7.6 Increasing and decreasing functions — Practise

Work through these on your own. Differentiate, factorise, and use a sign table or an inequality to state the intervals. Write each interval in the form $a < x < b$ or as two separate pieces.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

Find the interval on which $y = x^2 - 10x$ is decreasing.
Find the intervals on which $y = x^3 - 3x^2$ is increasing.
Find the interval on which $y = x^3 - 12x + 1$ is decreasing.
Find the intervals on which $y = 2x^3 - 3x^2 - 12x$ is increasing.
A function has $f'(x) = (x - 1)(x - 5)$ . Find the intervals on which $f$ is increasing.
Find the interval on which $y = x^2 + 6x + 1$ is increasing.
A function has $f'(x) = 3(x - 1)(x - 3)$ . Find the interval on which $f$ is decreasing.
Find the interval on which $y = x^3 - 6x^2 + 9x$ is decreasing.
Find the interval on which $y = 2x^2 - 8x + 3$ is increasing.
A function has $f'(x) = 6(x + 4)(x - 2)$ . Find the intervals on which $f$ is increasing.

Answers and concise solutions

1. The interval

Answer: $\dfrac{dy}{dx} = 2x - 10 < 0$ gives $x < 5$ . The curve is decreasing for $x < 5$ and increasing for $x > 5$ , with a minimum at $x = 5$ .

2. The intervals

Answer: $\dfrac{dy}{dx} = 3x^2 - 6x = 3x(x - 2) > 0$ . The product is positive outside the roots, so the curve is increasing for $x < 0$ and for $x > 2$ .

3. The interval

Answer: $\dfrac{dy}{dx} = 3x^2 - 12 = 3(x - 2)(x + 2) < 0$ . The product is negative between the roots, so the curve is decreasing for $-2 < x < 2$ .

4. The intervals

Answer: $\dfrac{dy}{dx} = 6x^2 - 6x - 12 = 6(x - 2)(x + 1) > 0$ . The product is positive outside the roots, so the curve is increasing for $x < -1$ and for $x > 2$ .

5. The intervals

Answer: The product $(x - 1)(x - 5)$ is positive outside the roots, so $f$ is increasing for $x < 1$ and for $x > 5$ .

6. The interval

Answer: $\dfrac{dy}{dx} = 2x + 6 > 0$ gives $x > -3$ . The curve is increasing for $x > -3$ , with a minimum at $x = -3$ .

7. The interval

Answer: The product is negative between the roots, so $f$ is decreasing for $1 < x < 3$ . It is increasing for $x < 1$ and for $x > 3$ .

8. The interval

Answer: $\dfrac{dy}{dx} = 3x^2 - 12x + 9 = 3(x - 1)(x - 3) < 0$ , so the curve is decreasing for $1 < x < 3$ . The gradient is zero at $x = 1$ and $x = 3$ .

9. The interval

Answer: $\dfrac{dy}{dx} = 4x - 8 > 0$ gives $x > 2$ . The minimum is at $x = 2$ .

10. The intervals

Answer: The product is positive outside the roots $x = -4$ and $x = 2$ , so $f$ is increasing for $x < -4$ and for $x > 2$ .