}{dx}[x^2 - 2x + 5] \\
&= 2x - 2
\end{align}



$$\begin{align}$$
& f'(x) \\
&= \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \\
&= \frac{\sec^2(x)(x^2 - 2x + 5) - \tan(x)(2x - 2)}{(x^2 - 2x + 5)^2}
\end{align}


When evaluating at x = 0, remember that $\tan(0) = 0$ and $\sec^2(0) = 1$, which simplifies our work considerably!



At $x = 0$:


$$\begin{align}$$
& f'(0) \\
&= \frac{\sec^2(0)(0^2 - 2(0) + 5) - \tan(0)(2(0) - 2)}{(0^2 - 2(0) + 5)^2} \\
&= \frac{1 \cdot 5 - 0 \cdot (-2)}{5^2} \\
&= \frac{5}{25} \\
&= \frac{1}{5}
\end{align}


Therefore:

$$\boxed{f'(0) = \frac{1}{5}}$$

Let $u(x) = (x - 1)^2$ and $v(x) = x + 2$

$$\begin{align}$$
& \frac{dy}{dx} \\
&= \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}
\end{align}

Finding the derivatives:

$$\begin{align}$$
& u'(x) \\
&= \frac{d}{dx}[(x - 1)^2] \\
&= 2(x - 1)
\end{align}

$$\begin{align}$$
& v'(x) \\
&= \frac{d}{dx}[x + 2] \\
&= 1
\end{align}

Substituting into the quotient rule:

$$\begin{align}$$
& \frac{dy}{dx} \\
&= \frac{2(x - 1)(x + 2) - (x - 1)^2 \cdot 1}{(x + 2)^2}
\end{align}



At $x = 2$:

$$\begin{align}$$
& \frac{dy}{dx}\bigg|_{x=2} \\
&= \frac{2(2 - 1)(2 + 2) - (2 - 1)^2}{(2 + 2)^2} \\
&= \frac{2(1)(4) - (1)^2}{4^2} \\
&= \frac{8 - 1}{16} \\
&= \frac{7}{16}
\end{align}



The slope of the tangent line at $(2, 1)$ is $m = \frac{7}{16}$

Using the point-slope form of a line:

$$\begin{align}$$
& y - y_1 = m(x - x_1) \\
&y - 1 = \frac{7}{16}(x - 2) \\
&y - 1 = \frac{7}{16}x - \frac{7}{8} \\
&y = \frac{7}{16}x - \frac{7}{8} + 1 \\
&y = \frac{7}{16}x - \frac{7}{8} + \frac{8}{8} \\
&y = \frac{7}{16}x + \frac{1}{8}
\end{align}



Let's find a point with a simple x-value, say $x = 0$:

$$\begin{align}$$
& y = \frac{7}{16} \cdot 0 + \frac{1}{8} \\
&y = 0 + \frac{1}{8} \\
&y = \frac{1}{8}
\end{align}

Therefore, the point $(0, \frac{1}{8})$ lies on the tangent line.

$$\boxed{(x,y) = \left(0, \frac{1}{8}\right)}$$

We already found the derivative of this function in our previous question:

$$\begin{align}$$
& \frac{dy}{dx} \\
&= \frac{2(x - 1)(x + 2) - (x - 1)^2}{(x + 2)^2}
\end{align}

To find critical points, we set this equal to zero:

$$\begin{align}$$
& \frac{2(x - 1)(x + 2) - (x - 1)^2}{(x + 2)^2} = 0
\end{align}

Since the denominator cannot be zero on our interval (as $x + 2 > 0$ when $x \geq -1.9$), we only need:

$$\begin{align}$$
& 2(x - 1)(x + 2) - (x - 1)^2 = 0 \\
&(x - 1)[2(x + 2) - (x - 1)] = 0 \\
&(x - 1)[2x + 4 - x + 1] = 0 \\
&(x - 1)[x + 5] = 0
\end{align}

So $x = 1$ or $x = -5$ are our critical points.



Our interval is $[-8, -4]$, and we found one critical point at $x = -5$ which is within this interval.

Extreme Value Theorem tells us that the maximum must occur either at a critical point within the interval or at an endpoint.

We need to evaluate the function at $x = -8$, $x = -5$, and $x = -4$.



$$\begin{align}$$
& f(-8) \\
&= \frac{(-8 - 1)^2}{-8 + 2} \\
&= \frac{(-9)^2}{-6} \\
&= \frac{81}{-6} \\
&= -13.5
\end{align}



$$\begin{align}$$
& f(-5) \\
&= \frac{(-5 - 1)^2}{-5 + 2} \\
&= \frac{(-6)^2}{-3} \\
&= \frac{36}{-3} \\
&= -12
\end{align}



$$\begin{align}$$
& f(-4) \\
&= \frac{(-4 - 1)^2}{-4 + 2} \\
&= \frac{(-5)^2}{-2} \\
&= \frac{25}{-2} \\
&= -12.5
\end{align}

When finding extreme values on a closed interval, always check both the critical points inside the interval AND the endpoint values!



Comparing all three values:
- $f(-8) = -13.5$
- $f(-5) = -12$
- $f(-4) = -12.5$

The maximum value is $-12$ which occurs at $x = -5$.

$$\boxed{\text{Maximum value} = -12}$$