Question

Let $f\left(x\right)=\sqrt{x+4}$. Then the point $c$ that satisfies the mean value theorem for the function on the interval $[0,5]$, is

• A. $\frac{1}{4}$
• B. $\frac{25}{4}$
• C. $\frac{9}{16}$
• D. $\frac{9}{4}$

Answer 1

The correct option is D $\frac{9}{4}$

The function is continuous on the closed interval $[0,5]$ and differentiable on the open interval $(0,5)$, so the LMVT is applicable for this function

$f^{\prime }\left(x\right)=\left(\sqrt{x+4}\right)=\frac{1}{2\sqrt{x+4}}$
$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a},$
$\Rightarrow \frac{1}{2\sqrt{c+4}}=\frac{\sqrt{5+4}-\sqrt{0+4}}{5-0}$
$\Rightarrow \frac{1}{2\sqrt{c+4}}=\frac{1}{5}$
$\Rightarrow \sqrt{c+4}=\frac{5}{2}\Rightarrow c+4=\frac{25}{4}$
$\Rightarrow c=\frac{9}{4}$