 # find the absolute maximum and absolute minimum values of f over the interval. ƒ(x) = (4/x) + ln (x^2), 1<=x <= 4 Brittney Lord 2020-10-26 Answered
find the absolute maximum and absolute minimum values of f over the interval.
$ƒ\left(x\right)=\left(\frac{4}{x}\right)+\mathrm{ln}\left({x}^{2}\right),1\le x\le 4$
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Step 1
Given,
$f\left(x\right)=\left(\frac{4}{x}\right)+\mathrm{ln}\left({x}^{2}\right),1\le x\le 4$
Step 2
The absolute maximum or absolute minimum values of a function exist at a point where its first derivative is zero or at the end points of the given interval.
Now differentiating given function with respect to x, we get
${f}^{\prime }\left(x\right)=-\frac{4}{{x}^{2}}+\frac{1}{{x}^{2}}\cdot 2x$
$⇒{f}^{\prime }\left(x\right)=-\frac{4}{{x}^{2}}+\frac{2}{x}$
Now f'(x)=0
$⇒-\frac{4}{{x}^{2}}+\frac{2}{x}=0$
$⇒-4+2x=0$
$⇒2x=4$
$⇒x=2$
Step 3
Now computing the value of given function at x = 1, x = 4 (end points) and at x = 2 (point at which first derivative is zero).
$f\left(1\right)=\frac{4}{1}+\mathrm{ln}\left({1}^{2}\right)=4$
$f\left(2\right)=\frac{4}{2}+\mathrm{ln}\left({2}^{2}\right)=3.386294$
$f\left(4\right)=\frac{4}{4}+\mathrm{ln}\left({4}^{2}\right)=3.772588$
Step 4
Therefore absolute minimum value is 3.386294 and absolute maximum value is 4.
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