Absolute Minimums and Maximums — Question 7

Question 7 – Solution

Strategy. Reduce the function to the radial variable $s=x^2+y^2\ge 0$ and study a one-variable function on an unbounded interval.

Step 1: Radial reduction Let $$g(s)=s+\frac 1{1+s},\qquad s\ge 0.$$ Then $$g'(s)=1-\frac 1{(1+s)^2}.$$ At $s=0$, $g'(0)=0$, and for every $s>0$, $(1+s)^2>1$, so $g'(s)>0$. Therefore $g$ increases strictly away from $s=0$.

It follows that $$\boxed{f_{\min}=1\text{ at }(0,0)}.$$

Step 2: No maximum As $x^2+y^2=s\to\infty$, $$g(s)=s+\frac 1{1+s}\longrightarrow\infty.$$ Thus $f$ is unbounded above and has .

Step 3: Interpretation The domain $\mathbb R^2$ is closed but not bounded, so the Extreme Value Theorem does not apply. Nevertheless, direct analysis proves that an absolute minimum is attained. Compactness guarantees both extrema for every continuous function, but a particular continuous function may still attain one or both extrema on a noncompact domain.