Global extrema of $x+1/x$ on a closed interval¶

Math I: Calculus · Week 3 test, open question · 7 points · Lectures 5–6

Problem¶

Let $f\colon[1/2,3]\to\mathbb{R}$ be defined by $f(x)=x+1/x\text{.}$

Explain why $f$ must attain both a global minimum and a global maximum on $[1/2,3]\text{.}$ [2 points]
Find every stationary point of $f$ in $(1/2,3)$ and classify each one as a local minimum, a local maximum, or neither. [2 points]
Determine the global minimum and global maximum of $f$ on $[1/2,3]\text{,}$ including the points where they are attained. [3 points]

What you need¶

Extreme value theorem: a continuous function on a closed, bounded interval attains a maximum and a minimum.
Stationary point and the first-order condition.
Local extremum and global extremum.
Closed and bounded intervals, whose endpoints are candidates too.

Plan¶

The extreme value theorem says the extrema exist, and the first-order condition says where they can be: at an interior stationary point or at an endpoint. So the method is a short list of candidates and a comparison of values.

Step-by-step solution¶

Try each step yourself before you open it.

Part 1 · Existence (2 points)

$f$ is continuous on $[1/2,3]$ (it is a sum of continuous functions and $0\notin[1/2,3]$ $[1/2, 3]$ ), and $[1/2,3]$ is closed and bounded. By the extreme value theorem, $f$ attains a global minimum and a global maximum there.

Both hypotheses must be named. On $(0,3]$ the same formula has no maximum.

Concept: Extreme value theorem

Part 2, Step 1 · Find the stationary points

For $x\in(1/2,3)\text{,}$

f'(x)=1-\frac{1}{x^2}.

$f'(x)=0$ means $x^2=1\text{,}$ and only $x=1$ lies in the interval.

Part 2, Step 2 · Classify x = 1 (with Step 1: 2 points)

For every $x>0\text{,}$

f(x)-f(1)=x+\frac{1}{x}-2=\frac{(x-1)^2}{x}\geq0,

with equality only at $x=1\text{.}$ So $x=1$ is a strict local minimum.

A second-derivative check works too: $f''(x)=2/x^3\text{,}$ so $f''(1)=2>0\text{.}$

Part 3 · Compare candidates (3 points)

The global extrema can only occur at the stationary point or at the endpoints:

f\left(\tfrac{1}{2}\right)=\frac{5}{2},\quad f(1)=2,\quad f(3)=\frac{10}{3}.

The global minimum is $2\text{,}$ attained at $x=1\text{.}$ The global maximum is $10/3\text{,}$ attained at $x=3\text{.}$

Concept: Global extremum

Where marks are lost¶

Naming the theorem without its hypotheses. Say that $f$ is continuous and the interval is closed and bounded.
Forgetting the endpoints. Here the global maximum is at an endpoint, where $f'\neq0\text{.}$
Keeping $x=-1\text{.}$ It solves $f'(x)=0$ but is outside the domain.

Try next¶

Stationary points, convexity and an alternating series
A function with three local extrema: the same tools with a parameter.

Symbols: element of, real numbers

Global extrema of x+1/xx+1/xx+1/x on a closed interval¶