 # Determine the location and value of the absolute extreme value of f on the given interval, if they exist. f(x)=x\ \ln \frac{x}{5}\ on\ \left[0.1, 5\right] Cem Hayes 2020-11-17 Answered
Determine the location and value of the absolute extreme value of f on the given interval, if they exist.
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Step 1
Given Data
The function is $f\left(x\right)=x\frac{\mathrm{ln}\left(x\right)}{5}$.
The interval is $\left[0.1,5\right]$.
Differentiate the function $f\left(x\right)=x\frac{\mathrm{ln}\left(x\right)}{5}$ with respect to x and equate to zero to evaluate the location of the absolute extreme,
$\frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(x\frac{\mathrm{ln}\left(x\right)}{5}\right)$
${f}^{prime}\left(x\right)=\frac{\mathrm{ln}\left(x\right)}{5}\frac{d}{dx}\left(x\right)+x\frac{d}{dx}\left(\frac{\mathrm{ln}\left(x\right)}{5}\right)$
$0=\frac{\mathrm{ln}\left(x\right)}{5}\cdot 1+x\left(\frac{1}{x}\right)\left(\frac{1}{5}\right)$
$\frac{\mathrm{ln}\left(x\right)}{5}+\frac{1}{5}=0$
$\frac{\mathrm{ln}\left(x\right)+1}{5}=0$
$\mathrm{ln}\left(x\right)=-1$
$x=-{e}^{-1}$
$x=\frac{1}{e}$
The value of absolute extreme $x=\frac{1}{e}$ lie in given interval $\left[0.1,5\right]$.
Hence the location of the absolute extreme value of f on the given interval is $x=\frac{1}{e}$.
Step 2
Evaluate the function $f\left(x\right)=x\frac{\mathrm{ln}\left(x\right)}{5}$ by substituting the value $x=\frac{1}{e}$,
$f\left(x\right)=x\frac{\mathrm{ln}\left(x\right)}{5}$
$f\left(\frac{1}{e}\right)=\frac{1}{e}\frac{\mathrm{ln}\left(\frac{1}{e}\right)}{5}$
$=-0.0735$
$<0$ Hence the minimum value of the absolute extreme value of f on the given interval is -0.0735, which is less than zero.

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