Question

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

Answer 1

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

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

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

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

⇒ $x=\sqrt{\frac{b}{a}}$ (x > 0)

Now,

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

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

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

So, $x=\sqrt{\frac{b}{a}}$ is the point of local minimum of f(x).

Thus, the function takes the least value at $x=\sqrt{\frac{b}{a}}$.


The function f(x) = ax + $\frac{b}{x}$ , a, b, x > 0 takes on the least value at x equal to $\underline{\sqrt{\frac{b}{a}}}$.