Increasing and decreasing functions

AS · P1 · 15 min

The sign of the derivative tells you which way a curve is heading. Where $\dfrac{dy}{dx}$ is positive the curve rises from left to right; where it is negative the curve falls. Paper 1 asks you to find the interval on which a function is increasing or decreasing, to show that a function is increasing (or decreasing) for every value of $x$ , and to use that fact to decide whether a function has an inverse. These are short questions, usually 3 to 5 marks, but they reward precise algebra and a clearly stated reason.

The sign of the gradient

Walk along the curve $y = x^3 - 3x^2 - 9x + 2$ from left to right. You go uphill until $x = -1$ , downhill from $x = -1$ to $x = 3$ , then uphill again. Uphill means the tangent slopes upwards, so the gradient is positive; downhill means the gradient is negative.

y = x^3 - 3x^2 - 9x + 2 (-1, 7) (3, -25) x = -1 x = 3

The derivative is $f'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1)$ . It is positive for $x < -1$ and $x > 3$ , and negative for $-1 < x < 3$ , exactly matching the picture.

Definition

A function $f$ is increasing on an interval if $f(x)$ gets larger as $x$ gets larger throughout that interval. It is decreasing on an interval if $f(x)$ gets smaller as $x$ gets larger.

For a differentiable function:

if $f'(x) > 0$ for every $x$ in an interval, $f$ is increasing on that interval;
if $f'(x) < 0$ for every $x$ in an interval, $f$ is decreasing on that interval.

The boundaries between increasing and decreasing sections are the points where $f'(x) = 0$ : the stationary points. This is why the two topics go together.

Strict or not strict

The function $f(x) = x^3$ has $f'(0) = 0$ , yet it is increasing everywhere: it never goes down. A gradient that is zero at a single isolated point does not stop a function being increasing. In exam answers, showing $f'(x) > 0$ is the cleanest argument. If the derivative is a perfect square, such as $3(x - 2)^2$ , say that $f'(x) \ge 0$ with equality only at $x = 2$ , so $f$ is increasing. Mark schemes for interval answers normally accept either strict or non-strict inequalities at the end points, such as $-1 < x < 3$ or $-1 \le x \le 3$ .

Finding where a function is increasing or decreasing

This is an inequality problem. Differentiate, then solve $f'(x) > 0$ or $f'(x) < 0$ . When $f'(x)$ is a quadratic, use the method from quadratic inequalities: find the roots, then decide which side of them the inequality holds.

Interval where f is increasing or decreasing

Find $f'(x)$ , rewriting as powers first if needed.
Solve $f'(x) = 0$ to find the critical values.
Decide where $f'(x) > 0$ (increasing) or $f'(x) < 0$ (decreasing). For a quadratic with positive $x^2$ coefficient, it is negative between the roots and positive outside them. A quick sketch of $y = f'(x)$ helps.
Respect the domain of $f$ : an answer of $x < 2$ is wrong if $f$ is only defined for $x > 0$ , where it should be $0 < x < 2$ .
State the answer as an interval or set of values.

Showing a function is increasing for all x

To prove that $f$ is increasing for every $x$ , you must show $f'(x) > 0$ for every $x$ , not just test a few values. Three standard arguments do this.

Completing the square. If $f'(x)$ is a quadratic, write it as $p(x - q)^2 + r$ with $p > 0$ and $r > 0$ . Since $(x - q)^2 \ge 0$ , the whole expression is at least $r$ , which is positive.

Discriminant. A quadratic with positive leading coefficient and negative discriminant never meets the $x$ -axis, so it is positive for all $x$ . This is the method to use when the quadratic contains an unknown constant.

Signs of terms. If every term of $f'(x)$ is positive (or every term negative) for all $x$ in the domain, you are done. For example, $f'(x) = 3 + \dfrac{2}{(x + 1)^2}$ is a positive number plus a square, so it is positive for $x \neq -1$ . Even powers like $(x + 1)^2$ and $(x + 1)^4$ are never negative; odd powers like $(2x - 1)^3$ take the sign of the bracket, so you need the domain to fix their sign.

Key result

$f'(x) > 0$ throughout an interval: $f$ is increasing there.
$f'(x) < 0$ throughout an interval: $f$ is decreasing there.
To show $f'(x) > 0$ for all $x$ : complete the square, use the discriminant, or argue from the signs of the terms.
A function that is increasing (or decreasing) on its whole domain is one-one, so it has an inverse.

Increasing functions and inverses

A function has an inverse only if it is one-one: no two inputs give the same output. A function that is increasing throughout its domain can never return to a value it has already taken, so it is automatically one-one. The same is true of a function that is decreasing throughout its domain.

This gives a calculus route to two common questions:

"Determine whether $f$ has an inverse." Show that $f'(x)$ is always positive (or always negative) on the domain, and conclude that $f$ is one-one.
"Find the least value of $a$ for which $f$ , with domain $x \ge a$ , has an inverse." Find where $f$ turns, using $f'(x) = 0$ . The domain must start at or beyond the last turning point so that $f$ only increases (or only decreases) on it.

Worked examples

Interval where a cubic decreases

Find the set of values of $x$ for which $f(x) = x^3 - 3x^2 - 9x + 2$ is decreasing.

Solution

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

The critical values are $x = -1$ and $x = 3$ . The quadratic $3(x - 3)(x + 1)$ has a positive $x^2$ coefficient, so it is negative between its roots.

$f$ is decreasing for $-1 < x < 3$ .

Increasing for every x

Show that $f(x) = x^3 - 6x^2 + 15x - 4$ is an increasing function.

Solution

f'(x) = 3x^2 - 12x + 15 = 3(x^2 - 4x + 5)

Complete the square:

x^2 - 4x + 5 = (x - 2)^2 - 4 + 5 = (x - 2)^2 + 1

f'(x) = 3(x - 2)^2 + 3

Since $(x - 2)^2 \ge 0$ for all $x$ , $f'(x) \ge 3 > 0$ for all $x$ . Therefore $f$ is an increasing function.

(Alternatively: the discriminant of $x^2 - 4x + 5$ is $16 - 20 = -4 < 0$ and the leading coefficient is positive, so $f'(x) > 0$ for all $x$ .)

A chain rule function and its inverse

The function $f$ is defined by $f(x) = \dfrac{1}{(2x - 1)^2} - x$ for $x > \tfrac{1}{2}$ .

(a) Find $f'(x)$ and hence determine whether $f$ is increasing or decreasing.

(b) State, with a reason, whether $f$ has an inverse.

Solution

(a) Write $f(x) = (2x - 1)^{-2} - x$ . By the chain rule,

f'(x) = -2(2x - 1)^{-3} \times 2 - 1 = -\frac{4}{(2x - 1)^3} - 1

For $x > \tfrac{1}{2}$ , $2x - 1 > 0$ , so $(2x - 1)^3 > 0$ and $-\dfrac{4}{(2x - 1)^3} < 0$ . Both terms of $f'(x)$ are negative, so $f'(x) < 0$ for all $x$ in the domain. Hence $f$ is decreasing.

(b) $f$ is decreasing throughout its domain, so it is one-one, and therefore $f$ has an inverse.

Note the role of the domain: the cube $(2x - 1)^3$ could be negative if $x < \tfrac{1}{2}$ were allowed, and then the argument would fail.

Restricting a domain for an inverse

The function $f$ is defined by $f(x) = 2x^3 - 9x^2 + 12x$ for $x \ge a$ . Find the least value of $a$ for which $f$ has an inverse.

Solution

f'(x) = 6x^2 - 18x + 12 = 6(x - 1)(x - 2)

So $f$ is increasing for $x < 1$ , decreasing for $1 < x < 2$ and increasing for $x > 2$ . There are turning points at $x = 1$ and $x = 2$ .

For $f$ to be one-one on $x \ge a$ , the domain must not contain a turning point in its interior, so it must lie entirely in the region $x \ge 2$ , where $f$ is increasing. The least value is $a = 2$ .

y = 2x^3 - 9x^2 + 12x (1, 5) (2, 4) x = 2

The graph shows why $a = 1$ would fail: on $x \ge 1$ the curve falls from $(1, 5)$ to $(2, 4)$ and then rises back through $y = 5$ , so some outputs are taken twice.

An unknown constant

Find the set of values of the constant $k$ for which the function $f(x) = x^3 - 6x^2 + kx + 1$ has $f'(x) > 0$ for all $x$ , so that $f$ is increasing.

Solution

f'(x) = 3x^2 - 12x + k

This quadratic has a positive $x^2$ coefficient, so it is positive for all $x$ exactly when it has no real roots:

b^2 - 4ac = 144 - 12k < 0 \quad\Rightarrow\quad k > 12

(At $k = 12$ , $f'(x) = 3(x - 2)^2$ , which is zero at $x = 2$ only; $f$ is still increasing in the everyday sense, but $f'(x) > 0$ fails at one point, so $k = 12$ is excluded from this answer.)

A bracket to the power three

Find the set of values of $x$ for which $f(x) = (3x - 2)^3 - 9x$ is decreasing.

Solution

By the chain rule,

f'(x) = 3(3x - 2)^2 \times 3 - 9 = 9(3x - 2)^2 - 9

$f$ is decreasing where $f'(x) < 0$ :

9(3x - 2)^2 < 9 \quad\Rightarrow\quad (3x - 2)^2 < 1 \quad\Rightarrow\quad -1 < 3x - 2 < 1

Adding $2$ and dividing by $3$ : $\tfrac{1}{3} < x < 1$ .

Do not expand the cube here. Keeping $(3x - 2)^2$ as a single square makes the inequality a one-line job.

Watch out

Testing a few values. Showing $f'(0) > 0$ , $f'(1) > 0$ and $f'(5) > 0$ does not prove $f'(x) > 0$ for every $x$ . You need an algebraic argument: completed square, discriminant or signs of terms.

Getting the side of the roots wrong. For $3(x - 3)(x + 1) < 0$ the answer is between the roots, $-1 < x < 3$ , not $x < -1$ or $x > 3$ . Sketch the parabola $y = f'(x)$ if unsure.

Square roots with one sign. $(3x - 2)^2 < 1$ means $-1 < 3x - 2 < 1$ , not just $3x - 2 < 1$ .

Assuming an odd power is positive. $\dfrac{4}{(2x - 1)^3}$ is positive only when $2x - 1 > 0$ . Quote the domain when you use it.

Answering in terms of $f(x)$ . " $f$ is increasing for $f(x) > 7$ " is meaningless; the answer is a set of $x$ -values.

Exam tip

"Show that $f$ is increasing" needs the derivative, an argument valid for all $x$ in the domain, and a concluding sentence. The final mark is for the conclusion with its reason, for example " $f'(x) = 3(x - 2)^2 + 3 > 0$ for all $x$ , so $f$ is increasing."
"Determine whether $f$ is increasing, decreasing or neither": give the sign of $f'(x)$ with a reason, then the word. If $f'(x)$ changes sign in the domain, the answer is "neither".
Inverse questions: link the steps explicitly. "Decreasing, therefore one-one, therefore has an inverse." Examiners look for "one-one".
Least value of $a$ : identify the turning points, then choose the end point that leaves no turning point inside the domain. Students often give the other turning point by mistake.
Keep squares together. A derivative like $9(3x - 2)^2 - 9$ is easier to handle without expanding.

Summary

$f'(x) > 0$ on an interval means $f$ is increasing there; $f'(x) < 0$ means decreasing.
To find where $f$ increases or decreases, solve the inequality $f'(x) > 0$ or $f'(x) < 0$ , respecting the domain.
A positive quadratic is negative between its roots and positive outside them.
To show $f'(x) > 0$ for all $x$ : complete the square, show the discriminant is negative, or argue that every term is positive.
An isolated zero of $f'(x)$ (as for $x^3$ at $0$ ) does not stop $f$ being increasing.
A function that is increasing or decreasing on its whole domain is one-one and has an inverse.
For the least $a$ with domain $x \ge a$ , start the domain at the last turning point.

Practice questions

Question

Find the set of values of $x$ for which $f(x) = x^2 - 8x + 3$ is increasing.
Find the set of values of $x$ for which $f(x) = 2x^3 + 3x^2 - 36x$ is decreasing.
Show that $f(x) = x^3 + 3x^2 + 5x - 1$ is an increasing function.
The function $f$ is defined by $f(x) = x + \dfrac{4}{x}$ for $x > 0$ . Find the set of values of $x$ for which $f$ is increasing and the set for which it is decreasing.
The function $f$ is defined by $f(x) = \dfrac{1}{x + 2} - 3x$ for $x > -2$ . Determine whether $f$ is increasing or decreasing, and state whether $f$ has an inverse.
The function $f$ is defined by $f(x) = (2x + 1)^{\frac{3}{2}} - 3x$ for $x \ge -\tfrac{1}{2}$ . Find the set of values of $x$ for which $f$ is increasing.
The function $f$ is defined by $f(x) = 2x^3 + 3x^2 - 12x$ for $x \le a$ . Find the greatest value of $a$ for which $f$ has an inverse.
Find the set of values of $k$ for which $f(x) = x^3 + kx^2 + 27x$ satisfies $f'(x) > 0$ for all $x$ .
The function $f$ is defined by $f(x) = x + \dfrac{9}{x - 1}$ for $x > 1$ . (a) Find the set of values of $x$ for which $f$ is increasing. (b) The function $g$ is defined by $g(x) = x + \dfrac{9}{x - 1}$ for $x \ge a$ . Find the least value of $a$ for which $g$ has an inverse.

Answers

$f'(x) = 2x - 8 > 0$ when $x > 4$ .
$f'(x) = 6x^2 + 6x - 36 = 6(x + 3)(x - 2)$ . Negative between the roots: $-3 < x < 2$ .
$f'(x) = 3x^2 + 6x + 5 = 3(x^2 + 2x) + 5 = 3(x + 1)^2 - 3 + 5 = 3(x + 1)^2 + 2$ . Since $(x + 1)^2 \ge 0$ , $f'(x) \ge 2 > 0$ for all $x$ , so $f$ is increasing.
$f'(x) = 1 - \dfrac{4}{x^2} = \dfrac{x^2 - 4}{x^2}$ . For $x > 0$ the denominator is positive, so the sign is that of $x^2 - 4$ . Increasing for $x > 2$ ; decreasing for $0 < x < 2$ .
$f(x) = (x + 2)^{-1} - 3x$ , so $f'(x) = -(x + 2)^{-2} - 3 = -\dfrac{1}{(x + 2)^2} - 3$ . Both terms are negative, so $f'(x) < 0$ for all $x > -2$ : $f$ is decreasing. It is therefore one-one and has an inverse.
$f'(x) = \tfrac{3}{2}(2x + 1)^{\frac{1}{2}} \times 2 - 3 = 3\sqrt{2x + 1} - 3$ . This is positive when $\sqrt{2x + 1} > 1$ , so $2x + 1 > 1$ , $x > 0$ . $f$ is increasing for $x > 0$ (and decreasing for $-\tfrac{1}{2} < x < 0$ ).
$f'(x) = 6x^2 + 6x - 12 = 6(x + 2)(x - 1)$ . $f$ is increasing for $x < -2$ , decreasing for $-2 < x < 1$ . For the domain $x \le a$ to contain no turning point in its interior, $a \le -2$ . The greatest value is $a = -2$ .
$f'(x) = 3x^2 + 2kx + 27$ . Positive for all $x$ when the discriminant is negative: $4k^2 - 4(3)(27) < 0$ , so $k^2 < 81$ , giving $-9 < k < 9$ .
(a) $f'(x) = 1 - \dfrac{9}{(x - 1)^2}$ . This is positive when $(x - 1)^2 > 9$ . For $x > 1$ , $x - 1 > 0$ , so $x - 1 > 3$ and $x > 4$ . $f$ is increasing for $x > 4$ (and decreasing for $1 < x < 4$ ). (b) $g$ has a minimum at $x = 4$ (where $f'(x) = 0$ ). For $g$ to be one-one the domain must not contain points on both sides of $x = 4$ , so the least value is $a = 4$ .

The sign of the gradient

Finding where a function is increasing or decreasing

Showing a function is increasing for all x

Increasing functions and inverses

Worked examples

Practice questions

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