The Function f(x) = x + a²/x · Little Blue Book, Visualized

Try this

Drag $P$ . The two colored vertical bars are $x$ and $\frac{a ^{2}}{x}$ . When one gets longer, what does the other do? What about their product?
When does $P$ land on the dashed “floor” $y = 2 a$ ? Are the two bars equally long then? What shape is the rectangle?
Drag $P$ far to the right: which line does the curve get closer and closer to? And when you drag it left, close to the $y$ -axis?
Switch on “Restrict to [m, n]” and move the whole interval to the left or to the right of $x = a$ . Is the minimum still on the floor?

Let $a > 0$ . For $x > 0$ , the numbers $x$ and $\frac{a ^{2}}{x}$ are the sides of a rectangle of area $a^{2}$ , and $f (x)$ is its half-perimeter. For a fixed area, the square has the smallest perimeter:

f (x) = x + \frac{a ^{2}}{x} = 2 a + (x - \frac{a}{x})^{2} \geq 2 a, with equality if and only if x = a .

$f (x)$ is an odd function, so $f (x) \leq - 2 a$ for $x < 0$ , and the range is $(- \infty, - 2 a] \cup [2 a, + \infty)$ . The graph looks like a “check mark” (✓), with asymptotes the $y$ -axis and the line $y = x$ .

Predict first

For $x > 0$ , what is the minimum of $x + \frac{9}{x}$ ?

Choose an answer first; then you can see the result.

Why: the minimum and the monotonicity

f(x) = x + 4/x (a = 2) on (0, +∞)

Step 1 / 6

Two terms with a fixed product
At $x = 1$ the two terms are $1$ and $4$ , with sum $5$ . Whatever $x$ is, the product of the two terms is $a^{2} = 4$ .

$x \cdot \frac{4}{x} = 4$
The rectangle
Smallest for the square
Decreasing on the left part
Increasing on the right part
Max and min on an interval

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: $x + \frac{a ^{2}}{x} \geq 2 a$

On the right branch alone, the curve always stays above the floor $y = 2 a$ (or just touches it).

Practice

Exercise 1Basic

For $x > 0$ , find the minimum of $x + \frac{16}{x}$ .

Exercise 2Basic

Find the minimum of $f (x) = x + \frac{4}{x}$ on $[3, 6]$ .

Exercise 3Advanced

For $x > - 1$ , find the minimum of $y = \frac{x ^{2} + x + 4}{x + 1}$ .

Exercise 4Challenge

The minimum of $f (x) = x + \frac{1}{x}$ on $(0, t]$ is $\frac{5}{2}$ . Find $t$ .

Mistake Museum · Exhibits from this stop

Before you leave, look at the traps other people fell into here.See every exhibit in the museum →

Exhibit

Equality cannot be reached, yet the minimum is still given as $2a$

The minimum of

x + \frac{4}{x}

[3, 6]

24 = 4

What went wrong? Think first, then open

Diagnosis Equality in $x + \frac{4}{x} \geq 4$ holds at $x = 2$ , and $2 \in / [3, 6]$ . On this interval the value is always greater than $4$ , so $4$ is not the minimum.

[3, 6]

lies entirely to the right of the turning point

x = 2

, where the function is increasing, so the minimum is

f (3) = 3 + \frac{4}{3} = \frac{13}{3}

“When you use AM–GM to find an extreme value, first ask whether equality can actually be reached”

Exhibit

Substituting without checking the range of the new variable

y = x^{2} + 5 + \frac{1}{x ^{2} + 5} \geq 2

, so the minimum is

2

What went wrong? Think first, then open

Diagnosis Equality needs $x^{2} + 5 = 1$ , which is impossible. Let $t = x^{2} + 5$ ; then $t \geq 5 > 1$ , which is on the increasing part of $t + \frac{1}{t}$ .

The minimum is reached at

t = 5

(that is,

x = 0

5 + \frac{1}{5} = \frac{6 5}{5}

“After a substitution, write down the range of the new variable first”