1.07PureStretch

Increasing and Decreasing Functions

Q: Does increasing mean strictly increasing?

OCR uses increasing for f^ (x) 0 and strictly increasing for f^ (x) > 0. Follow the wording of the question.

Knowing the gradient formula tells you more than the steepness at a single point. Its $\textbf{sign}$ tells you where a function is going up, where it is going down, and — where the gradient is zero — where it pauses. Those pauses, the $\textbf{stationary points}$ , are the peaks and troughs that sketching and optimisation depend on.

40 min Differentiation

The big picture

A positive derivative means the function is increasing; negative means decreasing. So solving the inequality $\mathrm{f}^{\prime}(x) > 0$ turns a question about the shape of a graph into an algebra problem you already know how to solve. The boundaries between increasing and decreasing regions are where $\mathrm{f}^{\prime}(x) = 0$ , the stationary points, and checking the sign of the gradient either side classifies each as a maximum, a minimum or a point of inflection. That sign test always works, even when the second-derivative test in the next lesson fails, and it underpins curve sketching, proving that a function is one-to-one, and justifying where iterative methods will converge.

What you'll be able to do

Use the sign of $\mathrm{f}^{\prime}(x)$ to identify where a function is increasing or decreasing
Solve inequalities to find the intervals on which a function increases or decreases
Find stationary points by solving $\mathrm{f}^{\prime}(x) = 0$
Classify stationary points using the sign of the gradient on either side

Increasing and decreasing

A function is $\textbf{increasing}$ on an interval if $\mathrm{f}^{\prime}(x) \ge 0$ throughout it, and $\textbf{strictly increasing}$ if $\mathrm{f}^{\prime}(x) > 0$ . Similarly for decreasing with $\le 0$ and $< 0$ .

To find where a function is increasing, differentiate, then solve $\mathrm{f}^{\prime}(x) > 0$ as an inequality. Quadratic inequalities need a sketch of the derivative.

Example 1Find the set of values of

x

for which

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

is decreasing.

\mathrm{f}^{\prime}(x) = 3x^{2} - 12x + 9 = 3(x - 1)(x - 3)

2This positive-quadratic derivative is negative between its roots.

Answer

1 < x < 3

(or

1 \le x \le 3

for non-strict decreasing)

Why this works

Solving $\mathrm{f}^{\prime}(x) < 0$ is just a quadratic inequality. Sketch the parabola $y = \mathrm{f}^{\prime}(x)$ and read off where it is below the axis — the same skill from the quadratics topic, applied to a derivative.

Showing a function is always increasing

To prove a function is increasing for all $x$ , show that its derivative is positive for all $x$ . Completing the square is the standard way to do this for a quadratic derivative.

An increasing function is one-to-one, so this is also a way to prove that an inverse exists.

Example 1Show that

\mathrm{f}(x) = x^{3} + 3x^{2} + 5x - 1

is increasing for all real

x

\mathrm{f}^{\prime}(x) = 3x^{2} + 6x + 5

2Complete the square:

3(x^{2} + 2x) + 5 = 3(x + 1)^{2} + 2

(x + 1)^{2} \ge 0

, so

\mathrm{f}^{\prime}(x) \ge 2 > 0

for all

x

AnswerSince

\mathrm{f}^{\prime}(x) > 0

for all

x

\mathrm{f}

is increasing everywhere.

Tip — A conclusion sentence is required: "since $\mathrm{f}^{\prime}(x) > 0$ for all $x$ , $\mathrm{f}$ is an increasing function". The proof is not complete without it.

Stationary points

A $\textbf{stationary point}$ is where $\mathrm{f}^{\prime}(x) = 0$ — the tangent is horizontal.

There are three kinds. A $\textbf{local maximum}$ : the gradient changes from positive to negative. A $\textbf{local minimum}$ : from negative to positive. A $\textbf{stationary point of inflection}$ : the gradient has the same sign on both sides.

To find them, solve $\mathrm{f}^{\prime}(x) = 0$ for $x$ , then substitute each root into $\mathrm{f}$ for the $y$ -coordinate.

Example 1Find and classify the stationary points of

y = x^{3} - 6x^{2} + 9x + 2

\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3(x - 1)(x - 3) = 0

x = 1

and

x = 3

2At

x = 1

y = 1 - 6 + 9 + 2 = 6

. At

x = 3

y = 27 - 54 + 27 + 2 = 2

3Test

x = 0

2

4

: gradients

9

-3

9

. So the sign goes

+, -, +

4At

x = 1

the sign changes

+

-

: maximum. At

x = 3

it changes

-

+

: minimum.

AnswerMaximum

(1, 6)

; minimum

(3, 2)

Inflection and the sign test in general

For $y = x^{3}$ , $\dfrac{\mathrm{d}y}{\mathrm{d}x} = 3x^{2}$ is zero at $x = 0$ but positive on both sides. The curve flattens momentarily and carries on rising: a stationary point of inflection.

When testing signs, choose test values between consecutive stationary points (and not beyond the next one), so that you are genuinely looking at the gradient just either side.

The sign test works for any differentiable function, including cases where the second derivative is zero or awkward to compute.

Example 1Classify the stationary point of

y = (x - 2)^{4} + 1

\dfrac{\mathrm{d}y}{\mathrm{d}x} = 4(x - 2)^{3} = 0

x = 2

, where

y = 1

2At

x = 1

: gradient

-4

. At

x = 3

: gradient

4

AnswerMinimum at

(2, 1)

— the second derivative is zero here, but the sign test still decides it.

Tip — A table with columns for $x$ slightly less than, at, and slightly more than the stationary point, and rows for the sign of $\dfrac{\mathrm{d}y}{\mathrm{d}x}$ and a sketch of the slope, is clear and earns method marks.

Think like an examiner

Factorise the derivative before solving $\mathrm{f}^{\prime}(x) = 0$ or an inequality.
Give both coordinates of each stationary point.
Show the gradient values used in a sign test.
End "show that" proofs with a clear conclusion sentence.

Common misconceptions

Giving only the

x

-coordinate of a stationary point.

Substitute back into

\mathrm{f}(x)

for the

y

-coordinate.

Concluding a maximum because

\mathrm{f}^{\prime}(x) = 0

\mathrm{f}^{\prime}(x) = 0

only locates the point; the sign change classifies it.

Testing a point beyond the next stationary point.

Choose test values between consecutive stationary points.

Solving

3(x-1)(x-3) < 0

x < 1

x > 3

A positive quadratic is negative between its roots:

1 < x < 3

Gradient and shape

\mathrm{f}^{\prime}(x) > 0 \Rightarrow \text{increasing}

Increasing.

\mathrm{f}^{\prime}(x) < 0 \Rightarrow \text{decreasing}

Decreasing.

\mathrm{f}^{\prime}(x) = 0 \Rightarrow \text{stationary point}

Stationary.

Stretch yourself

The function $\mathrm{f}(x) = x^{3} + kx^{2} + 12x$ is increasing for all real $x$ . Find the set of possible values of $k$ .

Hint — The derivative is a quadratic. For it to be non-negative everywhere, what must be true of its discriminant?

Reveal solution

Solution

\mathrm{f}^{\prime}(x) = 3x^{2} + 2kx + 12

. This positive quadratic is never negative exactly when it has no two distinct real roots, i.e. its discriminant is at most zero:

(2k)^{2} - 4(3)(12) \le 0

, so

4k^{2} \le 144

and

k^{2} \le 36

. Hence

-6 \le k \le 6

. At the boundary, e.g.

k = 6

\mathrm{f}^{\prime}(x) = 3(x + 2)^{2} \ge 0

, zero only at

x = -2

, which is a stationary point of inflection — the function is still increasing. So

-6 \le k \le 6

\blacksquare

Questions students ask

What is the difference between a local and a global maximum?

A local maximum is higher than nearby points. A global maximum is the highest value over the whole domain, which may occur at a local maximum or at an end of the domain.

Is every stationary point a turning point?

No. A stationary point of inflection has zero gradient but the curve does not turn.

Should I use the sign test or the second derivative?

Either, unless told. The second derivative is usually faster, but when it is zero the sign test is needed.

Does increasing mean strictly increasing?

OCR uses increasing for

\mathrm{f}^{\prime}(x) \ge 0

and strictly increasing for

\mathrm{f}^{\prime}(x) > 0

. Follow the wording of the question.

Key takeaways

Increasing where $\mathrm{f}^{\prime}(x) > 0$ ; decreasing where $\mathrm{f}^{\prime}(x) < 0$ .
Solve inequalities in $\mathrm{f}^{\prime}(x)$ to find intervals; complete the square to prove a sign for all $x$ .
Stationary points satisfy $\mathrm{f}^{\prime}(x) = 0$ .
Sign change $+ \to -$ is a maximum, $- \to +$ a minimum, no change a point of inflection.
The sign test works even when the second derivative is zero.