Question

Let $f\left(x\right)=\sqrt{3x-3}$ and $c$ be the number that satisfies the Mean value theorem for $f$ on the interval $[4,13]$. What is the value of $c$ ?

• A. $11.5$
• B. $7.75$
• C. $7.5$
• D. $5.5$

Answer 1

Answer: (B)

$7.75$

Given : $f\left(x\right)=\sqrt{3x-3}(4,13)$

According to mean value theorem $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{(b-a)}$

$\frac{3}{2\sqrt{3\times -3}}=\frac{(6-3)}{(13-4)}=\frac{3}{9}=\frac{1}{3}9=2\sqrt{3c-3}81=4(3c-3)81=12c-1281+12=12c93=12cc=7.75$

Hence the correct answer is $c=7.75$