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Let
f(x)=sqrt(2x+1) and let
c be the number that satisfies the Mean Value Theorem for
f on the interval
4 <= x <= 12.
What is
c ?
Choose 1 answer:
(A) 0
(B) 1.5
(C) 4
(D) 7.5

Let $f(x)=\sqrt{2 x+1}$ and let $c$ be the number that satisfies the Mean Value Theorem for $f$ on the interval $4 \leq x \leq 12$ . $\newline$ What is $c$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $0$ $\newline$ (B) $1$ . $5$ $\newline$ (C) $4$ $\newline$ (D) $7$ . $5$

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Solve quadratic inequalities

Full solution

Q. Let

f(x)=\sqrt{2 x+1}

and let

c

be the number that satisfies the Mean Value Theorem for

f

on the interval

4 \leq x \leq 12

.

\newline

What is

c

?

\newline

Choose

1

answer:

\newline

(A)

0

\newline

(B)

1

.

5

\newline

(C)

4

\newline

(D)

7

.

5

Understand Mean Value Theorem: 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 $f'(c) = \frac{f(b) - f(a)}{b - a}$ .
Apply MVT to Function: Apply the MVT to the function $f(x) = \sqrt{2x + 1}$ on the interval $[4, 12]$ . $\newline$ First, calculate $f(4)$ and $f(12)$ . $\newline$ $f(4) = \sqrt{2\cdot4 + 1} = \sqrt{9} = 3$ $\newline$ $f(12) = \sqrt{2\cdot12 + 1} = \sqrt{25} = 5$
Calculate Difference Quotient: Calculate the difference quotient $(f(b) - f(a)) / (b - a)$ . $(f(12) - f(4)) / (12 - 4) = (5 - 3) / (12 - 4) = 2 / 8 = 1/4$
Find Derivative of $f(x)$ : Find the derivative of $f(x)$ .
$f'(x) = \frac{d}{dx} [\sqrt{2x + 1}]$
To differentiate $\sqrt{2x + 1}$ , use the chain rule.
$f'(x) = \frac{1}{2\sqrt{2x + 1}} \cdot \frac{d}{dx} [2x + 1]$
$f'(x) = \frac{1}{2\sqrt{2x + 1}} \cdot 2$
$f'(x) = \frac{1}{\sqrt{2x + 1}}$
Set Derivative Equal to Quotient: Set the derivative equal to the difference quotient and solve for $c$ . $\frac{1}{\sqrt{2c + 1}} = \frac{1}{4}$ Cross-multiply to solve for $c$ . $4 = \sqrt{2c + 1}$ Square both sides to eliminate the square root. $16 = 2c + 1$ Subtract $1$ from both sides. $15 = 2c$ Divide by $2$ . $c = \frac{15}{2}$ $c = 7.5$

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\newline

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\newline

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\newline

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for at least one

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0

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g(x)=\tan (x)

\newline

Can we use the intermediate value theorem to say the equation

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\newline

1

\newline

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

\newline

0

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

\newline

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

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\newline

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\newline

1

\newline

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\newline

0

8

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\newline

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