The minimum value of frac{(5 + x)(2 + x)}{1 + | vedclass

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(C) Let $f(x) = \frac{(5 + x)(2 + x)}{1 + x}$.We can rewrite the expression as:$f(x) = \frac{x^2 + 7x + 10}{x + 1} = \frac{x(x + 1) + 6(x + 1) + 4}{x

Vedclass · Accepted Answer

$9$

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(C) Let $f(x) = \frac{(5 + x)(2 + x)}{1 + x}$.We can rewrite the expression as:$f(x) = \frac{x^2 + 7x + 10}{x + 1} = \frac{x(x + 1) + 6(x + 1) + 4}{x + 1} = x + 6 + \frac{4}{x + 1}$.To find the minimum,we differentiate $f(x)$ with respect to $x$:$f'(x) = 1 - \frac{4}{(x + 1)^2}$.Setting $f'(x) = 0$ for critical points:$1 = \frac{4}{(x + 1)^2} \implies (x + 1)^2 = 4 \implies x + 1 = \pm 2$.Since $x$ is non-negative,$x + 1 = 2$,which gives $x = 1$.Using the second derivative test:$f''(x) = \frac{8}{(x + 1)^3}$.At $x = 1$,$f''(1) = \frac{8}{8} = 1 > 0$,so $x = 1$ is a point of local minima.The minimum value is $f(1) = \frac{(5 + 1)(2 + 1)}{1 + 1} = \frac{6 	imes 3}{2} = 9$.

The minimum value of $\frac{(5 + x)(2 + x)}{1 + x}$ for non-negative real $x$ is

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