# Find the absolute extrema on the given interval. f(x)=x-\sqrt[3]{x}, [-1,4]

Find the absolute extrema on the given interval.
$f\left(x\right)=x-\sqrt{3}\left\{x\right\},\left[-1,4\right]$
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Given function is $f\left(x\right)=x-\sqrt{3}\left\{x\right\}$ and interval is [-1,4].
We need to find critical points of the function.
Differentiating the given function with respect to x,
${f}^{\prime }\left(x\right)=\frac{d}{dx}\left(x-\sqrt{3}\left\{x\right\}\right)$
$=\frac{d}{dx}\left(x\right)-\frac{d}{dx}\left({x}^{\frac{1}{3}}\right)$
$=1-\frac{1}{3}{x}^{\frac{1}{3}-1}$
$=1-\frac{1}{3}{x}^{-\frac{2}{3}}$
Putting first derivative equals to zero.
$1-\frac{1}{3}{x}^{-\frac{2}{3}}=0$
$\frac{1}{3}{x}^{-\frac{2}{3}}=1$
${x}^{-\frac{2}{3}}=3$
$x={3}^{-\frac{3}{2}}$
$x=\frac{1}{{3}^{\frac{3}{2}}}$
$x=\frac{1}{3\sqrt{3}}$
Therefore, critical value of the function is $\frac{1}{3\sqrt{3}}$.
Step 5
$\frac{1}{3\sqrt{3}}$ lies in the interval [-1,4].
Now we will evaluate the function at the endpoints of the interval and critical value.
$f\left(-1\right)=1-\sqrt{3}\left\{1\right\}$
=1-1
=0
$f\left(\frac{1}{3\sqrt{3}}\right)=1-\sqrt{3}\left\{\left(3\sqrt{3}\right)\right\}$
$=1-\sqrt{3}\left\{5.19\right\}$
=1-2.27=-1.27
Step 6
$f\left(4\right)=4-\sqrt{3}\left\{4\right\}$
=4-1.58
=2.42
The absolute maximum value of the function is 2.42 at x=4.
The absolute minimum value of the function is -1.27 at $x=\frac{1}{3\sqrt{3}}$.