The function
Question:

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

Solution:

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

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

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

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

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

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