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Plot
y=(sqrt(20+x)-5)/(x-5). Ohoose the correct graph beluk
A.
tion 5
Use rHospitar's Rule to find the limit. Select the correct choice
astion 6
A.
lim_(x rarr5)(sqrt(20+x)-5)/(x-5)=◻ (Simplify your answer.)
B. The limit does not exist.
Question 9
Question 10
Question 11
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Plot $\newline$ $y=\frac{\sqrt{20+x}-5}{x-5}$ . Choose the correct graph below $\newline$ A. $\newline$ tion $5$ $\newline$ Use L'Hospital's Rule to find the limit. Select the correct choice $\newline$ $\lim_{x \to 5}\frac{\sqrt{20+x}-5}{x-5}=\square$ (Simplify your answer.) $\newline$ B. The limit does not exist.

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One-step inequalities: word problems

Full solution

Q. Plot

\newline

y=\frac{\sqrt{20+x}-5}{x-5}

. Choose the correct graph below

\newline

A.

\newline

tion

5

\newline

Use L'Hospital's Rule to find the limit. Select the correct choice

\newline

\lim_{x \to 5}\frac{\sqrt{20+x}-5}{x-5}=\square

(Simplify your answer.)

\newline

B. The limit does not exist.

Plug in $x = 5$ : First, let's try to plug in $x = 5$ directly into the equation and see what happens. $\newline$ $y = \frac{\sqrt{20+5}-5}{5-5}$ $\newline$ $y = \frac{\sqrt{25}-5}{0}$ $\newline$ $y = \frac{5-5}{0}$ $\newline$ $y = \frac{0}{0}$ $\newline$ We get an indeterminate form $0/0$ , so we can't determine the limit this way.
Apply L'Hôpital's Rule: Since we have an indeterminate form, let's apply L'Hôpital's Rule, which says we can take the derivative of the numerator and the denominator separately and then take the limit. $\newline$ First, find the derivative of the numerator, $\frac{d}{dx}[\sqrt{20+x}]$ . $\newline$ Using the chain rule, the derivative of $\sqrt{20+x}$ is $\frac{1}{2\sqrt{20+x}}$ .
Find Derivatives: Now, find the derivative of the denominator, $\frac{d}{dx}[x-5]$ . $\newline$ The derivative of $x-5$ is simply $1$ .
Take the Limit: Now we take the limit of the derivatives as $x$ approaches $5$ .
$\lim_{x \to 5} \left(\frac{1}{2\sqrt{20+x}}\right) / (1)$
$\lim_{x \to 5} \frac{1}{2\sqrt{20+x}}$
Now plug in $x = 5$ .
$\lim_{x \to 5} \frac{1}{2\sqrt{25}}$
$\lim_{x \to 5} \frac{1}{2\cdot 5}$
$\lim_{x \to 5} \frac{1}{10}$
So, the limit as $x$ approaches $5$ is $5$ $0$ .

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-11 x \geq-33

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Which of the following best describes the solutions to the inequality shown?

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Question

g(x)=5-e^{x}

\newline

Below is Sean's attempt to write a formal justification for the fact that the equation

g(x)=0

has a solution where

1 \leq x \leq 4

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Is Sean's justification complete? If not, why?

\newline

Sean's justification:

\newline

g

is defined for all real numbers, and exponential functions are continuous at all points in their domains.

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So, according to the intermediate value theorem,

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must have a solution somewhere between

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(A) Yes, Sean's justification is complete.

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(B) No, Sean didn't establish that

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f(x)=\sqrt[3]{\cos (\pi \cdot x)}

\newline

Below is Issac's attempt to write a formal justification for the fact that the equation

f(x)=0.1

has a solution where

1.5 \leq x \leq 2

\newline

Is Issac's justification complete? If not, why?

\newline

Issac's justification:

\newline

\begin{array}{l} f(1.5)=0 \text { and } \\ f(2)=1 \text {, so } \\ f(1.5)<0.1<f(2) . \end{array}

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So, according to the intermediate value theorem,

f(x)=0.1

must have a solution when

x

x=1.5

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(A) Yes, Issac's justification is complete.

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(B) No, Issac didn't establish that

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(C) No, Issac didn't establish that

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h(x)=5 \cdot \tan (x)

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Below is Bridget's attempt to write a formal justification for the fact that the equation

h(x)=2

has a solution where

0 \leq x \leq \frac{\pi}{4}

\newline

Is Bridget's justification complete? If not, why?

\newline

Bridget's justification:

\newline

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\newline

So, according to the intermediate value theorem,

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\newline

(A) Yes, Bridget's justification is complete.

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(B) No, Bridget didn't establish that

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\newline

(C) No, Bridget didn't establish that

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f(x)=2^{5-2 x}

\newline

Below is Ibrahim's attempt to write a formal justification for the fact that the equation

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has a solution where

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\newline

Is Ibrahim's justification complete? If not, why?

\newline

Ibrahim's justification:

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f

is defined for all real numbers, and exponential functions are continuous at all points in their domains. Also,

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\newline

So, according to the intermediate value theorem,

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must have a solution at some point between

x=-1

x=4

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1

\newline

(A) Yes, Ibrahim's justification is complete.

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(B) No, Ibrahim didn't establish that

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(c) No, Ibrahim didn't establish that

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g(x)=\cos \left(\pi x^{2}\right)

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Below is Amrita's attempt to write a formal justification for the fact that the equation

g^{\prime}(x)=0.4

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\newline

Is Amrita's justification complete?

\newline

\newline

Amrita's justification:

\newline

Polynomial and trigonometric functions are differentiable and continuous at all points in their domain, and

[-1,4]

g^{\prime}

s domain. Also,

\newline

\begin{array}{l} g(-1)=-1 \text { and } \\ g(4)=1 \text {, so } \\ \frac{g(4)-g(-1)}{4-(-1)}=0.4 . \end{array}

\newline

So, according to the mean value theorem,

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must have a solution somewhere between

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1

\newline

(A) Yes, Amrita's justification is complete.

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(B) No, Amrita didn't establish that the average rate of change of

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(C) No, Amrita didn't establish that

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Below is Omar's attempt to write a formal justification for the fact that there exists a value

c

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-2<c<2 \text { such that } h^{\prime}(c)=6 \text {. }

\newline

Is Omar's justification complete? If not, why?

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Omar's justification:

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\begin{array}{l} h(-2)=-4 \text { and } \\ h(2)=20 \text {, so } \\ \frac{h(2)-h(-2)}{2-(-2)}=6 . \end{array}

\newline

So, according to the mean value theorem, there exists a value

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somewhere in the interval

-2<c<2

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(A) Yes, Omar's justification is complete.

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(B) No, Omar didn't establish that the average rate of change of

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f(x)=\sqrt{x+\sqrt{x^{3}}}

\newline

Below is Lena's attempt to write a formal justification for the fact that the equation

f^{\prime}(x)=\frac{2}{3}

has a solution where

0<x<9

\newline

Is Lena's justification complete? If not, why?

\newline

Lena's justification:

\newline

f(0)=0

f(9)=6

, so the average rate of change of

f

over the interval

0 \leq x \leq 9

\frac{2}{3}

\newline

So, according to the mean value theorem,

f^{\prime}(x)=\frac{2}{3}

must have a solution somewhere in the interval

0<x<9

\newline

1

\newline

(A) Yes, Lena's justification is complete.

\newline

(B) No, Lena didn't establish that the average rate of change of

f

[0,9]

\frac{2}{3}

\newline

(C) No, Lena didn't establish that

f

is differentiable.

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Question

h(x)=3^{x}-x^{2}

\newline

Below is Uriah's attempt to write a formal justification for the fact that the equation

h^{\prime}(x)=8

has a solution where

1<x<3

\newline

Is Uriah's justification complete? If not, why?

\newline

Uriah's justification:

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Polynomial and exponential functions are differentiable and continuous at all points in their domain, and

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\frac{h(3)-h(1)}{3-1}=8 \text {. }

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So, according to the mean value theorem,

h^{\prime}(x)=8

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x=1

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1

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(A) Yes, Uriah's justification is complete.

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(B) No, Uriah didn't establish that the average rate of change of

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(C) No, Uriah didn't establish that

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