Find the absolute extreme values of the function.
$F(x)=\sqrt[3]{x}; -3 \leq x \leq 8$
Hi everyone, I have a calc test tomorrow and I am still confused on this question. We have some study questions for the exam, and this question is one of them.
The answer key says the Minimum is at $x=0$ and Maximum is at $x=8$. I found 0 as a critical point since $f '(0)$ does not exist, but shouldn't the absolute minimum be at $x=-3$, $y=\sqrt[3]{-3}$ since it is less than $y=\sqrt[3]{0}$ ? I don't understand how $x=0$ would even be a local min since the slope of $F(x)$ is always positive on the interval.
Was this a typo, or am I missing something about odd roots of negative numbers?
Thanks.