Graphs of f, f′ and f″
Read one from another
- In 2 of 20 recent exams
- Paper 1, no calculator
- 1 to 3 marks
How it is set in the exam
Paper 1 (no calculator), 1–3 marks per part. Rare (2 of 20 exam versions) as a whole question, but the same translation is how you read any GDC graph of f′ (graph of the derivative).
Exam phrases to spot (3)
“the diagram shows the graph of y=f′(x)”
“sketch the graph of y=f′′(x)”
“write down the x-coordinate of the local minimum point of f; justify your answer”
In disguise (1)
- a sign table of f′ instead of a graph
Looks like it, but isn't (3)
- regions between the graph of f′ and the x-axis, with their areas given, used to find values of f: integration (Definite integrals without the function)
- the graph of f itself is given and only its own features are asked, with no f′ or f′′ to read off or sketch: read them directly (Stationary points)
- a velocity–time graph of a moving object: its gradient and where it crosses the axis are Velocity and acceleration: at rest, direction, speed; its areas are Displacement, position and distance travelled
The graph of f′ is given
The graph of y=f′(x), for −3≤x≤6, is a smooth curve. The table describes it from left to right, where −3<p<q<r<s<t<6. It has no other zeros or turning points.
- −3≤x<p
- The graph of f′
- below the x-axis, rising
- Sign of f′
- −
- p
- The graph of f′
- crosses the x-axis, going up
- Sign of f′
- 0
- p<x<q
- The graph of f′
- above the x-axis, rising
- Sign of f′
- +
- q
- The graph of f′
- a local maximum
- Sign of f′
- +
- q<x<r
- The graph of f′
- above the x-axis, falling
- Sign of f′
- +
- r
- The graph of f′
- touches the x-axis (a local minimum)
- Sign of f′
- 0
- r<x<s
- The graph of f′
- above the x-axis, rising
- Sign of f′
- +
- s
- The graph of f′
- a local maximum
- Sign of f′
- +
- s<x<t
- The graph of f′
- above the x-axis, falling
- Sign of f′
- +
- t
- The graph of f′
- crosses the x-axis, going down
- Sign of f′
- 0
- t<x≤6
- The graph of f′
- below the x-axis, falling
- Sign of f′
- −
| x | The graph of f′ | Sign of f′ |
|---|---|---|
| −3≤x<p | below the x-axis, rising | − |
| p | crosses the x-axis, going up | 0 |
| p<x<q | above the x-axis, rising | + |
| q | a local maximum | + |
| q<x<r | above the x-axis, falling | + |
| r | touches the x-axis (a local minimum) | 0 |
| r<x<s | above the x-axis, rising | + |
| s | a local maximum | + |
| s<x<t | above the x-axis, falling | + |
| t | crosses the x-axis, going down | 0 |
| t<x≤6 | below the x-axis, falling | − |
To see the picture: plot y=−(x+2)(x−1)2(x−4) for −3≤x≤6, with y from −40 to 25. The curve runs off the bottom of this window near both ends (for x below about −2.5 and above about 4.5), beyond all the features, which is fine. Then p=−2, q=1−232≈−1.12, r=1, s=1+232≈3.12 and t=4. The exam gives only the picture.
(a) Find the values of x for which f is decreasing. Justify your answer. [2]
(b) Find the x-coordinates of the local maximum and local minimum points of f. Justify your answers. [4]
(c) Find the x-coordinates of the points of inflexion of the graph of f. Justify your answer. [3]
Try it yourself first, then compare.
Now try a question
Questions of this type are on their way. Try another one meanwhile: write your working on paper or on screen and see where each mark went. Free, with no account.
- 1
Read the axis label first:
is it the graph of f, f′ or f′′?
- 2
Translate each feature one level up:
On the graph of f′ For the graph of f above the x-axis (f′>0) increasing below the x-axis (f′<0) decreasing crosses the x-axis from + to − local maximum crosses the x-axis from − to + local minimum touches the x-axis without crossing stationary point of inflexion (not a turning point) a local maximum or minimum (a peak or a trough) point of inflexion rising (f′ increasing) concave up falling (f′ decreasing) concave down Going the other way (sketching f′ from f): each turning point of f is a zero of f′, each point of inflexion of f is a turning point of f′, and where f is steepest the graph of f′ is furthest from the x-axis.
- 3
Justify with the feature you read, naming the function:
"f′ changes from negative to positive at x=p", or "f′(x)<0 for x<p". "The gradient is positive" doesn't say whose gradient.
Reading the graph of f′ as if it were f
the peak at x=q is a point of inflexion of f, not a maximum.
Counting every zero of f′ as a turning point.
At x=r the graph of f′ touches the axis without crossing it: no turn.
"The gradient is positive" as a reason
write "f′(x)>0", or "the graph of f′ is above the x-axis".
More mistakes (1)
An extra wrong value
in a list of x-coordinates scores for the list.