7.4 The second derivative, convexity and points of inflection — Practise with help

Work each question closed book first. For every suspected point of inflection, check the sign of the second derivative on both sides.

Warm-up: remember

Differentiate $y = 5x^3$ .
Differentiate $y = \cos 3x$ .
Differentiate $g(x) = 8x - 5$ .

Answers

$\dfrac{dy}{dx} = 15x^2$ .
$\dfrac{dy}{dx} = -3\sin 3x$ .
$g'(x) = 8$ .

If question 2 was hard, revise the standard derivatives in Section 7.3 before you continue.

Practise with help

1. Find $f''(x)$ for $f(x) = x^5 - 2x^3 + 7$ .

Hint

Differentiate once, then differentiate the answer again. The constant $7$ has already gone after the first differentiation.

Next step

The first derivative is $f'(x) = 5x^4 - 6x^2$ . Differentiate that expression term by term.

Full working

f'(x) = 5x^4 - 6x^2, \qquad f''(x) = 20x^3 - 12x.

The second derivative is a cubic, because the original function was a quintic.

2. Find $\dfrac{d^2y}{dx^2}$ for $y = e^{3x}$ .

Hint

The derivative of $e^{3x}$ is $3e^{3x}$ . Differentiate that result once more, remembering the factor $3$ again.

Next step

$\dfrac{dy}{dx} = 3e^{3x}$ . Now differentiate $3e^{3x}$ .

Full working

\frac{dy}{dx} = 3e^{3x}, \qquad \frac{d^2y}{dx^2} = 9e^{3x}.

Each differentiation multiplies by another $3$ . After two differentiations the factor is $3 \times 3 = 9$ .

3. For $f(x) = x^3 - 6x^2 + 9x$ , find $f''(x)$ , find where $f''(x) = 0$ , and test whether that point is a point of inflection.

Hint

Differentiate twice to find $f''$ , then solve $f''(x) = 0$ . Choose a value below and a value above your answer and compare the signs of $f''$ .

Next step

$f'(x) = 3x^2 - 12x + 9$ , so $f''(x) = 6x - 12$ . This is zero at $x = 2$ . Now test $x = 1$ and $x = 3$ .

Full working

f'(x) = 3x^2 - 12x + 9, \qquad f''(x) = 6x - 12.

Set $f''(x) = 0$ : $6x - 12 = 0$ , so $x = 2$ .

At $x = 1$ , $f''(1) = -6 < 0$ , so the curve is concave. At $x = 3$ , $f''(3) = 6 > 0$ , so the curve is convex.

The second derivative changes from negative to positive, so $x = 2$ is a point of inflection.

4. For $f(x) = x^3 + 3x^2$ , state the intervals where the curve is convex and where it is concave.

Hint

Find $f''(x)$ and solve $f''(x) = 0$ . The sign of $f''$ on each side of that value gives the bending.

Next step

$f'(x) = 3x^2 + 6x$ , so $f''(x) = 6x + 6$ . This is zero at $x = -1$ . Test $x = -2$ and $x = 0$ .

Full working

f''(x) = 6x + 6.

$f''(x) = 0$ at $x = -1$ .

For $x < -1$ , $f''(x) < 0$ , so the curve is concave. For $x > -1$ , $f''(x) > 0$ , so the curve is convex.

The curve is concave on $x < -1$ and convex on $x > -1$ .

5. A student says that $f(x) = x^5$ does not have a point of inflection at $x = 0$ , because $f''(0) = 0$ . Check the claim by finding $f''(x)$ and testing its sign either side of $x = 0$ .

Hint

Find $f''(x)$ and look at its sign for a positive and a negative value of $x$ .

Next step

$f'(x) = 5x^4$ and $f''(x) = 20x^3$ . At $x = 0$ this is $0$ . Test $x = -1$ and $x = 1$ .

Full working

f''(x) = 20x^3.

At $x = 0$ , $f''(0) = 0$ . At $x = -1$ , $f''(-1) = -20 < 0$ , so the curve is concave. At $x = 1$ , $f''(1) = 20 > 0$ , so the curve is convex.

The second derivative does change sign here, so $x = 0$ is a point of inflection. The statement in the question is wrong: $y = x^5$ does have a point of inflection at the origin. Compare $y = x^4$ , where $f''(x) = 12x^2$ never changes sign.

6. Use the second derivative to classify the turning points of $f(x) = x^3 - 3x$ .

Hint

Find the turning points from $f'(x) = 0$ . Then substitute each value of $x$ into $f''(x)$ . A positive value means a minimum, and a negative value means a maximum.

Next step

$f'(x) = 3x^2 - 3$ , which is zero at $x = 1$ and $x = -1$ . Now work out $f''(1)$ and $f''(-1)$ .

Full working

f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1), \qquad f''(x) = 6x.

The turning points are at $x = 1$ and $x = -1$ .

$f''(1) = 6 > 0$ , so $x = 1$ is a minimum. $f''(-1) = -6 < 0$ , so $x = -1$ is a maximum.