Find the absolute maximum value and the absol | vedclass

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(A) The given function is $f(x) = x^{3}$.$	herefore f^{\prime}(x) = 3x^{2}$.Now,setting $f^{\prime}(x) = 0$ gives $3x^{2} = 0$,which implies $x = 0$.

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Absolute maximum value is $8$ and absolute minimum value is $-8$.

Vedclass · Answer

(A) The given function is $f(x) = x^{3}$.$	herefore f^{\prime}(x) = 3x^{2}$.Now,setting $f^{\prime}(x) = 0$ gives $3x^{2} = 0$,which implies $x = 0$.We evaluate the value of $f$ at the critical point $x = 0$ and at the endpoints of the interval $[-2, 2]$:$f(0) = 0^{3} = 0$.$f(-2) = (-2)^{3} = -8$.$f(2) = (2)^{3} = 8$.Comparing these values,the absolute maximum value of $f$ on $[-2, 2]$ is $8$ (at $x = 2$) and the absolute minimum value of $f$ on $[-2, 2]$ is $-8$ (at $x = -2$).

Find the absolute maximum value and the absolute minimum value of the function given by $f(x) = x^{3}, x \in [-2, 2]$.

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