If f(x)=3 sqrt[3]{x^2}-x^2, then | vedclass

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(B) Given $f(x) = 3x^{2/3} - x^2$. To find the extrema, we differentiate $f(x)$ with respect to $x$: $f'(x) = 3 \cdot \frac{2}{3} x^{-1/3} - 2x = 2x^{

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$f$ is maximum at two points $x=1$ and $x=-1$

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(B) Given $f(x) = 3x^{2/3} - x^2$. To find the extrema, we differentiate $f(x)$ with respect to $x$: $f'(x) = 3 \cdot \frac{2}{3} x^{-1/3} - 2x = 2x^{-1/3} - 2x = 2 \left( \frac{1 - x^{4/3}}{x^{1/3}} \right)$. Setting $f'(x) = 0$, we get $1 - x^{4/3} = 0$, which implies $x^{4/3} = 1$, so $x = 1$ or $x = -1$. Also, $f'(x)$ is undefined at $x = 0$. We analyze the sign of $f'(x)$ around the critical points $x = -1, 0, 1$: For $x  0$. For $-1 For $0  0$. For $x > 1$, $f'(x) At $x = -1$, $f'(x)$ changes from positive to negative, so $f$ has a local maximum at $x = -1$. At $x = 0$, $f'(x)$ changes from negative to positive, so $f$ has a local minimum at $x = 0$. At $x = 1$, $f'(x)$ changes from positive to negative, so $f$ has a local maximum at $x = 1$. Thus, $f$ is maximum at two points $x = 1$ and $x = -1$.

If $f(x)=3 \sqrt[3]{x^2}-x^2$, then

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