## Related questions with answers

Determine the location and value of the absolute extreme values of f on the given interval, if they exist. $f(x)=3 x^{2 / 3}-x \text { on }[0,27]$

Solution

VerifiedDifferentiating $f(x)=3x^{\frac{2}{3}}-x$ on $[0,27]$, we have

$\begin{align*} f'(x)&= 3\cdot \frac{2}{3}x^{\frac{2}{3}-1}-1\\ &=2x^{-\frac{1}{3}}-1\\ &=\dfrac{2}{\sqrt[3]{x}}-1 \end{align*}$

To find the critical points, we solve the equation $f'(x)=0$ and we search for values where $f'(x)$ is undefined.

$\begin{gather*} f'(x)=0\\ \dfrac{2}{\sqrt[3]{x}}-1=0\\ \dfrac{2}{\sqrt[3]{x}}=1\\ \sqrt[3]{x}=2\\ x=8 \end{gather*}$

Also, at $x=0$, $f'(x)$ is undefined. [10pt] These two critical points and the endpoints are candidates for the location of the absolute extrema.

The next step is to evaluate $f$ at the critical points and endpoints.

$\begin{align*} f(0)=3\cdot0^{\frac{2}{3}}-0=0\\ f(8)=3\cdot8^{\frac{2}{3}}-8=4\\ f(27)=3\cdot27^{\frac{2}{3}}-27=0\\ \end{align*}$

The largest of these function values is $f(8)=4$, which is the absolute maximum value of $f$ on $[0,27]$. The least of these values is $f(0)=f(27)=0$, which is the absolute minimum value of $f$ on $[0,27]$.

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