The function f(x) = ax + frac{b}{x} where a, | vedclass

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(D) Given the function $f(x) = ax + \frac{b}{x}$.To find the critical points,we differentiate $f(x)$ with respect to $x$:$f'(x) = a - \frac{b}{x^2}$.S

Vedclass · Accepted Answer

$\sqrt{\frac{b}{a}}$

Vedclass · Answer

(D) Given the function $f(x) = ax + \frac{b}{x}$.To find the critical points,we differentiate $f(x)$ with respect to $x$:$f'(x) = a - \frac{b}{x^2}$.Setting $f'(x) = 0$ for critical points:$a - \frac{b}{x^2} = 0 \Rightarrow a = \frac{b}{x^2} \Rightarrow x^2 = \frac{b}{a} \Rightarrow x = \sqrt{\frac{b}{a}}$ (since $x > 0$).Now,we find the second derivative to check for minima:$f''(x) = \frac{d}{dx}(a - bx^{-2}) = 0 - b(-2)x^{-3} = \frac{2b}{x^3}$.Since $b > 0$ and $x > 0$,$f''(x) = \frac{2b}{x^3} > 0$ for all $x > 0$.Since $f''(x) > 0$ at $x = \sqrt{\frac{b}{a}}$,the function attains its least value at $x = \sqrt{\frac{b}{a}}$.

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

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