Question

The least value of the function $f(x)=ax+b/x,a>0,b>0,x>0$ is

• A. $\sqrt{ab}$
• B. $2\sqrt{\frac{a}{b}}$
• C. $2\sqrt{\frac{b}{a}}$
• D. $2\sqrt{ab}$

Answer 1

Answer: (D)

Given, $f(x)=ax+\frac{b}{x}$
On differentiating w.r.t. ' $x$ ', we get
$f^{\prime }(x)=a-\frac{b}{x^2}$
Put $f^{\prime }(x)=0$,
$\Rightarrow a-\frac{b}{x^2}=0$
$\Rightarrow x=\pm \sqrt{\frac{b}{a}}$
$\therefore f^{\prime \prime }(x)=\frac{2b}{x^3}$
At $x=\sqrt{\frac{b}{a}}$,
$f^{\prime \prime }(x)=\frac{2b}{{\left(\sqrt{\frac{b}{a}}\right)}^3}>0,\text{ minima }$
$\therefore$ Minimum value of $f(x)$ is
$f\left(\sqrt{\frac{b}{a}}\right)=a\sqrt{\frac{b}{a}}+\frac{b}{\sqrt{\frac{b}{a}}}$
$=\sqrt{ab}+\sqrt{ab}=2\sqrt{ab}$