Second derivatives and concavity | Year 12

01 / LEARN

Understand the idea.

Differentiate the gradient

The second derivative $f^{''}$ measures how the tangent gradient changes. Where $f^{''} > 0$ , gradients increase and the curve is concave up; where $f^{''} < 0$ , gradients decrease and the curve is concave down.

Keep height and gradient separate. A decreasing curve can be concave up when its negative gradients move towards zero. Use $f^{'}$ to decide whether the function rises and $f^{''}$ to decide how it bends.

Use intervals, not a single point

Find possible concavity boundaries from zeros or undefined values of $f^{''}$ and from domain boundaries. Determine its sign on each resulting interval; an isolated zero alone does not describe the curve's bend on either side.

When $s (t)$ is position in metres and time is in seconds, $s^{'}$ is velocity in $m s^{- 1}$ and $s^{''}$ is acceleration in $m s^{- 2}$ . Differentiating twice changes both the quantity and its units.

Labelled graph of y=x cubed minus 3x, showing roots negative square root 3, zero and square root 3, maximum (-1,2), minimum (1,-2), with numbered axes.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Concavity of a cubic

Find the concavity of $f (x) = x^{3} - 3 x$ .

Worked solution

1
Differentiate twice.
$\begin{aligned} f (x) & = x^{3} - 3 x \\ f^{'} (x) & = 3 x^{2} - 3 \\ f^{''} (x) & = 6 x \end{aligned}$
Concavity depends on how the first derivative changes.
2
Find the candidate boundary.
$6 x = 0 ⟹ x = 0$
The second derivative can change sign only at zero here because it is continuous.
3
Determine signs on both intervals.
$\begin{aligned} x < 0 & ⟹ f^{''} (x) < 0 \\ x > 0 & ⟹ f^{''} (x) > 0 \end{aligned}$
Multiplication by positive 6 preserves the sign of x.
4
Translate the signs into concavity.
$Concave down on (- \infty, 0); concave up on (0, \infty)$
Negative second derivative means decreasing gradients; positive means increasing gradients.

Answer

Concave down on $x < 0$ ; concave up on $x > 0$ .

EXAMPLE 02

Acceleration as a second derivative

A position is $s (t) = t^{3} - 3 t^{2} + 2$ metres, with t in seconds. Find acceleration at $t = 2$ seconds.

Worked solution

1
Find velocity from position.
$v (t) = s^{'} (t) = 3 t^{2} - 6 t$
The first time derivative of position is velocity.
2
Differentiate velocity.
$a (t) = v^{'} (t) = 6 t - 6$
The second derivative of position is acceleration.
3
Evaluate at two seconds.
$a (2) = 6 (2) - 6 = 6 m s^{- 2}$
Acceleration has position units divided by time squared.

Answer

$6 {m/s}^{2}$ when position is in metres and time in seconds.

03 / YOUR TURN

Try it yourself.

Check your understanding

Find the second derivative of $2 x^{4} - 3 x^{2}$ .

Show answer & reasoning

$24 x^{2} - 6$

First differentiate $2 x^{4} - 3 x^{2}$ to obtain $8 x^{3} - 6 x$ . Differentiating again gives $24 x^{2} - 6$ ; the linear term contributes the constant $- 6$ .

Keep practising

6 worksheet questions · 15 marks

Open worksheet Worked solutions

Read the questions online

Find $f^{''} (x)$ for $f (x) = x^{5} - 2 x^{3} + x$ .
2 marks
State the concavity of $y = - 2 x^{2} + 3 x + 4$ .
2 marks
For $f (x) = x^{4} - 4 x^{2}$ , find the intervals of concavity.
3 marks
Give an example of a function that decreases and is concave up for $x > 0$ .
3 marks
A population model is $P (t) = 100 e^{0.2 t}$ . Find $P^{''} (0)$ .
2 marks
If $f^{''} (x) > 0$ , does it follow that $f^{'} (x) > 0$ ? Explain.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Find $f^{''} (x)$ for $f (x) = x^{5} - 2 x^{3} + x$ .
2 marks
Differentiate to obtain $5 x^{4} - 6 x^{2} + 1$ .
Differentiate once more.
$20 x^{3} - 12 x$
State the concavity of $y = - 2 x^{2} + 3 x + 4$ .
2 marks
The first derivative is $- 4 x + 3$ .
The second derivative is $- 4 < 0$ .
Concave down for all real $x$ .
For $f (x) = x^{4} - 4 x^{2}$ , find the intervals of concavity.
3 marks
The second derivative is $12 x^{2} - 8$ .
Solve $12 x^{2} - 8 = 0$ and test signs.
Up for $x < - \sqrt{\frac{2}{3}}$ or $x > \sqrt{\frac{2}{3}}$ ; down between these values.
Give an example of a function that decreases and is concave up for $x > 0$ .
3 marks
Its first derivative is $- e^{- x} < 0$ .
Its second derivative is $e^{- x} > 0$ .
$f (x) = e^{- x}$ works.
A population model is $P (t) = 100 e^{0.2 t}$ . Find $P^{''} (0)$ .
2 marks
The first derivative is $20 e^{0.2 t}$ .
The second is $4 e^{0.2 t}$ .
$4$ people per time-unit squared
If $f^{''} (x) > 0$ , does it follow that $f^{'} (x) > 0$ ? Explain.
3 marks
For $f (x) = e^{- x}$ the second derivative is positive.
Its first derivative is negative.
No. It says the gradient is increasing, not necessarily positive.

Checking saved progress…

Curriculum reference & lesson scope

SACE Stage 2 Mathematical Methods

Further differentiation and applications from the current SACE Methods outline. Logarithmic derivatives have their own chapter; integration follows later. Trigonometric calculus uses radians.

Official curriculum source ↗

In this lesson

Understand the idea.

Differentiate the gradient

Use intervals, not a single point

See the method in action.

Concavity of a cubic

Differentiate twice.

Find the candidate boundary.

Determine signs on both intervals.

Translate the signs into concavity.

Acceleration as a second derivative

Find velocity from position.

Differentiate velocity.

Evaluate at two seconds.

Try it yourself.

Check your understanding

Keep practising