Question

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

Answer 1

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

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

Differentiating both sides with respect to x, we get

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

For maxima or minima,

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

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

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

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

Now,

$f\text{'}\text{'}\left(x\right)=\frac{2b}{x^3}$

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

$f\text{'}\text{'}\left(\sqrt{\frac{b}{a}}\right)=\frac{2b}{{\left(\sqrt{{\displaystyle \frac{b}{a}}}\right)}^3}=2a\sqrt{\frac{a}{b}}>0$ (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

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

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

$=\sqrt{ab}+\sqrt{ab}$

$=2\sqrt{ab}$

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


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