Answer
Absolute maximum: $f(0.2)=5.2$
Absolute minimum: $f(1)=2$
Work Step by Step
Using closed interval method, as $f$ in continuous over the given range, let's first calculate the value of $f$ at critical numbers followed by values of $f$ at endpoints of the given range.
For critical numbers, $f'(x)=0$ or $f'(x)$ should not exist. Let's differentiate $f(x)=x+\frac1x$ with respect to $x$ keeping $[0.2,4]$ as interval.
$$\frac{df}{dx}=1-\frac{1}{x^2}$$ $\frac{df}{dx}$ exists on the given interval. Thus, $\frac{df}{dx}=0$, $$1-\frac{1}{x^2}=0\implies x^2-1=0\implies x=-1, 1.$$
Thus, $f(1)=2$. At endpoints,
$f(0.2)=5.2$ & $f(4)=4.25$. Thus, absolute maximum $=f(0.2)=5.2$ and absolute minimum $=f(1)=2$.
