The minimum value of left( x^2 + frac{250}{x} | vedclass

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(A) Let $f(x) = x^2 + \frac{250}{x}$.To find the minimum value,we first find the derivative $f'(x)$:$f'(x) = 2x - \frac{250}{x^2}$.Set $f'(x) = 0$ to

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(A) Let $f(x) = x^2 + \frac{250}{x}$.To find the minimum value,we first find the derivative $f'(x)$:$f'(x) = 2x - \frac{250}{x^2}$.Set $f'(x) = 0$ to find the critical points:$2x - \frac{250}{x^2} = 0 \implies 2x^3 = 250 \implies x^3 = 125 \implies x = 5$.Now,we check the second derivative $f''(x)$ to confirm the nature of the extremum:$f''(x) = 2 + \frac{500}{x^3}$.At $x = 5$,$f''(5) = 2 + \frac{500}{125} = 2 + 4 = 6$.Since $f''(5) > 0$,the function has a local minimum at $x = 5$.The minimum value is $f(5) = 5^2 + \frac{250}{5} = 25 + 50 = 75$.

The minimum value of $\left( x^2 + \frac{250}{x} \right)$ is

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