Step 1

Given,

\(f(x)=\frac{4}{x}+\ln(x^{2}), 1\leq x\leq 4\)

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

\(f'(x)=-\frac{4}{x^{2}}+\frac{1}{x^{2}} \cdot 2x\)

\(\Rightarrow f'(x)=-\frac{4}{x^{2}}+\frac{2}{x}\)

Now \(f'(x) = 0\)

\(\Rightarrow -\frac{4}{x^{2}}+\frac{2}{x}=0\)

\(\Rightarrow -4+2x=0\)

\(\Rightarrow 2x=4\)

\(\Rightarrow x = 2\)

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).

\(f(1)=\frac{4}{1}+\ln(1^{2})=4\)

\(f(2)=\frac{4}{2}+\ln(2^{2})=3.386294\)

\(f(4)=\frac{4}{4}+\ln(4^{2})=3.772588\)

Step 4

Therefore absolute minimum value is 3.386294 and absolute maximum value is 4.