Graph of the Derivative f'(x) | IEB Grade 12 Maths

Q: What does the turning point of f'(x) tell me?

It gives the $x$-value of the point of inflection of $f$, where the concavity of $f$ changes.

Q: Can I find the turning point values of f from the graph of f'?

You can find their $x$-values. For the $y$-values you need the equation of $f$ itself.

IEB examiners like to give you the graph of $f^{'} (x)$ instead of $f (x)$ and ask what it says about $f$ . These questions test understanding rather than algebra, and they are where many strong candidates lose easy marks.

Sketching f'(x) from f(x)

The derivative of a cubic is a quadratic, so the graph of $f^{'}$ is a parabola. Match features like this:

Turning points of $f$ are the $x$ -intercepts of $f^{'}$ .
Point of inflection of $f$ is the turning point of $f^{'}$ .
Where $f$ is increasing, $f^{'}$ is above the $x$ -axis.
Where $f$ is decreasing, $f^{'}$ is below the $x$ -axis.

Example. $f (x) = x^{3} - 6 x^{2} + 9 x$ .

$f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3)$

The parabola cuts the $x$ -axis at 1 and 3, cuts the $y$ -axis at 9, and has its turning point at $x = 2$ , where $f^{'} (2) = - 3$ .

Reading f(x) from the graph of f'(x)

When you are given the graph of $f^{'}$ :

$x$ -intercepts of $f^{'}$ give the $x$ -values of the stationary points of $f$ .
Sign change from positive to negative means $f$ has a local maximum there. Negative to positive means a local minimum.
The turning point of $f^{'}$ gives the $x$ -value of the point of inflection of $f$ .
Where $f^{'}$ is increasing, $f$ is concave up. Where $f^{'}$ is decreasing, $f$ is concave down.

Example. $f^{'}$ is a parabola with $x$ -intercepts $- 1$ and $3$ and a minimum turning point at $x = 1$ .

$f$ has a local maximum at $x = - 1$ ( $f^{'}$ changes from positive to negative)
$f$ has a local minimum at $x = 3$
$f$ is decreasing for $- 1 < x < 3$
$f$ has a point of inflection at $x = 1$ and is concave up for $x > 1$

You cannot read the $y$ -values of $f$ from the graph of $f^{'}$ . You need the equation of $f$ for those.

Key terms

Derivative graph: The graph of $y = f^{'} (x)$ , showing the gradient of $f$ at each $x$ .
Sign of the derivative: Positive means $f$ is increasing; negative means $f$ is decreasing.
Concavity: The direction a graph bends, decided by whether $f^{'}$ is increasing or decreasing.

Common mistakes

Reading the turning point of $f^{'}$ as a turning point of $f$ . It is the point of inflection.
Reading $y$ -values of $f$ off the graph of $f^{'}$ .
Mixing up 'above the $x$ -axis' (increasing $f$ ) with 'increasing $f^{'}$ ' (concave up $f$ ).
Giving intervals for increasing or decreasing in terms of $y$ .

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

3 marks total

knowledge

Given $f^{'} (x) = 3 x^{2} - 12$ . For which values of $x$ is $f$ increasing?

Worked solution

Hidden - try the question first

Solution

$f$ is increasing where $f^{'} (x) > 0$ : $3 x^{2} - 12 > 0$
$3 (x - 2) (x + 2) > 0$
$x < - 2$ or $x > 2$

Question 2

5 marks total

routine

Given $f (x) = x^{3} - 6 x^{2} + 9 x$ .

Determine $f^{'} (x)$ and give the $x$ -intercepts, the $y$ -intercept and the turning point of the graph of $f^{'}$ .

Worked solution

Hidden - try the question first

Solution

$f$
$3 (x - 1) (x - 3) = 0$ , so the $x$ -intercepts are $x = 1$ and $x = 3$
$y$ -intercept: $(0; 9)$
Turning point at $x = \frac{1 + 3}{2} = 2$
$f^{'} (2) = 12 - 24 + 9 = - 3$ , so $(2; - 3)$

Question 3

6 marks total

complex

The graph of $y = f^{'} (x)$ is a parabola with $x$ -intercepts at $x = - 1$ and $x = 3$ and a minimum turning point at $(1; - 4)$ .

(a) Write down the $x$ -values of the turning points of $f$ and state whether each is a maximum or a minimum. (3)

(b) For which values of $x$ is $f$ decreasing? (1)

(d) For which values of $x$ is $f$ concave up? (1)

Worked solution

Hidden - try the question first

Solution

(a) $f^{'}$ changes from positive to negative at $x = - 1$ : local maximum
$3 x^{2} - 12 > 0$ changes from negative to positive at $x = 3$ : local minimum
Both identified from the $3 (x - 2) (x + 2) > 0$ -intercepts of $f^{'}$
(b) $f^{'} (x) < 0$ for $- 1 < x < 3$
(c) $x = 1$ , the turning point of $f^{'}$
(d) $f^{'}$ is increasing for $x > 1$ , so $f$ is concave up for $x > 1$

Question 4

8 marks total

problem-solving

$f (x) = a x^{3} + b x^{2} + c x$ . The graph of $f^{'}$ is a parabola that cuts the $x$ -axis at $x = - 1$ and $x = 3$ and passes through $(0; - 9)$ .

(a) Determine the values of $a$ , $b$ and $c$ . (5)

(b) Calculate the $y$ -coordinate of the local minimum of $f$ . (3)

Worked solution

Hidden - try the question first

Solution

(a) $f^{'}$
$f^{'} (0) = - 3 k = - 9 \Rightarrow k = 3$ , so $f^{'} (x) = 3 x^{2} - 6 x - 9$
From $f (x) = a x^{3} + b x^{2} + c x$ : $f^{'} (x) = 3 a x^{2} + 2 b x + c$
Comparing: $3 a = 3$ , $2 b = - 6$ , $c = - 9$
$f$ , $b = - 3$ , $c = - 9$
(b) The minimum is at $x = 3$ ( $f$ changes from negative to positive)
$f (3) = 27 - 27 - 27$
$= - 27$

FAQs

Why does the IEB give the graph of f'(x) instead of f(x)?

It tests whether you understand what the derivative means, not just how to calculate it. These questions are often short but high-scoring.

What does the turning point of f'(x) tell me?

It gives the $x$ -value of the point of inflection of $f$ , where the concavity of $f$ changes.

Can I find the turning point values of f from the graph of f'?

You can find their $x$ -values. For the $y$ -values you need the equation of $f$ itself.

Drawing and interpreting the graph of the derivative

Sketching f'(x) from f(x)

Reading f(x) from the graph of f'(x)

Key terms

Common mistakes

Practice questions

FAQs

Need a tutor for this subject?