Final, Winter 2018

To evaluate this limit, start by factoring both the numerator and the denominator. Factoring can often help us cancel out terms, especially if direct substitution initially results in an indeterminate form like $\frac{0}{0}$.

The numerator $x^2 - 8x + 15$ factors as:

$$x^2-8x+15=(x-5)(x-3)$$
x^2 - 8x + 15 = (x - 5)(x - 3)



The denominator $x^2 - 25$ is a difference of squares, which factors as:

$$x^2-25=(x-5)(x+5)$$
x^2 - 25 = (x - 5)(x + 5)


So, the expression becomes:

$$\frac{(x-5)(x-3)}{(x-5)(x+5)}$$
\frac{(x - 5)(x - 3)}{(x - 5)(x + 5)}




Now, cancel the common factor $(x - 5)$ from both the numerator and the denominator:

$$\frac{(x-3)}{(x+5)}$$
\frac{(x - 3)}{(x + 5)}




Now that we’ve simplified the expression, we can substitute $x = 5$ directly:

$$\frac{5-3}{5+5}=\frac{2}{10}=\frac{1}{5}$$
\frac{5 - 3}{5 + 5} = \frac{2}{10} = \frac{1}{5}


Key Tip: When faced with indeterminate forms, factor and cancel common terms to simplify the expression.

$$\boxed{{\displaystyle \frac{1}{5}}}$$
\boxed{\frac{1}{5}}

To find the derivative, apply the rules of differentiation to each term individually. We’ll use the following key rules:

Derivative of $\log_a(x)$: $\frac{1}{x \ln(a)}$ and Derivative of $a^x$: $a^x \ln(a)$

**Derivative of** $\log_3(x)$:

$$\frac{d}{dx}({\log }_3(x))=\frac{1}{x\ln (3)}$$
\frac{d}{dx} \left( \log_3(x) \right) = \frac{1}{x \ln(3)}


**Derivative of** $4^x$:

$$\frac{d}{dx}\left(4^x\right)=4^x\ln (4)$$
\frac{d}{dx} \left( 4^x \right) = 4^x \ln(4)


**Derivative of** $-x^2$:

$$\frac{d}{dx}(-x^2)=-2x$$
\frac{d}{dx} \left( -x^2 \right) = -2x


**Derivative of** $17$ (a constant):

$$\frac{d}{dx}\left(17\right)=0$$
\frac{d}{dx} \left( 17 \right) = 0




Now, we can combine each of these derivatives to find the overall derivative:




$$\boxed{\frac{1}{x \ln(3)} + 4^x \ln(4) - 2x}$$

To differentiate $\frac{1 + e^x}{x^2 + 9}$, we use the quotient rule, which states:

Quotient Rule: If $f(x) = \frac{u}{v}$, then $f'(x) = \frac{u'v - uv'}{v^2}$

In this case:
- $u = 1 + e^x$
- $v = x^2 + 9$



**Derivative of** $u = 1 + e^x$:

$$u^{\mathrm{\prime }}=\frac{d}{dx}(1+e^x)=e^x$$
u' = \frac{d}{dx} (1 + e^x) = e^x


**Derivative of** $v = x^2 + 9$:

$$v^{\mathrm{\prime }}=\frac{d}{dx}(x^2+9)=2x$$
v' = \frac{d}{dx} (x^2 + 9) = 2x




Now, substitute $u$, $u'$, $v$, and $v'$ into the quotient rule formula:







Distribute $e^x$ and $2x$:

$$=\frac{e^x\cdot x^2+9e^x-2x-2x{e}^x}{(x^2+9{)}^2}$$
= \frac{e^x \cdot x^2 + 9e^x - 2x - 2x e^x}{(x^2 + 9)^2}


$$\boxed{{\displaystyle \frac{e^x\cdot x^2+9e^x-2x-2x{e}^x}{(x^2+9{)}^2}}}$$
\boxed{\frac{e^x \cdot x^2 + 9e^x - 2x - 2x e^x}{(x^2 + 9)^2}}

To differentiate $\sin \left( e^{3x} + 17 \ln(x^2 + 1) \right)$, we’ll apply the chain rule. The chain rule states:

Chain Rule: If $f(x) = g(h(x))$, then $f'(x) = g'(h(x)) \cdot h'(x)$

In this case:
- $g(x) = \sin(x)$
- $h(x) = e^{3x} + 17 \ln(x^2 + 1)$



First, differentiate the outer function $g(x) = \sin(x)$:

$$g^{\mathrm{\prime }}(h(x))=\cos (e^{3x}+17\ln (x^2+1))$$
g'(h(x)) = \cos \left( e^{3x} + 17 \ln(x^2 + 1) \right)




Now, we differentiate the inner function $h(x) = e^{3x} + 17 \ln(x^2 + 1)$:

**Derivative of** $e^{3x}$:

$$\frac{d}{dx}\left(e^{3x}\right)=3e^{3x}$$
\frac{d}{dx} \left( e^{3x} \right) = 3e^{3x}


**Derivative of** $17 \ln(x^2 + 1)$:

Using the chain rule again on $\ln(x^2 + 1)$:

$$\frac{d}{dx}(17\ln (x^2+1))=17\cdot \frac{1}{x^2+1}\cdot 2x=\frac{34x}{x^2+1}$$
\frac{d}{dx} \left( 17 \ln(x^2 + 1) \right) = 17 \cdot \frac{1}{x^2 + 1} \cdot 2x = \frac{34x}{x^2 + 1}


Combining these, we have:

$$h^{\mathrm{\prime }}(x)=3e^{3x}+\frac{34x}{x^2+1}$$
h'(x) = 3e^{3x} + \frac{34x}{x^2 + 1}




Now, we can combine $g'(h(x))$ and $h'(x)$ to find the derivative:





This question checks your ability to apply the chain rule in complex scenarios involving trigonometric, exponential, and logarithmic functions. The professor likely chose this question to ensure you can differentiate nested functions accurately, a skill that is crucial for handling advanced calculus problems.

To evaluate this limit, start by simplifying the expression within the tangent function. We’ll try factoring the numerator and the denominator.
The numerator $x^3 - x^2 - x - 2$ can be factored by grouping terms.



To factor $x^3 - x^2 - x - 2$, we group the terms in pairs:

$$x^3-x^2-x-2=(x^3-x^2)-(x+2)$$
x^3 - x^2 - x - 2 = (x^3 - x^2) - (x + 2)


Now, factor out $x^2$ from the first group and factor out $-1$ from the second group:

$$=x^2(x-1)-1(x-1)$$
= x^2(x - 1) - 1(x - 1)


Now we have a common factor of $(x - 2)$:

$$=(x-2)(x^2+x+1)$$
= (x - 2)(x^2 + x + 1)


The denominator $x^2 + 17x - 38$ factors as:

$$x^2+17x-38=(x-2)(x+19)$$
x^2 + 17x - 38 = (x - 2)(x + 19)


So, the expression becomes:

$$\tan \left(\frac{(x-2)(x^2+x+1)}{(x-2)(x+19)}\pi \right)$$
\tan \left( \frac{(x - 2)(x^2 + x + 1)}{(x - 2)(x + 19)} \pi \right)




Now, cancel the common factor $(x - 2)$ from both the numerator and the denominator:

$$\tan \left(\frac{x^2+x+1}{x+19}\pi \right)$$
\tan \left( \frac{x^2 + x + 1}{x + 19} \pi \right)




Now that we’ve simplified the expression, we can substitute $x = 2$ directly:

$$\tan \left(\frac{2^2+2+1}{2+19}\pi \right)=$$
\tan \left( \frac{2^2 + 2 + 1}{2 + 19} \pi \right) =


$$\tan \left(\frac{4+2+1}{21}\pi \right)=$$
\tan \left( \frac{4 + 2 + 1}{21} \pi \right) =


$$\tan \left(\frac{7}{21}\pi \right)=$$
\tan \left( \frac{7}{21} \pi \right) =


$$\tan \left(\frac{\pi }{3}\right)$$
\tan \left( \frac{\pi}{3} \right)


Tip: You need to know the unit circle and the values of common trigonometric functions at special angles, such as $\frac{\pi}{3}$, to solve problems like this.


Since $\tan \left( \frac{\pi}{3} \right) = \sqrt{3}$, we have:

$$\boxed{{\displaystyle \sqrt{3}}}$$
\boxed{\sqrt{3}}




This question is designed to test your ability to handle limits involving trigonometric functions, especially by simplifying rational expressions through factoring. The professor likely included this question to emphasize the importance of **factoring and cancelling terms** before evaluating limits, a common approach in calculus to avoid indeterminate forms.

The derivative of $f(x) = x^3 + 17x + 9$ from first principles is defined as:

$$f^{\mathrm{\prime }}(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$
f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}




In this case, $f(x) = x^3 + 17x + 9$, so let’s compute $f(x + h)$:



Expanding $(x + h)^3$:

Then:

Now, substitute into the definition:




Cancel out the terms $x^3$, $17x$, and $9$ in the numerator:



Now, factor $h$ from each term in the numerator:

Cancel $h$ in the numerator and denominator:



Now, substitute $h = 0$:

$$f^{\mathrm{\prime }}(x)=3x^2+17$$
f'(x) = 3x^2 + 17


$$\boxed{{\displaystyle 3x^2+17}}$$
\boxed{3x^2 + 17}

For the natural logarithm $\ln(y)$, the argument $y$ must be greater than zero. In this case that means:

$${\cot }^{-1}(x^2)-\frac{\pi }{6}>0$$
\cot^{-1}(x^2) - \frac{\pi}{6} > 0


which is therefore our first condition that we need to satisfy.



The function $\cot^{-1}(x^2)$ is defined for all $x$ because $x^2 \geq 0$ for all real values of $x$.

So there are no added restrictions on the domain from $\cot^{-1}(x^2)$.





To satisfy the condition for the logarithm, we need:

$${\cot }^{-1}(x^2)-\frac{\pi }{6}>0$$
\cot^{-1}(x^2) - \frac{\pi}{6} > 0


This simplifies to:

$${\cot }^{-1}(x^2)>\frac{\pi }{6}$$
\cot^{-1}(x^2) > \frac{\pi}{6}




Now, we find the values of $x$ such that $\cot^{-1}(x^2) > \frac{\pi}{6}$.

Since $\cot^{-1}(x^2)$ decreases as $x^2$ increases, we need:

$$x^2<\cot \left(\frac{\pi }{6}\right)$$
x^2 < \cot\left( \frac{\pi}{6} \right)


From the unit circle, we know $\cot\left( \frac{\pi}{6} \right) = \sqrt{3}$, so:

$$x^2<\sqrt{3}$$
x^2 < \sqrt{3}


Taking the square root of both sides:

$$\mathrm{\mid }x\mathrm{\mid }<\sqrt[4]{3}$$
|x| < \sqrt[4]{3}


Thus, the domain of the function is:

$$\boxed{{\displaystyle -\sqrt[4]{3}<x<\sqrt[4]{3}}}$$
\boxed{-\sqrt[4]{3} < x < \sqrt[4]{3}}




This problem tests your understanding of the domain of composite functions, especially when involving inverse trigonometric functions and logarithms. The professor likely included this question to ensure you know how to work with domain restrictions and analyze inequalities involving inverse trigonometric expressions.

First, let’s substitute $x = 1$ into the expression:

Numerator:

$$4{\tan }^{-1}(1)-\pi$$
4 \tan^{-1}(1) - \pi


$$=4\cdot \frac{\pi }{4}-\pi =\pi -\pi =0$$
= 4 \cdot \frac{\pi}{4} - \pi = \pi - \pi = 0


Denominator:

$$1^2+8\cdot 1-9=1+8-9=0$$
1^2 + 8 \cdot 1 - 9 = 1 + 8 - 9 = 0


Since both the numerator and denominator are zero, we have an indeterminate form $\frac{0}{0}$. Therefore, we can apply **L'Hôpital's Rule**, which states:

L'Hôpital's Rule: If $\lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0}$ or $\frac{\pm \infty}{\pm \infty}$, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$, provided the limit on the right exists.



**Differentiate the numerator** $4 \tan^{-1}(x) - \pi$:

Tip: Remember the derivatives of inverse trigonometric functions! For $\tan^{-1}(x)$, the derivative is $\frac{1}{1 + x^2}$.

$$\frac{d}{dx}(4{\tan }^{-1}(x)-\pi )$$
\frac{d}{dx} \left( 4 \tan^{-1}(x) - \pi \right)


$$=4\cdot \frac{1}{1+x^2}=\frac{4}{1+x^2}$$
= 4 \cdot \frac{1}{1 + x^2} = \frac{4}{1 + x^2}


**Differentiate the denominator** $x^2 + 8x - 9$:

$$\frac{d}{dx}(x^2+8x-9)=2x+8$$
\frac{d}{dx} \left( x^2 + 8x - 9 \right) = 2x + 8




Now, substitute these derivatives into the limit:

$$\underset{x\to 1}{\lim }\frac{\frac{4}{1+x^2}}{2x+8}$$
\lim_{x \to 1} \frac{\frac{4}{1 + x^2}}{2x + 8}


Substitute $x = 1$:

$$=\frac{\frac{4}{1+1^2}}{2\cdot 1+8}=\frac{\frac{4}{2}}{10}=\frac{2}{10}=\frac{1}{5}$$
= \frac{\frac{4}{1 + 1^2}}{2 \cdot 1 + 8} = \frac{\frac{4}{2}}{10} = \frac{2}{10} = \frac{1}{5}


$$\boxed{\frac{1}{5}}$$

To find the maximum, we need to calculate the derivative $f'(x)$ and set it to zero to find critical points. We’ll use the product rule since $f(x)$ is the product of $(x^2 - 6x + 9)$ and $e^x$.

**Using the product rule**:

$$f^{\mathrm{\prime }}(x)=\frac{d}{dx}((x^2-6x+9)e^x)$$
f'(x) = \frac{d}{dx} \left( (x^2 - 6x + 9)e^x \right)


Let:
- $u = x^2 - 6x + 9$
- $v = e^x$

Then:
- $u' = 2x - 6$
- $v' = e^x$

So,

$$f^{\mathrm{\prime }}(x)$$
f'(x)

$$=u^{\mathrm{\prime }}v+u{v}^{\mathrm{\prime }}$$
= u'v + uv'

$$=(2x-6)e^x+(x^2-6x+9)e^x$$
= (2x - 6)e^x + (x^2 - 6x + 9)e^x


Combine terms:



Simplify the expression inside the parentheses:

$$f^{\mathrm{\prime }}(x)=(x^2-4x+3)e^x$$
f'(x) = \left( x^2 - 4x + 3 \right) e^x


Now, set $f'(x) = 0$:

$$(x^2-4x+3)e^x=0$$
(x^2 - 4x + 3)e^x = 0


Since $e^x \neq 0$, we only need to solve:

$$x^2-4x+3=0$$
x^2 - 4x + 3 = 0


Factoring gives:

$$(x-1)(x-3)=0$$
(x - 1)(x - 3) = 0


So, $x = 1$ and $x = 3$. However, since we are only interested in the interval $[0, 2]$, we discard $x = 3$ and keep $x = 1$.



Now, evaluate $f(x)$ at $x = 0$, $x = 1$, and $x = 2$:

**At** $x = 0$:

$$f(0)=(0^2-6\cdot 0+9)e^0=9\cdot 1=9$$
f(0) = (0^2 - 6 \cdot 0 + 9)e^0 = 9 \cdot 1 = 9


**At** $x = 1$:

$$f(1)=(1^2-6\cdot 1+9)e^1=(1-6+9)e=4e$$
f(1) = (1^2 - 6 \cdot 1 + 9)e^1 = (1 - 6 + 9)e = 4e


**At** $x = 2$:

$$f(2)=(2^2-6\cdot 2+9)e^2=(4-12+9)e^2=1\cdot e^2=e^2$$
f(2) = (2^2 - 6 \cdot 2 + 9)e^2 = (4 - 12 + 9)e^2 = 1 \cdot e^2 = e^2




Now, compare the values at $x = 0$, $x = 1$, and $x = 2$:

$$f(0)=9,f(1)=4e,f(2)=e^2$$
f(0) = 9, \quad f(1) = 4e, \quad f(2) = e^2


Since $e^2 \approx 7.39$ and $4e \approx 10.87$, the maximum value occurs at $x = 1$ with $f(1) = 4e$.

$$\boxed{{\displaystyle 4e}}$$
\boxed{4e}




This question is designed to test your understanding of the product rule, as well as evaluating functions at critical points and endpoints to find absolute maximums or minimums. The professor likely included this question to ensure you can handle more complex functions and apply calculus principles over a specified interval.

To determine where $f(x)$ is concave down, we need to find the second derivative $f''(x)$ and determine where it is negative.

Recall that $f(x) = (x^2 - 6x + 9)e^x$. We already found the first derivative:

$$f^{\mathrm{\prime }}(x)=(x^2-4x+3)e^x$$
f'(x) = \left( x^2 - 4x + 3 \right) e^x


Now, we’ll differentiate $f'(x)$ again to find $f''(x)$, using the product rule.

Let:
- $u = x^2 - 4x + 3$
- $v = e^x$

Then:
- $u' = 2x - 4$
- $v' = e^x$

Applying the product rule:

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=u^{\mathrm{\prime }}v+u{v}^{\mathrm{\prime }}$$
f''(x) = u'v + uv'


Substitute $u$, $u'$, $v$, and $v'$:




Combine terms:

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=(2x-4+x^2-4x+3)e^x$$
f''(x) = \left( 2x - 4 + x^2 - 4x + 3 \right)e^x


Simplify inside the parentheses:

$$f^{\mathrm{\prime }\mathrm{\prime }}(x)=(x^2-2x-1)e^x$$
f''(x) = \left( x^2 - 2x - 1 \right) e^x




To find where $f(x)$ is concave down, we need $f''(x) < 0$.

Since $e^x > 0$ for all $x$, the sign of $f''(x)$ depends only on $x^2 - 2x - 1$:

$$x^2-2x-1<0$$
x^2 - 2x - 1 < 0


Solve this inequality by finding the roots of $x^2 - 2x - 1 = 0$ using the quadratic formula:

$$x=\frac{2\pm \sqrt{(2{)}^2-4\cdot 1\cdot (-1)}}{2\cdot 1}$$
x = \frac{2 \pm \sqrt{(2)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1}


$$=\frac{2\pm \sqrt{4+4}}{2}$$
= \frac{2 \pm \sqrt{4 + 4}}{2}


$$=\frac{2\pm 2\sqrt{2}}{2}$$
= \frac{2 \pm 2\sqrt{2}}{2}


$$=1\pm \sqrt{2}$$
= 1 \pm \sqrt{2}




Now, we test the sign of $x^2 - 2x - 1$ in the intervals around these roots, namely, in the intervals $(-\infty, 1 - \sqrt{2})$, $(1 - \sqrt{2}, 1 + \sqrt{2})$, and $(1 + \sqrt{2}, \infty)$.

** Interval 1: $(-\infty, 1 - \sqrt{2})$
Choose a test point $x = -2$ in this interval.

Substitute $x = -2$ into $x^2 - 2x - 1$:

$$(-2{)}^2-2(-2)-1=4+4-1=7$$
(-2)^2 - 2(-2) - 1 = 4 + 4 - 1 = 7


Since $7 > 0$, the function is positive in this interval.

**Interval 2: $(1 - \sqrt{2}, 1 + \sqrt{2})$
Choose a test point $x = 1$ in this interval.

Substitute $x = 1$ into $x^2 - 2x - 1$:

$$(1{)}^2-2(1)-1=1-2-1=-2$$
(1)^2 - 2(1) - 1 = 1 - 2 - 1 = -2


Since $-2 < 0$, the function is negative in this interval.

** Interval 3: $(1 + \sqrt{2}, \infty)$
Choose a test point $x = 3$ in this interval.

Substitute $x = 3$ into $x^2 - 2x - 1$:

$$(3{)}^2-2(3)-1=9-6-1=2$$
(3)^2 - 2(3) - 1 = 9 - 6 - 1 = 2


Since $2 > 0$, the function is positive in this interval.





Since the function is negative in the interval $(1 - \sqrt{2}, 1 + \sqrt{2})$, this is the interval where the function is concave down.

$$\boxed{{\displaystyle (1-\sqrt{2},1+\sqrt{2})}}$$
\boxed{(1 - \sqrt{2}, 1 + \sqrt{2})}

To find horizontal asymptotes, we examine the behavior of $f(x) = (x^2 - 6x + 9)e^x$ as $x$ approaches positive and negative infinity.

**As $x \to \infty$:**

The exponential term $e^x$ grows extremely fast as $x \to \infty$. Therefore, $f(x)$ will also grow without bound because of $e^x$, making $f(x) \to \infty$.

**Conclusion**: There is no horizontal asymptote as $x \to \infty$.

**As $x \to -\infty$:**

When $x \to -\infty$, $e^x$ approaches $0$ because exponential functions decay towards zero for large negative inputs. Thus:

$$f(x)=(x^2-6x+9)e^x\approx 0\text{ as }x\to -\mathrm{\infty }$$
f(x) = (x^2 - 6x + 9)e^x \approx 0 \text{ as } x \to -\infty




Since $f(x)$ approaches $0$ as $x \to -\infty$, there is a horizontal asymptote at:

$$\boxed{{\displaystyle y=0}}$$
\boxed{y = 0}

To show that $f(x)$ has at least one zero in $[0, 1]$, we can apply the **Intermediate Value Theorem**, which states:

Intermediate Value Theorem: If $f$ is continuous on $[a, b]$ and $f(a)$ and $f(b)$ have opposite signs, then there exists at least one $c \in (a, b)$ such that $f(c) = 0$.



The function $f(x) = 5x^{10} + 3x^3 - 1$ is a polynomial, and polynomials are continuous everywhere. Therefore, $f(x)$ is continuous on $[0, 1]$.



Evaluate $f(0)$:

$$f(0)=5\cdot 0^{10}+3\cdot 0^3-1=-1$$
f(0) = 5 \cdot 0^{10} + 3 \cdot 0^3 - 1 = -1


Evaluate $f(1)$:

$$f(1)$$
f(1)

$$=5\cdot 1^{10}+3\cdot 1^3-1$$
= 5 \cdot 1^{10} + 3 \cdot 1^3 - 1

$$=5+3-1=7$$
= 5 + 3 - 1 = 7

$$=7$$
= 7




Since $f(0) = -1$ and $f(1) = 7$, $f(0)$ and $f(1)$ have opposite signs. By the Intermediate Value Theorem, there must be at least one $c \in (0, 1)$ such that $f(c) = 0$.

$$\boxed{{\displaystyle \text{Therefore, }f(x)\text{ has at least one zero in }[0,1]\mathrm{.}}}$$
\boxed{\text{Therefore, } f(x) \text{ has at least one zero in } [0, 1].}

To show that $f(x)$ has only one zero in $[0, 1]$, we can examine the behavior of the function by analyzing its derivative $f'(x)$. If $f'(x)$ does not change sign in $[0, 1]$, then $f(x)$ is either strictly increasing or strictly decreasing on that interval, implying at most one zero.

Given:

$$f(x)=5x^{10}+3x^3-1$$
f(x) = 5x^{10} + 3x^3 - 1


Differentiate $f(x)$ to find $f'(x)$:

$$\begin{array}{rllll}& {\displaystyle f^{\mathrm{\prime }}(x)}& {\displaystyle =10\cdot 5x^9+3\cdot 3x^2}& & \\ & {\displaystyle }& {\displaystyle =50x^9+9x^2}& & \end{array}$$
\begin{align}
f'(x) &= 10 \cdot 5x^9 + 3 \cdot 3x^2 \\
&= 50x^9 + 9x^2
\end{align}




Since $f'(x) = 50x^9 + 9x^2$, note that each term $50x^9$ and $9x^2$ is non-negative for $x \geq 0$. Therefore, $f'(x) \geq 0$ for all $x \in [0, 1]$, and $f'(x) = 0$ only at $x = 0$.

Since $f'(x) \geq 0$ on $[0, 1]$, $f(x)$ is non-decreasing on this interval. This implies that $f(x)$ can have at most one zero in $[0, 1]$, as a non-decreasing function cannot cross the x-axis more than once without changing direction.



Since $f(x)$ is continuous on $[0, 1]$ and has a zero in this interval (from part (a)), and since $f(x)$ is non-decreasing on $[0, 1]$, we conclude that the zero is unique.

$$\boxed{{\displaystyle \text{The root of }f(x)\text{ in }[0,1]\text{ is unique.}}}$$
\boxed{\text{The root of } f(x) \text{ in } [0, 1] \text{ is unique.}}