Question
In Exercises $17-36,$ locate the absolute extrema of the function on the closed interval.$$g(x)=\sqrt[3]{x},[-1,1]$$
Step 1
To do this, we need to find the derivative of the function. The derivative of $g(x)=\sqrt[3]{x}$ is $g'(x)=\frac{1}{3}x^{-\frac{2}{3}}$. Show more…
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