find the absolute maximum and absolute minimum values of f over the interval. ƒ(x) = (4/x) + ln (x^2), 1<=x <= 4

Step 1
Given,
${\displaystyle f\left(x\right)=\left(\frac{4}{x}\right)+\ln \left(x^2\right),1\le x\le 4}$
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
${\displaystyle f^{\prime }\left(x\right)=-\frac{4}{x^2}+\frac{1}{x^2}\cdot 2x}$
${\displaystyle \Rightarrow f^{\prime }\left(x\right)=-\frac{4}{x^2}+\frac{2}{x}}$
Now f'(x)=0
${\displaystyle \Rightarrow -\frac{4}{x^2}+\frac{2}{x}=0}$
${\displaystyle \Rightarrow -4+2x=0}$
${\displaystyle \Rightarrow 2x=4}$
${\displaystyle \Rightarrow x=2}$
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).
${\displaystyle f\left(1\right)=\frac{4}{1}+\ln \left(1^2\right)=4}$
${\displaystyle f\left(2\right)=\frac{4}{2}+\ln \left(2^2\right)=3.386294}$
${\displaystyle f\left(4\right)=\frac{4}{4}+\ln \left(4^2\right)=3.772588}$
Step 4
Therefore absolute minimum value is 3.386294 and absolute maximum value is 4.