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Let
f(x)=sqrt(4x-3) and let
c be the number that satisfies the Mean Value Theorem for
f on the interval
1 <= x <= 3.
What is
c ?
Choose 1 answer:
(A)
1.5
(B)
1.75
(C) 2
(D) 2.25

Let $f(x)=\sqrt{4 x-3}$ and let $c$ be the number that satisfies the Mean Value Theorem for $f$ on the interval $1 \leq x \leq 3$ . $\newline$ What is $c$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\mathbf{1 . 5}$ $\newline$ (B) $\mathbf{1 . 7 5}$ $\newline$ (C) $2$ $\newline$ (D) $2$ . $25$

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Solve quadratic inequalities

Full solution

Q. Let

f(x)=\sqrt{4 x-3}

and let

c

be the number that satisfies the Mean Value Theorem for

f

on the interval

1 \leq x \leq 3

.

\newline

What is

c

?

\newline

Choose

1

answer:

\newline

(A)

\mathbf{1 . 5}

\newline

(B)

\mathbf{1 . 7 5}

\newline

(C)

2

\newline

(D)

2

.

25

Check Function Continuity: To apply the Mean Value Theorem (MVT), we need to ensure that the function $f(x)$ is continuous on the closed interval $[1, 3]$ and differentiable on the open interval $(1, 3)$ . The function $f(x) = \sqrt{4x - 3}$ is continuous and differentiable where 4x - 3 > 0, which is when x > \frac{3}{4}. Since 1 > \frac{3}{4}, the function is continuous on $[1, 3]$ and differentiable on $(1, 3)$ .
Mean Value Theorem Statement: The Mean Value Theorem states that there exists at least one number $c$ in the open interval $(1, 3)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$ , where $a = 1$ and $b = 3$ .
Calculate $f(a)$ and $f(b)$ : First, we calculate $f(a)$ and $f(b)$ :
$f(a) = f(1) = \sqrt{4\cdot 1 - 3} = \sqrt{1} = 1$
$f(b) = f(3) = \sqrt{4\cdot 3 - 3} = \sqrt{9} = 3$
Calculate Difference Quotient: Now, we calculate the difference quotient $[f(b) - f(a)] / (b - a)$ : $\frac{f(3) - f(1)}{3 - 1} = \frac{3 - 1}{3 - 1} = \frac{2}{2} = 1$
Find Derivative of $f(x)$ : Next, we find the derivative of $f(x)$ , $f'(x)$ :
$f'(x) = \frac{d}{dx} [\sqrt{4x - 3}]$
To differentiate $\sqrt{4x - 3}$ , we use the chain rule:
$f'(x) = \frac{1}{2\sqrt{4x - 3}} \cdot \frac{d}{dx} [4x - 3]$
$f'(x) = \frac{1}{2\sqrt{4x - 3}} \cdot 4$
$f'(x) = \frac{2}{\sqrt{4x - 3}}$
Set $f'(c)$ equal to Difference Quotient: Now, we set $f'(c)$ equal to the difference quotient we found earlier, which is $1$ : $\frac{2}{\sqrt{4c - 3}} = 1$ To solve for $c$ , we multiply both sides by $\sqrt{4c - 3}$ and then divide by $2$ : $2 = \sqrt{4c - 3}$ $2^2 = (\sqrt{4c - 3})^2$ $4 = 4c - 3$ $4 + 3 = 4c$ $7 = 4c$ $c = \frac{7}{4}$ $c = 1.75$

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