Question

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

Answer 1

Given: $f\left(x\right)=ax+\frac{b}{x}$
$\Rightarrow f^{\prime }\left(x\right)=a-\frac{b}{x^2}$
For critical points, $f^{\prime }\left(x\right)=0$
$\Rightarrow a-\frac{b}{x^2}=0$
$\Rightarrow x=\sqrt{\frac{b}{a}}$ $\because x>0$
$f^{\prime \prime }\left(x\right)=\frac{2b}{x^3}$
$\Rightarrow f^{\prime \prime }\left(\sqrt{\frac{b}{a}}\right)=2b\times \frac{a^{\frac{3}{2}}}{b^{\frac{3}{2}}}>0$
$(\because a>0,b>0)$
Hence, $x=\sqrt{\frac{b}{a}}$ is a point of minima
Min. value of f(x)= $f\left(\sqrt{\frac{b}{a}}\right)=a\sqrt{\frac{b}{a}}+b\sqrt{\frac{a}{b}}=2\sqrt{ab}$
Hence, the least value of $f\left(x\right)$is $2\sqrt{ab}$.