Answer
$f(0.2)=5.2$ is the absolute maximum value.
$f(1)=2$ is the absolute minimum value.
Work Step by Step
The Closed Interval Method
1) Function $f$ must be continuous on the closed interval $[a, b]$
2) Find the values of $f$ at the critical points in $(a,b)$ and the values of $f$ at the endpoints.
3) Compare the values.
- The largest is the absolute maximum value.
- The smallest is the absolute minimum value.
$$f(x)=x+\frac{1}{x,}\hspace{1.5cm}[0.2, 4]$$
First, $f(x)$ is continuous on the interval $[0.2,4]$, so the Closed Interval Method can be applied here.
1) Find the critical points of $f$
$$f'(x)=1+\frac{0x-1\times1}{x^2}$$
$$f'(x)=1-\frac{1}{x^2}$$
At $x=0$, $f'(x)$ does not exist. But $x=0$ does not lie in the domain of $f(x)$, which is $D_f=\{x|x\ne0\}$, so $x=0$ is not a critical point of $f$.
Also, $$f'(x)=0$$
$$\frac{1}{x^2}=1$$
$$x^2=1$$
$$x=\pm1$$
$x=\pm1$ all belong to the domain of $f$, so they are all critical points. However, only $x=1$ lies in interval $(0.2, 4)$ given, so only $x=1$ will move to the next sections.
2) Calculate the values of $f$ at critical points and endpoints.
- Critical points: $x=1$
$$f(1)=1+\frac{1}{1}=2$$
- End points: $x=0.2$ and $x=4$
$$f(0.2)=0.2+\frac{1}{0.2}=5.2$$
$$f(4)=4+\frac{1}{4}=4.25$$
3) Comparison
Among the results:
- $f(0.2)=5.2$ is the largest, so $f(0.2)=5.2$ is the absolute maximum value.
- $f(1)=2$ is the smallest, so $f(1)=2$ is the absolute minimum value.