The minimum value of f(x) = x + frac{4}{x + 2 | vedclass

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(D) Given function: $f(x) = x + \frac{4}{x + 2}$ for $x > -2$.To find the critical points,we calculate the derivative $f'(x)$ and set it to $0$:$f'(x)

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$2$

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(D) Given function: $f(x) = x + \frac{4}{x + 2}$ for $x > -2$.To find the critical points,we calculate the derivative $f'(x)$ and set it to $0$:$f'(x) = 1 - \frac{4}{(x + 2)^2}$Setting $f'(x) = 0$:$1 = \frac{4}{(x + 2)^2} \implies (x + 2)^2 = 4$Since $x > -2$,we have $x + 2 = 2$,which gives $x = 0$.Now,we check the second derivative $f''(x) = \frac{8}{(x + 2)^3}$.At $x = 0$,$f''(0) = \frac{8}{2^3} = 1 > 0$.Since $f''(0) > 0$,the function has a local minimum at $x = 0$.The minimum value is $f(0) = 0 + \frac{4}{0 + 2} = \frac{4}{2} = 2$.

The minimum value of $f(x) = x + \frac{4}{x + 2}$ for $x > -2$ is

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