If f(x) = ax + frac{b}{x} where a, b, x 0,t | vedclass

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(D) Given the function $f(x) = ax + \frac{b}{x}$.To find the critical points,we calculate the first derivative $f'(x)$:$f'(x) = \frac{d}{dx}(ax + bx^{

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

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(D) Given the function $f(x) = ax + \frac{b}{x}$.To find the critical points,we calculate the first derivative $f'(x)$:$f'(x) = \frac{d}{dx}(ax + bx^{-1}) = a - \frac{b}{x^2}$.Setting $f'(x) = 0$ for extrema:$a - \frac{b}{x^2} = 0 \Rightarrow a = \frac{b}{x^2} \Rightarrow x^2 = \frac{b}{a} \Rightarrow x = \sqrt{\frac{b}{a}}$ (since $x > 0$).Now,we check the second derivative $f''(x)$ to confirm the nature of the extremum:$f''(x) = \frac{d}{dx}(a - bx^{-2}) = 0 - b(-2)x^{-3} = \frac{2b}{x^3}$.Since $b > 0$ and $x > 0$,$f''(x) = \frac{2b}{x^3} > 0$.Since the second derivative is positive at $x = \sqrt{\frac{b}{a}}$,the function $f(x)$ attains its minimum value at $x = \sqrt{\frac{b}{a}}$.

If $f(x) = ax + \frac{b}{x}$ where $a, b, x > 0$,then the value of $x$ for which $f(x)$ is minimum is:

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