Answer
$f(1)=1$ is the absolute maximum value.
$f(0)=0$ 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.
$$f(x)=\frac{x}{x^2-x+1},\hspace{1.5cm}[0, 3]$$
First, $f(x)$ is continuous on the interval $[0,3]$, so the Closed Interval Method can be applied here.
1) Find critical points of $f$
$$f'(x)=\frac{1(x^2-x+1)-x(2x-1)}{(x^2-x+1)^2}$$
$$f'(x)=\frac{x^2-x+1-2x^2+x}{(x^2-x+1)^2}$$
$$f'(x)=\frac{-x^2+1}{(x^2-x+1)^2}$$
$f'(x)$ does not exist when $$x^2-x+1=0.$$ However, as $x^2-x+1\ne0$ for all real x, there are no values of $x$ so that $f'(x)$ does not exist.
Also, $$f'(x)=0$$
$$-x^2+1=0$$
$$x^2=1$$
$$x=\pm1$$
$x=\pm1$ all belong to the domain of $f$, so they are all critical points. However, only $x=1$ lies in interval $(0, 3)$ given, so only $x=1$ will move to the next sections.
2) Calculate the values of $f$ at critical points and endpoints.
- Critical points: $x=1$
$$f(1)=\frac{1}{1^2-1+1}=\frac{1}{1}=1$$
- End points: $x=0$ and $x=3$
$$f(0)=\frac{0}{0^2-0+1}=0$$
$$f(3)=\frac{3}{3^2-3+1}=\frac{3}{7}$$
3) Comparison
Among the results:
- $f(1)=1$ is the largest, so $f(1)=1$ is the absolute maximum value.
- $f(0)=0$ is the smallest, so $f(0)=0$ is the absolute minimum value.
