Question

Maximum value of the function $f\left(x\right)=\frac{x}{8}+\frac{2}{x}$ on the interval $[1,6]$ is

• A. $1$
• B. $\frac{9}{8}$
• C. $\frac{13}{12}$
• D. $\frac{17}{8}$

Answer 1

The correct option is D $\frac{17}{8}$

$f\left(x\right)=\frac{x}{8}+\frac{2}{x}$
$\therefore f^{\prime }\left(x\right)=\frac{1}{8}-\frac{2}{x^2}=\frac{x^2-16}{8x^2}$

For maximum or minimum, $f^{\prime }\left(x\right)$ must be vanish.

$\therefore f^{\prime }\left(x\right)=0$

$\Rightarrow \frac{x^2-16}{8x^2}=0\Rightarrow x=4,-4$

$\because x\in [1,6]$

$\therefore x\ne -4$

Also, in $[1,4],f^{\prime }\left(x\right)<0\Rightarrow f\left(x\right)$ is decreasing.

In $[4,6],f^{\prime }\left(x\right)>0\Rightarrow f\left(x\right)$ is increasing.

$f\left(1\right)=\frac{1}{8}+\frac{2}{1}=\frac{17}{8}$

$f\left(8\right)=\frac{6}{8}+\frac{2}{6}=\frac{3}{4}+\frac{1}{3}=\frac{13}{12}$

Hence, maximum value of $f\left(x\right)$ in $[1,6]$ is $\frac{17}{8}$.