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

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(C) Given function is $f(x) = \frac{x}{4 + x + x^2}$.To find the maximum value,we differentiate $f(x)$ with respect to $x$ using the quotient rule:$f'

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$\frac{1}{6}$

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(C) Given function is $f(x) = \frac{x}{4 + x + x^2}$.To find the maximum value,we differentiate $f(x)$ with respect to $x$ using the quotient rule:$f'(x) = \frac{(4 + x + x^2)(1) - x(1 + 2x)}{(4 + x + x^2)^2}$$f'(x) = \frac{4 + x + x^2 - x - 2x^2}{(4 + x + x^2)^2} = \frac{4 - x^2}{(4 + x + x^2)^2}$.For critical points,set $f'(x) = 0$,which implies $4 - x^2 = 0$,so $x = \pm 2$.Since $x = \pm 2$ are not in the interval $[-1, 1]$,the function has no critical points in the given interval.Since $4 - x^2 > 0$ for all $x \in [-1, 1]$,$f'(x) > 0$,meaning $f(x)$ is strictly increasing on $[-1, 1]$.The maximum value occurs at the right endpoint $x = 1$:$f(1) = \frac{1}{4 + 1 + 1^2} = \frac{1}{6}$.

The maximum value of $f(x) = \frac{x}{4 + x + x^2}$ on $[-1, 1]$ is

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