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

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

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.