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Let
g(x)=sqrt(2x-4) and let
c be the number that satisfies the Mean Value Theorem for
g on the interval
2 <= x <= 10.
What is
c ?
Choose 1 answer:
(A) 2.25
(B) 3.75
(C) 4
(D) 6

Let $g(x)=\sqrt{2 x-4}$ and let $c$ be the number that satisfies the Mean Value Theorem for $g$ on the interval $2 \leq x \leq 10$ . $\newline$ What is $c$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $2$ . $25$ $\newline$ (B) $3$ . $75$ $\newline$ (C) $4$ $\newline$ (D) $6$

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Solve quadratic inequalities

Full solution

Q. Let

g(x)=\sqrt{2 x-4}

and let

c

be the number that satisfies the Mean Value Theorem for

g

on the interval

2 \leq x \leq 10

.

\newline

What is

c

?

\newline

Choose

1

answer:

\newline

(A)

2

.

25

\newline

(B)

3

.

75

\newline

(C)

4

\newline

(D)

6

Understand the MVT: Understand the Mean Value Theorem (MVT). The MVT states that if a function $f$ is continuous on the closed interval $[a, b]$ and differentiable on the open interval $(a, b)$ , then there exists at least one number $c$ in $(a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$ .
Verify Conditions for MVT: Verify that $g(x)$ satisfies the conditions of the MVT on $[2, 10]$ . $g(x) = \sqrt{2x - 4}$ is continuous and differentiable on $[2, 10]$ because the square root function is continuous and differentiable wherever its argument is positive, and $2x - 4$ is positive for all $x$ in $[2, 10]$ .
Calculate $g(10)$ and $g(2)$ : Calculate $g(10)$ and $g(2)$ .
$g(10) = \sqrt{2\times 10 - 4} = \sqrt{20 - 4} = \sqrt{16} = 4$
$g(2) = \sqrt{2\times 2 - 4} = \sqrt{4 - 4} = \sqrt{0} = 0$
Calculate Average Rate of Change: Calculate the average rate of change of $g(x)$ on $[2, 10]$ . The average rate of change is $(g(10) - g(2)) / (10 - 2) = (4 - 0) / (8) = 4 / 8 = 1/2$ .
Find $g'(x)$ : Find $g'(x)$ , the derivative of $g(x)$ .
$g'(x) = \frac{d}{dx} [\sqrt{2x - 4}]$
To differentiate $\sqrt{2x - 4}$ , we use the chain rule.
$g'(x) = \frac{1}{2\sqrt{2x - 4}} \cdot \frac{d}{dx} [2x - 4]$
$g'(x) = \frac{1}{2\sqrt{2x - 4}} \cdot 2$
$g'(x) = \frac{1}{\sqrt{2x - 4}}$
Set $g'(c)$ equal to Average Rate: Set $g'(c)$ equal to the average rate of change and solve for $c$ . $\newline$ $\frac{1}{\sqrt{2c - 4}} = \frac{1}{2}$ $\newline$ To solve for $c$ , we square both sides of the equation to get rid of the square root. $\newline$ $\left(\frac{1}{\sqrt{2c - 4}}\right)^2 = \left(\frac{1}{2}\right)^2$ $\newline$ $\frac{1}{2c - 4} = \frac{1}{4}$ $\newline$ Now, cross-multiply to solve for $c$ . $\newline$ $4 = 2c - 4$ $\newline$ $8 = 2c$ $\newline$ $g'(c)$ $0$

More problems from Solve quadratic inequalities

Question

-10 x-3 \geq 8 x+4

\newline

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

\newline

1

\newline

x \leq-\frac{7}{18}

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x \leq-\frac{7}{2}

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

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x>\frac{1}{6}

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8 x+6<4 x+10

\newline

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

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1

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x<\frac{1}{3}

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x<4

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x<1

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6 x-2<2 x+3

\newline

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

\newline

1

\newline

x<\frac{5}{4}

\newline

x>\frac{5}{4}

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x<\frac{1}{4}

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Question

16<-4 x

\newline

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

\newline

1

\newline

x<-4

\newline

x>-4

\newline

x<-\frac{1}{4}

\newline

x>-\frac{1}{4}

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Question

\lim _{x \rightarrow 4} \frac{2-\sqrt{4 x-12}}{x-4}

\newline

1

\newline

2

\newline

1

\newline

-1

\newline

(D) The limit doesn't exist

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Question

\lim _{x \rightarrow 2} \frac{x^{4}-4 x^{3}+4 x^{2}}{x-2}

\newline

1

\newline

-4

\newline

0

\newline

4

\newline

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Question

f

be a continuous function on the closed interval

[-5,0]

f(-5)=0

f(0)=5

\newline

Which of the following is guaranteed by the Intermediate Value Theorem?

\newline

1

\newline

f(c)=-2

for at least one

c

0

5

\newline

f(c)=2

for at least one

c

0

5

\newline

f(c)=-2

for at least one

c

-5

0

\newline

f(c)=2

for at least one

c

-5

0

Posted 9 months ago

Question

g(x)=\tan (x)

\newline

Can we use the intermediate value theorem to say the equation

g(x)=0

has a solution where

\frac{\pi}{4} \leq x \leq \frac{3 \pi}{4}

\newline

1

\newline

(A) No, since the function is not continuous on that interval.

\newline

0

g\left(\frac{\pi}{4}\right)

g\left(\frac{3 \pi}{4}\right)

\newline

(C) Yes, both conditions for using the intermediate value theorem have been met.

Posted 9 months ago

Question

g(x)=\cos (x)

\newline

Can we use the intermediate value theorem to say the equation

g(x)=0.8

has a solution where

0 \leq x \leq \frac{\pi}{2}

\newline

1

\newline

(A) No, since the function is not continuous on that interval.

\newline

0

8

g(0)

g\left(\frac{\pi}{2}\right)

\newline

(C) Yes, both conditions for using the intermediate value theorem have been met.

Posted 9 months ago

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