Question

let $f(x) = \sqrt{x + 9}$ and let $c$ be the number that satisfies the mean value theorem for $f$ on the interval $0, 16$.
what is $c$?
choose 1 answer:
a 16
b 43
c 55
d 7

Explanation:

Step1: Recall Mean Value Theorem (MVT)

The MVT states that if \( f(x) \) is continuous on \([a,b]\) and differentiable on \((a,b)\), then there exists \( c \in (a,b) \) such that \( f'(c)=\frac{f(b)-f(a)}{b - a} \). Here, \( a = 0 \), \( b = 16 \), and \( f(x)=\sqrt{x + 9}=(x + 9)^{\frac{1}{2}} \).

Step2: Compute \( f(a) \) and \( f(b) \)

• \( f(0)=\sqrt{0 + 9}=3 \)
• \( f(16)=\sqrt{16 + 9}=\sqrt{25}=5 \)

Step3: Compute the average rate of change

\(\frac{f(16)-f(0)}{16 - 0}=\frac{5 - 3}{16}=\frac{2}{16}=\frac{1}{8}\)

Step4: Find the derivative of \( f(x) \)

Using the power rule, \( f'(x)=\frac{1}{2}(x + 9)^{-\frac{1}{2}}=\frac{1}{2\sqrt{x + 9}} \)

Step5: Set \( f'(c)=\frac{1}{8} \) and solve for \( c \)

\(\frac{1}{2\sqrt{c + 9}}=\frac{1}{8}\)
Cross - multiply: \( 2\sqrt{c + 9}=8 \)
Divide both sides by 2: \( \sqrt{c + 9}=4 \)
Square both sides: \( c + 9 = 16 \)
Subtract 9 from both sides: \( c=16 - 9 = 7 \)

Answer:

D. 7