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

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

Vedclass · Accepted Answer

$1/6$

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(C) Given function: $f(x) = \frac{x}{4 + x + x^2}$.To find the critical points,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}$.Setting $f'(x) = 0$ gives $4 - x^2 = 0$,which implies $x = 2$ or $x = -2$.Since both $x = 2$ and $x = -2$ lie outside the given interval $[-1, 1]$,the maximum and minimum values must occur at the endpoints of the interval.Evaluating at the endpoints:$f(-1) = \frac{-1}{4 - 1 + 1} = \frac{-1}{4} = -0.25$.$f(1) = \frac{1}{4 + 1 + 1} = \frac{1}{6} \approx 0.1667$.Comparing these values,the maximum value is $1/6$.

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

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