## Calculus: Early Transcendentals 8th Edition

$f(0.2)=5.2$ is the absolute maximum value. $f(1)=2$ is the absolute minimum value.
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.