Mathematics · Complete guide

Maxima, minima, increasing and decreasing functions

Every question in this topic is answered by one idea: the slope of the curve. Where the slope is positive the function rises; where it is negative it falls; where it turns from one to the other there is a peak or a valley. This guide builds each test from that picture, shows exactly when the shortcuts fail, and finishes with the economics questions they are really about: profit, cost and revenue.

Start here

The whole topic is about the slope

The derivative $f^{'} (x)$ is the slope of the tangent line at x. Walk along the curve from left to right and the sign of that slope tells you everything:

f′(x) > 0

Going uphill

The function is increasing: a bit further right, it is higher.

f′(x) < 0

Going downhill

The function is decreasing: a bit further right, it is lower.

f′(x) = 0

Flat, for an instant

A stationary point. It might be a peak, a valley, or just a pause.

The graph below colours the curve by the sign of f′: green going up, red going down. Drag the blue point along it. The strips underneath are the sign charts of f′ and f″, the tool every method in this guide uses.

f (x) = x^{3} - 3 x^{2} - 9 x + 5

Increasing (f′ > 0)Decreasing (f′ < 0)Stationary pointInflection point

At x = 0.5

Falling here

The tangent has slope f′ = −11.25, so f is decreasing here. f″ = −3 < 0: the curve bends downwards (concave, ∩).

What the calculus says about this function

Derivatives

f^{'} (x) = 3 x^{2} - 6 x - 9

f^{''} (x) = 6 x - 6

Stationary points

x = −1: local maximum, f = 10f″ = −12 < 0
x = 3: local minimum, f = −22f″ = 12 > 0

Where it rises and falls

(−∞, −1): increasing
(−1, 3): decreasing
(3, ∞): increasing

How it bends

(−∞, 1): concave (∩, f″ < 0)
(1, ∞): convex (∪, f″ > 0)

Inflection at x = 1.

Key idea

Maxima and minima can only happen where the curve switches from going up to going down, or the reverse. For a smooth function that switch needs the slope to pass through zero, which is why we always start by solving $f^{'} (x) = 0$ .

Monotonicity

Increasing and decreasing functions

f is increasing on an interval if a larger x always gives a larger value: $x_{1} < x_{2} \Rightarrow f (x_{1}) < f (x_{2})$ . Decreasing is the reverse. The derivative test turns that definition into a calculation:

$f^{'} (x) > 0$ on an interval $\Rightarrow$ f is increasing there.
$f^{'} (x) < 0$ on an interval $\Rightarrow$ f is decreasing there.

The sign-chart method

Find f′(x) and factorise it.
Solve f′(x) = 0. These x values cut the number line into intervals.
Test the sign of f′ in each interval (one value per interval is enough).
Read off: + means increasing, − means decreasing.

x(−∞, 1)(1, 2)(2, ∞)

x − 1−++

x − 2−−+

f′ = 6(x−1)(x−2)+ ↗− ↘+ ↗

The sign chart for the worked question below. Multiply the signs of the factors column by column.

Worked question · increasing and decreasing

Find the intervals on which

f (x) = 2 x^{3} - 9 x^{2} + 12 x - 5

is increasing and decreasing.

Differentiate and factorise.
$f^{'} (x) = 6 x^{2} - 18 x + 12 = 6 (x - 1) (x - 2)$
Cut points.
f′(x) = 0 at x = 1 and x = 2, giving three intervals.
Sign in each interval.
Try x = 0: f′ = 12 > 0. Try x = 1.5: f′ = 6(0.5)(−0.5) = −1.5 < 0. Try x = 3: f′ = 6(2)(1) = 12 > 0.

Answer

Increasing on

(- \infty, 1)

and

(2, \infty)

; decreasing on

(1, 2)

A subtle point exams test

$f (x) = x^{3}$ is increasing everywhere, even though $f^{'} (0) = 0$ . A derivative that is positive except at isolated points where it touches zero still gives a strictly increasing function. So “f′ > 0” is sufficient for increasing, but not necessary.

Test 1

Stationary points and the first-derivative test

Local vs global

A local maximum at c means f(c) is the highest value nearby: the top of a hill, not necessarily the tallest mountain. A global (absolute) maximum is the highest value over the whole domain. The same goes for minima.

Critical points

c is a critical point if $f^{'} (c) = 0$ (a stationary point) or f′(c) does not exist, like the corner of $∣ x ∣$ at 0. Local extrema of a function can only occur at critical points (or at the ends of the domain).

The first-derivative test

+ \to -

local maximum

- \to +

local minimum

no change

neither

Look at the sign of f′ just to the left and just to the right of the critical point c.

Worked question · first-derivative test

Find the local maximum and minimum of

f (x) = 2 x^{3} - 9 x^{2} + 12 x - 5

Critical points (from chapter 2).
f′(x) = 6(x − 1)(x − 2) = 0 at x = 1 and x = 2.
At x = 1 the sign goes + → −.
Uphill then downhill: a local maximum, f(1) = 2 − 9 + 12 − 5 = 0.
At x = 2 the sign goes − → +.
Downhill then uphill: a local minimum, f(2) = 16 − 36 + 24 − 5 = −1.

Answer

Local maximum 0 at x = 1; local minimum −1 at x = 2. Neither is global: a cubic runs off to

\pm \infty

Test 2

The second-derivative test, and when it fails

Instead of checking signs on both sides, look at how the curve bends at the stationary point itself. If it is bending downwards ( $f^{''} < 0$ , shaped like ∩), a flat point must be the top of a hill; bending upwards ( $f^{''} > 0$ , ∪) means the bottom of a valley.

The second-derivative test, at a point where f′(c) = 0

f^{''} (c) < 0

local maximum

f^{''} (c) > 0

local minimum

f^{''} (c) = 0

no conclusion

When f″(c) = 0: the higher-order test

Keep differentiating until you reach the first derivative that is not zero at c; call its order n.

n even: an extremum. Maximum if $f^{(n)} (c) < 0$ , minimum if $f^{(n)} (c) > 0$ .
n odd: no extremum; it is a point of inflection.

The two classic traps

$x^{3}$ at 0: f′ = f″ = 0, f‴ = 6 ≠ 0. Order 3 is odd, so no extremum. $x^{4}$ at 0: f′ = f″ = f‴ = 0, $f^{(4)} = 24 > 0$ . Order 4 is even and positive, so a minimum, even though the second-derivative test said nothing. Try both presets in the graph above.

Worked question · both tests

Find and classify the stationary points of

f (x) = x^{4} - 4 x^{3}

Stationary points.
$f^{'} (x) = 4 x^{3} - 12 x^{2} = 4 x^{2} (x - 3) = 0 \Rightarrow x = 0 or x = 3$
Second derivative at each.
$f^{''} (x) = 12 x^{2} - 24 x$ . At x = 3: $f^{''} (3) = 108 - 72 = 36 > 0$ , a local minimum with f(3) = 81 − 108 = −27. At x = 0: $f^{''} (0) = 0$ , so the test is silent.
Go higher at x = 0.
$f^{'''} (x) = 24 x - 24$ , so $f^{'''} (0) = - 24 \neq = 0$ . The first non-zero derivative has order 3, which is odd: no extremum at 0. (The first-derivative test agrees: $4 x^{2} (x - 3)$ is negative on both sides of 0.)

Answer

A local (and global) minimum of −27 at x = 3; a flat inflection point at x = 0, which is neither a maximum nor a minimum.

Shape

Concavity, convexity and inflection points

f″(x) > 0: convex (concave up)

The slope is increasing. The curve holds water (∪) and lies above its tangents. Cost curves in economics usually look like this beyond some output.

f″(x) < 0: concave (concave down)

The slope is decreasing. The curve spills water (∩) and lies below its tangents. Diminishing returns: a production or utility function with falling marginal product is concave.

An inflection point is where the curve switches between the two, so where f″ changes sign. Solve $f^{''} (x) = 0$ for the candidates, then check the sign really changes: $x^{4}$ has $f^{''} (0) = 0$ but is convex on both sides, so 0 is not an inflection point.

Naming varies between books. “Concave up” = convex, “concave down” = concave. Economics papers use convex and concave, so that is what this guide uses too.

Worked question · inflection point

Find the point of inflection of

f (x) = x^{3} - 6 x^{2} + 9 x + 1

and where f is concave and convex.

Second derivative.
$f^{'} (x) = 3 x^{2} - 12 x + 9, f^{''} (x) = 6 x - 12$
Candidate.
f″(x) = 0 at x = 2, and f″ is negative for x < 2 and positive for x > 2: a genuine sign change.
Height.
f(2) = 8 − 24 + 18 + 1 = 3.

Answer

Inflection point (2, 3). Concave on

(- \infty, 2)

, convex on

(2, \infty)

. Its turning points are at x = 1 and x = 3, and the inflection sits exactly halfway between them, as it always does for a cubic.

Absolute extrema

Global maximum and minimum on a closed interval

On a closed, bounded interval [a, b] a continuous function always reaches a highest and a lowest value (the extreme value theorem). They can only be at a critical point inside the interval or at one of the two endpoints, so the method is just a comparison:

Find the critical points that lie strictly inside (a, b).
Evaluate f at each of them and at a and b.
The largest value is the global maximum, the smallest the global minimum.

No derivative tests are needed at all. Tick “Restrict to a closed interval” in the graph in chapter 1 to see the candidates compared live.

Worked question · closed interval

Find the global maximum and minimum of

f (x) = x^{3} - 3 x

[- 2, 3]

Critical points inside.
$f^{'} (x) = 3 x^{2} - 3 = 0 \Rightarrow x = \pm 1$ , both inside [−2, 3].
Evaluate all four candidates.
x −2 (end) −1 1 3 (end)
f(x) −2 2 −2 18
Compare.
The largest value is 18, at the endpoint x = 3. The smallest is −2, reached twice: at x = −2 and at x = 1.

x	−2 (end)	−1	1	3 (end)
f(x)	−2	2	−2	18

Answer

Global maximum 18 (at x = 3); global minimum −2 (at x = −2 and x = 1). The local maximum at x = −1 is not the global one.

Open intervals and one critical point

On an open interval or the whole real line there may be no global extremum at all. One useful shortcut: if a function has a single stationary point on an interval and it is a local maximum (or f is concave everywhere), it is also the global maximum there.

Applications

Economics: profit, average cost, revenue

Profit maximisation

$π (Q) = T R (Q) - T C (Q)$ . First-order condition $π^{'} = 0 ⟺ M R = M C$ . Second-order condition $π^{''} < 0 ⟺ M C^{'} > M R^{'}$ : MC must cut MR from below.

Minimum average cost

$A C = T C / Q$ . At its minimum, $A C^{'} = 0$ , which rearranges to $M C = A C$ : marginal cost cuts average cost at AC's lowest point.

Maximum revenue

$M R = 0$ . Since $M R = P (1 - 1/∣ e ∣)$ , total revenue peaks where demand has unit elasticity, |e| = 1.

Worked question · why the second-order condition matters

A firm has

T R = 60 Q - Q^{2}

and

T C = \frac{1}{3} Q^{3} - 6 Q^{2} + 69 Q + 20

. Find the profit-maximising output and the maximum profit.

MR = MC.
$60 - 2 Q = Q^{2} - 12 Q + 69 \Rightarrow Q^{2} - 10 Q + 9 = 0 \Rightarrow Q = 1 or Q = 9$
Two candidates, so check the second-order condition.
$π^{''} (Q) = M R^{'} - M C^{'} = - 2 - (2 Q - 12) = 10 - 2 Q$ . At Q = 1 it is +8 > 0: that is a profit minimum. At Q = 9 it is −8 < 0: the maximum.
Profit at Q = 9.
$T R = 540 - 81 = 459, T C = 243 - 486 + 621 + 20 = 398, π = 61$

Answer

Q* = 9, maximum profit 61. Stopping at “MR = MC” would have left a one-in-two chance of picking the worst output.

Quick one · minimum AC

T C = Q^{2} + 100

. Where is average cost lowest?

AC and its derivative.
$A C = Q + \frac{100}{Q}, A C^{'} = 1 - \frac{100}{Q ^{2}} = 0 \Rightarrow Q = 10$
Check MC = AC.
AC(10) = 20 and MC = 2Q = 20. They match, as they must.

Answer

Q = 10, AC = 20

Quick one · maximum revenue

Demand is

P = 100 - 2 Q

. Where is revenue highest?

MR = 0.
$T R = 100 Q - 2 Q^{2}, M R = 100 - 4 Q = 0 \Rightarrow Q = 25, P = 50$
Check |e| = 1.
$∣ e ∣ = \frac{d Q}{d P} \cdot \frac{P}{Q} = \frac{1}{2} \cdot \frac{50}{25} = 1$

Answer

Q = 25, P = 50, TR = 1,250

Going further

Two variables: the Hessian test

For $z = f (x, y)$ the idea is the same, one level up. First-order conditions: both partial derivatives are zero, $f_{x} = f_{y} = 0$ . For the second-order conditions, build the Hessian determinant from the second partials.

The second-order test at a stationary point

D = f_{xx} f_{x y} f_{x y} f_{y y} = f_{xx} f_{y y} - f_{x y}^{2}

D > 0, f_{xx} < 0

local maximum

D > 0, f_{xx} > 0

local minimum

D < 0

saddle point

D = 0

no conclusion

fxy is the cross partial ∂²f/∂x∂y. For smooth functions fxy = fyx.

Worked question · two variables

Find and classify the stationary point of

f (x, y) = x^{2} + x y + y^{2} - 3 x

First-order conditions.
$f_{x} = 2 x + y - 3 = 0, f_{y} = x + 2 y = 0 \Rightarrow x = 2, y = - 1$
Second partials.
$f_{xx} = 2, f_{y y} = 2, f_{x y} = 1, D = 2 \cdot 2 - 1^{2} = 3 > 0$
Classify.
D > 0 and f_xx > 0: a local minimum, with f(2, −1) = 4 − 2 + 1 − 6 = −3.

Answer

Minimum −3 at (2, −1). By contrast,

x^{2} - y^{2}

has D = −4 < 0 at the origin: a saddle, rising one way and falling the other.

Practice

Exercises, with answers

0 of 22 correct so far

A. Find and classify

Type numbers; fractions like 3/2 are fine.

Problem 1 · local extrema

f (x) = x^{3} - 12 x + 1

. Find the local maximum and the local minimum.

x at the local maxLocal maximum valuex at the local minLocal minimum value

Problem 2 · closed interval

Find the global maximum and minimum of

f (x) = x^{2} - 4 x + 5

[0, 5]

Global maximumGlobal minimum

Problem 3 · inflection

At what x does

f (x) = x^{3} - 3 x^{2} + 5

have its point of inflection?

B. Economics and two variables

Problem 4 · profit

T R = 40 Q - Q^{2}

and

T C = Q^{2} + 8 Q + 20

. Find the profit-maximising output and the maximum profit.

Q*Maximum profit

Problem 5 · average cost

T C = 2 Q^{2} + 50

. Find the output that minimises average cost, and the minimum AC.

QMinimum AC

Problem 6 · two variables

Find the stationary point of

f (x, y) = x^{2} + y^{2} - 2 x - 4 y + 10

and its value.

xyf at that point

C. Concept checks

One attempt each. Read the explanation even when you get it right.

1If f′(c) = 0 and f″(c) > 0, then at x = c the function has:

2For f(x) = x³, the point x = 0 is:

3For f(x) = x⁴, at x = 0 the second-derivative test gives f″(0) = 0. The point is:

4A continuous function on a closed interval [a, b] attains its global maximum:

5If f″(x) < 0 for every x and f′(c) = 0, then f(c) is:

6The second-order condition for profit maximisation requires that at the chosen output:

7At a stationary point of f(x, y), fxx = 1, fyy = 2 and fxy = 3. The point is:

8At the output where average cost is at its minimum:

Revise

One-page revision sheet

Increasing / decreasing

Factorise f′, cut at its zeros, test one point per interval. + ↗, − ↘.

First-derivative test

At a critical point: + → − is a max, − → + is a min, no sign change is neither.

Second-derivative test

f′(c) = 0 and f″(c) < 0: max. f″(c) > 0: min. f″(c) = 0: go to the first non-zero derivative; even order means extremum, odd means inflection.

Shape

f″ > 0 convex ∪, f″ < 0 concave ∩. Inflection needs f″ to change sign, not just to equal 0.

Closed interval

Global max/min: evaluate f at the interior critical points and both endpoints; pick the largest and smallest.

Economics

Profit: MR = MC with MC cutting MR from below. Min AC: MC = AC. Max TR: MR = 0, |e| = 1. Two variables: D = fxx·fyy − fxy².

Watch out for

The mistakes examiners are fishing for

Treating every stationary point as a max or min. x³ has a flat point at 0 that is neither.
Concluding “no extremum” when f″(c) = 0. The test is silent, not negative: x⁴ has a minimum there.
Calling every zero of f″ an inflection point. Check the sign actually changes.
Forgetting the endpoints in a closed-interval question. The global maximum is often at an end.
Stopping at MR = MC. There can be two solutions; only the one where MC cuts MR from below maximises profit.
Mixing up concave and convex. Concave is ∩ (f″ < 0), like diminishing returns; convex is ∪ (f″ > 0).

Try it on real questions

Twelve past-paper questions on this

From JNU, the University of Hyderabad, CUET-PG and IIT JAM: where a function rises and falls, local and global extrema, inflection points, and the profit, revenue and average-cost versions.

These are questions from the TestStop bank. A free account answers 10 a day, each with its worked solution; All-Access opens every set with no daily limit.

Practice these →All questions on this topic →More resources →

Maxima, minima, increasing and decreasing functions

The whole topic is about the slope

Falling here

Increasing and decreasing functions

The sign-chart method

Stationary points and the first-derivative test

The second-derivative test, and when it fails

Concavity, convexity and inflection points

Global maximum and minimum on a closed interval

Economics: profit, average cost, revenue

Two variables: the Hessian test

Exercises, with answers

A. Find and classify

B. Economics and two variables

C. Concept checks

One-page revision sheet

Increasing / decreasing

First-derivative test

Second-derivative test

Shape

Closed interval

Economics

The mistakes examiners are fishing for

Twelve past-paper questions on this

Read next

Maxima, minima, increasing and decreasing functions

The whole topic is about the slope

Falling here

Increasing and decreasing functions

The sign-chart method

Stationary points and the first-derivative test

The second-derivative test, and when it fails

Concavity, convexity and inflection points

Global maximum and minimum on a closed interval

Economics: profit, average cost, revenue

Two variables: the Hessian test

Exercises, with answers

A. Find and classify

B. Economics and two variables

C. Concept checks

One-page revision sheet

Increasing / decreasing

First-derivative test

Second-derivative test

Shape

Closed interval

Economics

The mistakes examiners are fishing for

Twelve past-paper questions on this

Read next