# The least value of the function

Question:

The least value of the function $f(x)=a x+\frac{b}{x}(a>0, b>0, x>0)$ is_____________

Solution:

The given function is $f(x)=a x+\frac{b}{x}(a>0, b>0, x>0)$.

$f(x)=a x+\frac{b}{x}$

Differentiating both sides with respect to x, we get

$f^{\prime}(x)=a-\frac{b}{x^{2}}$

For maxima or minima,

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

$\Rightarrow a-\frac{b}{x^{2}}=0$

$\Rightarrow x^{2}=\frac{b}{a}$

$\Rightarrow x=\sqrt{\frac{b}{a}} \quad(x>0)$

Now,

$f^{\prime \prime}(x)=\frac{2 b}{x^{3}}$

At $x=\sqrt{\frac{b}{a}}$, we have

$f^{\prime} \prime\left(\sqrt{\frac{b}{a}}\right)=\frac{2 b}{\left(\sqrt{\frac{b}{a}}\right)^{3}}=2 a \sqrt{\frac{a}{b}}>0 \quad(a>0, b>0)$

So, $x=\sqrt{\frac{b}{a}}$ is the point of local minimum of $f(x)$. Thus, the function takes the least value at $x=\sqrt{\frac{b}{a}}$.

∴ Least value of the given function

$=f\left(\sqrt{\frac{b}{a}}\right)$

$=a \times \sqrt{\frac{b}{a}}+\frac{b}{\sqrt{\frac{b}{a}}}$          $\left[f(x)=a x+\frac{b}{x}\right]$

$=\sqrt{a b}+\sqrt{a b}$

$=2 \sqrt{a b}$

Thus, the least value of the function $f(x)=a x+\frac{b}{x}(a>0, b>0, x>0)$ is $2 \sqrt{a b}$.

The least value of the function $f(x)=a x+\frac{b}{x}(a>0, b>0, x>0)$ is   $2 \sqrt{a b}$