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

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(C) Given $f(x) = \frac{x}{x^2 + x + 4}$.To find the critical points,we calculate the derivative $f'(x)$:$f'(x) = \frac{(x^2 + x + 4)(1) - x(2x + 1)}{

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$\frac{1}{6}$

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(C) Given $f(x) = \frac{x}{x^2 + x + 4}$.To find the critical points,we calculate the derivative $f'(x)$:$f'(x) = \frac{(x^2 + x + 4)(1) - x(2x + 1)}{(x^2 + x + 4)^2} = \frac{x^2 + x + 4 - 2x^2 - x}{(x^2 + x + 4)^2} = \frac{4 - x^2}{(x^2 + x + 4)^2}$.Setting $f'(x) = 0$ gives $4 - x^2 = 0$,so $x = 2$ or $x = -2$.Neither $x = 2$ nor $x = -2$ lies in the interval $[-1, 1]$.Therefore,the function is monotonic on $[-1, 1]$.We evaluate the function 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 the values,the maximum value is $\frac{1}{6}$.

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

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