Maxima and Minima of Functions

What this move does

Saying “the minimum of $f (x)$ is $m$ ” really says two things at once:

f (x) \geq m for every x in the domain and there is an x_{0} with f (x_{0}) = m

The first half is bounding; the second half is reaching equality. The six methods of this chapter (completing the square, the discriminant method, inequalities, substitution, construction and using properties of the function) are all different tools for the bound. The equality step you have to supply yourself, every time.

Bound: use some tool to prove $f (x) \geq m$ (or $f (x) \leq M$ ).
Check: find the $x_{0}$ where equality holds, and make sure it lies in the domain.
Only when both steps are done may you write “the minimum is $m$ ”. With only the first step, $m$ is just a bound.

The probe above draws exactly this. If the line $y = m$ stays below the graph without touching it, it is a lower bound; if it touches the graph, the value is reached. Only a line that stays on one side and touches gives an extreme value.

Predict first

What is the minimum of $y = \frac{2 x ^{2} + 2 x + 5}{x ^{2} + x + 1}$ ?

2

\frac{8}{3}

3

There is no minimum

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

Work through one: the discriminant method

When the function is “quadratic over quadratic”, the most reliable tool is the discriminant method: treat $y$ as a parameter and ask “when does the equation $f (x) = y$ have a real solution?”

Example: find the maximum and minimum of y = (x² + x + 1)/(x² + 1)

Step 1 / 6

Treat y as a parameter
A number $m$ is in the range exactly when the equation $f (x) = m$ has a solution. Clear the denominator (it is $x^{2} + 1 > 0$ ):

$(1 - m) x^{2} + x + (1 - m) = 0$
An m with no intersection
Solve Δ ≥ 0
Maximum: tangent where Δ = 0
Minimum
Do not forget m = 1

Six tools, one use each

Each example below comes with data where equality can be reached. Check the equality case yourself.

Tool	Example	Bound	Equality
Completing the square	$u = x^{2} + 2 y^{2} - 2 x y - 4 y + 7$	$u = (x - y)^{2} + (y - 2)^{2} + 3 \geq 3$	$x = y = 2$
Discriminant method	$y = \frac{x ^{2} + x + 1}{x ^{2} + 1}$	$Δ \geq 0 \Rightarrow \frac{1}{2} \leq y \leq \frac{3}{2}$	$x = \mp 1$
Inequalities	$y = x + \frac{4}{x - 1} (x > 1)$	$(x - 1) + \frac{4}{x - 1} + 1 \geq 2 \cdot 2 + 1 = 5$	$x = 3$
Substitution	$y = x - 2 x - 1$	let $t = 2 x - 1 \geq 0$ ; $y = \frac{( t - 1 ) ^{2}}{2} \geq 0$	$t = 1$ , i.e. $x = 1$
Construction	$y = x^{2} + 1 + (x - 3)^{2} + 4$	sum of distances from $(x, 0)$ to $(0, 1)$ and $(3, - 2)$ is $\geq 32$	$x = 1$
Properties of the function	$y = x + \frac{9}{x} (1 \leq x \leq 2)$	decreasing on $(0, 3]$ , so $y \geq f (2) = \frac{13}{2}$	$x = 2$

Often more than one tool works on the same problem, but they cost different amounts:

One problem, several solutions

Find the range of $y = \frac{2 x ^{2} + 2 x + 5}{x ^{2} + x + 1}$ .

Substitution + monotonicity

$t = x^{2} + x + 1 \geq \frac{3}{4}$ , and $y = 2 + \frac{3}{t}$ .

$\frac{3}{t}$ is decreasing on $[\frac{3}{4}, + \infty)$ , and takes the values $(0, 4]$ .

So $y \in (2, 6]$ .

CostYou see at a glance that 2 is never reached, but first you must spot that the numerator is a multiple of the denominator plus a constant.

Discriminant method

$(2 - y) x^{2} + (2 - y) x + (5 - y) = 0$ .

If $y \neq = 2$ : $Δ = (2 - y)^{2} - 4 (2 - y) (5 - y) = (2 - y) (3 y - 18) \geq 0$ , which gives $2 < y \leq 6$ .

If $y = 2$ : the equation is $3 = 0$ , with no solution. So the range is $(2, 6]$ .

CostNo trick needed, just mechanical work; but the case where the leading coefficient is 0 must be checked separately.

When it fails

Break it

Counterexample lab: remove one condition

Conclusion: $Δ \geq 0$ means that $m$ is reached

$m = 6$ : the equation $- 4 x^{2} - 4 x - 1 = 0$ is quadratic, $Δ = 0$ , and it has one solution $x = - \frac{1}{2}$ . The line is tangent to the graph.

Practice

Exercise 1Basic

Find the minimum of $u = x^{2} + y^{2} - 4 x + 2 y + 9$ ( $x, y$ real).

Exercise 2Advanced

Find the minimum of $y = \frac{x ^{2} + 5}{x ^{2} + 4}$ .

Exercise 3Advanced

Find the maximum of $y = \frac{x ^{2} - x + 1}{x ^{2} + x + 1}$ .

Exercise 4Challenge

Find the minimum of $y = x^{2} + 1 + x^{2} - 8 x + 20$ .

What this move does

Work through one: the discriminant method

Example: find the maximum and minimum of y = (x² + x + 1)/(x² + 1)

Six tools, one use each

Substitution + monotonicity

Discriminant method

When it fails

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Equality in AM–GM cannot hold, yet you still write “minimum”

The discriminant method, forgetting the leading coefficient can be 0