Where is f increasing or decreasing?

In 6 of 20 recent exams
Paper 1 or 2
1 to 4 marks

How it is set in the exam

Paper 1 (no calculator) and Paper 2 (calculator), 1–4 marks, usually one part of a longer question, straight after $f'(x)$ . Common (6 of 20 exam versions).

Spot it in the exam

Exam phrases to spot (4)

“find the values of $x$ for which $f$ is increasing”
“the interval on which $f$ is decreasing”
“show that $f$ is increasing for all $x$ ”
“show that $f$ is never increasing”

In disguise (3)

"show that $f$ has an inverse", or "… is one-to-one", on an interval domain: $f'(x) > 0$ (or $f'(x) < 0$ ) for every $x$ in it
a context: "when is the water level falling?", "in which months does the population go down?": the sign of the rate
$g'(x)$ is a quadratic $f(x)$ from an earlier part minus a constant $d$ , and $g$ must be increasing for all $x$ : $f(x) - d > 0$ everywhere, so $d$ is below the least value of $f$ , its vertex value (no $\Delta$ needed)

Looks like it, but isn't (3)

"is it increasing at $t = 2$ ?" (one instant): the sign of one number, in Rate of change in context
"the gradient is increasing", "concave up": the sign of $f''$ , in Concavity and points of inflexion
a quadratic $f$ that must be negative (or positive) and increasing: the interval between the roots and the vertex's side combine, with no derivative needed, in Quadratic inequalities

Worked example

The sign of f′ from its factors, and a gap in the domain

Paper 1 (no calculator)

Question

Let $f(x) = \dfrac{e^{3x}}{x}$ , for $x \ne 0$ .
(a) Show that $f'(x) = \dfrac{e^{3x}(3x - 1)}{x^2}$ . [3]
(b) Find the values of $x$ for which $f$ is decreasing. [3]

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Steps

Find $f'(x)$ and factorise it
(as one fraction, if there is a denominator).
Find the critical values:
where $f'(x) = 0$ , and where $f$ is undefined (an asymptote, a gap in the domain). On : the zeros of $y = f'(x)$ from its graph (zeros and intersections).
Find the sign of $f'$ between the critical values:
a sign table from the factors (labelled $f'$ ), or one test value in each interval. A factor that is always positive ( $e^{3x}$ , $x^2 + 4$ , a squared denominator) doesn't change the sign. When $f'$ is a quadratic, this step is a quadratic inequality: a U-shaped $f'$ is negative between its roots.
Answer with intervals:
increasing where $f'(x) > 0$ , decreasing where $f'(x) < 0$ . Strict or non-strict inequalities are both accepted at a critical value, but never include a value where $f$ is undefined. On each side of a gap or an asymptote, write a separate interval.
"For all $x$ ":
show that $f'(x) > 0$ for every $x$ in the domain. Complete the square, or show $\Delta < 0$ with a positive $x^2$ coefficient, or write $f'$ as a product of positive factors. Then conclude. An argument from a sketch of $f$ is not accepted.

Common mistakes

Giving the critical values as the answer.
$x = \frac{1}{3}$ is where the sign of $f'$ changes; the answer is an interval.
One interval across a gap or an asymptote.
In Example 1.1, " $x < \frac{1}{3}$ " includes $x = 0$ , where $f$ is undefined, and $f(-1) < f(0.1)$ . In the same way, $f(x) = \frac{2x + 1}{x - 1}$ has $f'(x) < 0$ on each side of $x = 1$ , but $f(0) = -1 < f(2) = 5$ : it is decreasing on $x < 1$ and on $x > 1$ , not overall.
Reading the sign of $f$ instead of $f'$
"the graph is above the x-axis" says $f(x) > 0$ , not that $f$ is increasing.

More mistakes (3)

"For all $x$ " from a sketch or from a few values.
Markschemes want $f'(x) > 0$ shown for every $x$ in the domain.
Confusing "increasing" with "concave up".
$f'(x) > 0$ means increasing; $f''(x) > 0$ means the gradient is increasing.
Degree mode
with a trig model on . Calculus with trig needs radians: check the settings before you start. This applies to every trig model in this guide, not only this type's.

Easy to confuse with

Next typeGraphs of f, f′ and f″

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