Question

Let $f\left(x\right)=\sqrt{2x+1}$ and let $c$ be the number that satisfies the Mean value theorem for $f$ on the interval $[4,12]$. What is $c$ ?

• A. $0$
• B. $1.5$
• C. $4$
• D. $7.5$

Answer 1

The correct option is D $7.5$

MVT states that there exists some $c$ such that $f^1\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Here $a=4$ and $b=12$

$f\left(a\right)=f\left(4\right)=\sqrt{9}=3$

$f\left(b\right)=f\left(12\right)=\sqrt{25}=5$

$f^1\left(c\right)=\frac{1}{\sqrt{2c+1}}$

Therefore $\frac{1}{\sqrt{2c+1}}=\frac{2}{8}=\frac{1}{4}$

$\Rightarrow 2c+1=16$

$\Rightarrow c=\frac{15}{2}=7.5$