Question

Find the absolute maximum and absolute minimum values of f over the interval. f(x)=(\frac{4}{x})+\ln(x^{2}), 1\leq x\leq 4

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asked 2021-05-21
Find the absolute maximum and absolute minimum values of f over the interval. \(f(x)=(\frac{4}{x})+\ln(x^{2}), 1\leq x\leq 4\)

Answers (1)

2021-05-22

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.

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