Second derivatives and concavity | Year 12

01 / LEARN

The idea, step by step

Differentiate the gradient

The second derivative $f^{''}$ is the derivative of $f^{'}$ . When it is positive, tangent gradients increase as $x$ increases: the curve is concave up. When it is negative, gradients decrease: the curve is concave down.

Concave up does not mean the original function is increasing. A decreasing curve can bend upwards while its negative gradients become less steep.

Use intervals, not a single point

Solve $f^{''} (x) = 0$ to find possible boundaries, add points where a derivative is undefined, and test its sign on each interval. A sign chart is often more informative than only substituting a stationary point.

For position measured in metres against seconds, the first derivative is m/s and the second is m/s squared. Name the units when interpreting an answer.

A common trap

Calling every concave-up part of a graph increasing.

Instead: Use the first derivative for increasing/decreasing height; use the second derivative for changing gradient.

02 / WORKED EXAMPLES

See the thinking

Example 1

Concavity of a cubic

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

$f^{'} = 3 x^{2} - 3$ and $f^{''} = 6 x$ .
The second derivative is negative for $x < 0$ and positive for $x > 0$ .

Concave down on

x < 0

; concave up on

x > 0

Example 2

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.

$v (t) = 3 t^{2} - 6 t$ and $a (t) = 6 t - 6$ .
At 2 seconds, $a = 6$ .

6 {m/s}^{2}

when position is in metres and time in seconds.

Check your understanding

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

Show answer

$24 x^{2} - 6$

Differentiate twice: $8 x^{3} - 6 x$ , then $24 x^{2} - 6$ .

03 / WORKSHEET

Your turn

6 questions · 15 marks · Work at your own pace.

Worksheet PDF 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 $x < - \sqrt{2 / 3}$ or $x > \sqrt{2 / 3}$ ; 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.

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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 ↗

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