The function f(x) = x^4 - frac{x^3}{3} is: | vedclass

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(A) Given the function $f(x) = x^4 - \frac{x^3}{3}$.To determine the intervals of increase and decrease,we find the first derivative $f'(x)$:$f'(x) =

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Increasing for $x > \frac{1}{4}$ and decreasing for $x < \frac{1}{4}$

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(A) Given the function $f(x) = x^4 - \frac{x^3}{3}$.To determine the intervals of increase and decrease,we find the first derivative $f'(x)$:$f'(x) = \frac{d}{dx}(x^4 - \frac{x^3}{3}) = 4x^3 - x^2$.Set $f'(x) = 0$ to find the critical points:$x^2(4x - 1) = 0$,which gives $x = 0$ and $x = \frac{1}{4}$.For the function to be increasing,$f'(x) > 0$:$x^2(4x - 1) > 0$.Since $x^2$ is always non-negative,$f'(x) > 0$ when $4x - 1 > 0$,which means $x > \frac{1}{4}$.For the function to be decreasing,$f'(x) $x^2(4x - 1) Thus,the function is increasing for $x > \frac{1}{4}$ and decreasing for $x < \frac{1}{4}$.

The function $f(x) = x^4 - \frac{x^3}{3}$ is:

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