The sum of the global minimum and global maxi | vedclass

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(A) Given the function $f(x) = \frac{4}{3}x^3 - 4x$ on the interval $[0, 2]$.First,find the critical points by setting the derivative to zero:$f'(x) =

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(A) Given the function $f(x) = \frac{4}{3}x^3 - 4x$ on the interval $[0, 2]$.First,find the critical points by setting the derivative to zero:$f'(x) = 4x^2 - 4 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$.Since we are considering the interval $[0, 2]$,we only consider $x = 1$ as a critical point within the interval.Now,evaluate the function at the critical point and the endpoints of the interval:$f(0) = \frac{4}{3}(0)^3 - 4(0) = 0$.$f(1) = \frac{4}{3}(1)^3 - 4(1) = \frac{4}{3} - 4 = -\frac{8}{3}$.$f(2) = \frac{4}{3}(2)^3 - 4(2) = \frac{32}{3} - 8 = \frac{32 - 24}{3} = \frac{8}{3}$.Comparing these values,the global minimum is $-\frac{8}{3}$ and the global maximum is $\frac{8}{3}$.The sum of the global minimum and global maximum values is $\frac{8}{3} + (-\frac{8}{3}) = 0$.

The sum of the global minimum and global maximum values of the function $f(x) = \frac{4}{3}x^3 - 4x$ in the interval $[0, 2]$ is:

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