Just need some help finding the absolute maximum and minimum of $f(t)=t-\sqrt[3]t$ for $t\in [-1,5]$.
So far, I have found the critical values of $0$, $\pm \frac{1}{3 \sqrt {3}}$ I just need to know where to go from here. Thanks.
Absolute maximum and minimum values
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calculus
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1Please use LaTex to write up questions in the future. – 2017-02-27
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0Just tried to format your question a bit in LaTeX. Please verify if your `1/3sqrt3` means `1/3 \sqrt 3` – $1/3 \sqrt 3$, or `1/(3 \sqrt 3)` — $1/(3 \sqrt 3)$. – 2017-02-27
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0thanks and it means 1/(3 \sqrt 3) – 2017-02-27
1 Answers
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now you know that $f$ increases on $(-\infty,-\frac{1}{3\sqrt{3}}]$, decreases on $[-\frac{1}{3\sqrt{3}},0)$ and $(0,\frac{1}{3\sqrt{3}}]$ and increases on $[\frac{1}{3\sqrt{3}},\infty)$.
also $-\frac{1}{3\sqrt{3}}>-1$.
By above results, the absolute maximum on $[-1,5]$ is $\max(f(-\frac{1}{3\sqrt{3}}),f(5))$ which turns out to be $f(5)$.