7.6 Increasing and decreasing functions — Practise with help

Work each question closed book first. Differentiate, factorise, then test the sign of the derivative in each interval between the roots.

Warm-up: remember

Solve $3x - 9 > 0$ .
Factorise $x^2 - x - 6$ .
Solve $(x - 1)(x + 4) < 0$ .

Answers

$3x > 9$ , so $x > 3$ .
$x^2 - x - 6 = (x - 3)(x + 2)$ .
A product is negative when the two factors have opposite signs. That happens between the roots, so $-4 < x < 1$ .

If question 3 was hard, revise quadratic inequalities in Chapter 2.

Practise with help

1. Find the interval on which $y = x^2 - 6x$ is increasing.

Hint

Differentiate, then solve $\dfrac{dy}{dx} > 0$ .

Next step

$\dfrac{dy}{dx} = 2x - 6$ . Solve $2x - 6 > 0$ .

Full working

\frac{dy}{dx} = 2x - 6 > 0 \quad \Rightarrow \quad x > 3.

The curve is increasing for $x > 3$ and decreasing for $x < 3$ . The turning point at $x = 3$ is a minimum.

2. Find the intervals on which $y = x^3 - 12x$ is increasing and those on which it is decreasing.

Hint

The derivative is a quadratic with two roots. Use a sign table with a test value in each of the three intervals.

Next step

$\dfrac{dy}{dx} = 3x^2 - 12 = 3(x - 2)(x + 2)$ . The roots are $x = 2$ and $x = -2$ . Test $x = -3$ , $x = 0$ and $x = 3$ .

Full working

Interval	Test value	Sign of $\dfrac{dy}{dx}$	Behaviour
$x < -2$	$x = -3$	$3(-5)(-1) = 15 > 0$	increasing
$-2 < x < 2$	$x = 0$	$3(-2)(2) = -12 < 0$	decreasing
$x > 2$	$x = 3$	$3(1)(5) = 15 > 0$	increasing

The curve is increasing for $x < -2$ and for $x > 2$ , and decreasing for $-2 < x < 2$ .

3. Find the intervals on which $y = 2x^3 + 3x^2 - 12x$ is increasing and those on which it is decreasing.

Hint

Differentiate and take out the common factor $6$ . Factorise the remaining quadratic.

Next step

$\dfrac{dy}{dx} = 6x^2 + 6x - 12 = 6(x^2 + x - 2) = 6(x + 2)(x - 1)$ . The roots are $x = -2$ and $x = 1$ .

Full working

Interval	Test value	Sign of $\dfrac{dy}{dx}$	Behaviour
$x < -2$	$x = -3$	$6(-1)(-4) = 24 > 0$	increasing
$-2 < x < 1$	$x = 0$	$6(2)(-1) = -12 < 0$	decreasing
$x > 1$	$x = 2$	$6(4)(1) = 24 > 0$	increasing

The curve is increasing for $x < -2$ and for $x > 1$ , and decreasing for $-2 < x < 1$ . It has a maximum at $x = -2$ and a minimum at $x = 1$ .

4. The derivative of a function is $f'(x) = (x - 3)(x + 1)$ . State the intervals on which $f$ is increasing and those on which it is decreasing.

Hint

The roots are $x = 3$ and $x = -1$ . Test one value in each of the three intervals and look at the sign of the product.

Next step

Test $x = -2$ , $x = 0$ and $x = 4$ . At $x = -2$ both factors are negative, so the product is positive. At $x = 0$ the factors have opposite signs.

Full working

Interval	Test value	Sign of $f'(x)$	Behaviour
$x < -1$	$x = -2$	$(-5)(-1) = 5 > 0$	increasing
$-1 < x < 3$	$x = 0$	$(-3)(1) = -3 < 0$	decreasing
$x > 3$	$x = 4$	$(1)(5) = 5 > 0$	increasing

The function is increasing for $x < -1$ and for $x > 3$ , and decreasing for $-1 < x < 3$ .

5. Show that $y = x^3$ is increasing for every value of $x$ , even though it has a stationary point at $x = 0$ .

Hint

Differentiate and look at the sign of $3x^2$ for positive and negative values of $x$ .

Next step

$\dfrac{dy}{dx} = 3x^2$ . A square is never negative, so the derivative is positive for every $x \neq 0$ .

Full working

\frac{dy}{dx} = 3x^2.

For every $x \neq 0$ , $3x^2 > 0$ , so the curve is increasing. At $x = 0$ the derivative is zero, so the curve is momentarily flat. It is still increasing through the origin: the gradient falls to zero and rises again without becoming negative.

This is a stationary point of inflection, not a maximum or a minimum.

6. A curve has $f'(x) < 0$ for $0 < x < 4$ and $f'(x) > 0$ for $x < 0$ and for $x > 4$ . Describe the curve between $x = -1$ and $x = 5$ .

Hint

Mark the sign of the derivative on the intervals $x < 0$ , $0 < x < 4$ and $x > 4$ , then translate the signs into rising and falling.

Next step

The curve rises until $x = 0$ , falls between $x = 0$ and $x = 4$ , then rises after $x = 4$ .

Full working

Between $x = -1$ and $x = 5$ the curve rises to a maximum at $x = 0$ , falls to a minimum at $x = 4$ , and then rises again. The two turning points are at $x = 0$ and $x = 4$ , because the derivative is zero there.