Question

The function $f\left(x\right)=ax+\frac{b}{x}$ ,$a,b,x>0$ takes on the least value at $x=$

• A. $b$
• B. $\sqrt{a}$
• C. $\sqrt{b}$
• D. $\sqrt{\frac{b}{a}}$

Answer 1

The correct option is D

$\sqrt{\frac{b}{a}}$

Explanation for correct option

Given function $f\left(x\right)=ax+\frac{b}{x}$

$\Rightarrow f\text{'}\left(x\right)=a-\frac{b}{x^2}$

The maximum and minimum values of $f\left(x\right)$ occurs at the critical point.

The critical point of $f\left(x\right)$ are the values of $x$ for which $f\text{'}\left(x\right)=0$.

Now,

$f\text{'}\left(x\right)=0$

$$\begin{gathered} \Rightarrow a-\frac{b}{x^2}=0 \\ \Rightarrow a=\frac{b}{x^2} \\ \Rightarrow x^2=\frac{b}{a} \\ \Rightarrow x=\sqrt{\frac{b}{a}} \end{gathered}$$

Thus, the given function $f\left(x\right)$ attains the least value at $x=\sqrt{\frac{b}{a}}$.

Hence, option $\mathrm{Option}\left(\mathrm{D}\right)$ is correct .