In our previous lesson we found key features using calculus and the equation of the function but we could be given pieces of information and be required to identify an appropriate graph or sketch a possible graph. Information given could include points the graph passes through, the derivative graph, the second derivative graph or features of these.

Worked examples

Example 1

Sketch a possible graph that has the following properties:

$f'(2)=0$f′(2)=0 and $f'(1)=0$f′(1)=0
$f(2)=0$f(2)=0
$f(1)=1$f(1)=1
$f(0)=-4$f(0)=−4
$f'(x)<0$f′(x)<0 for $11<x<2 and $f'(x)>0$f′(x)>0 elsewhere

Think: Carefully consider if the information given is about the function or its derivative.

The first clue tells us that the graph has stationary points at $x=2$x=2 and $x=1$x=1.

The second piece of information tells us the graph passes through $\left(2,0\right)$(2,0). This is also our stationary point from the first clue and an $x$x-intercept.

Third piece of information tells us that the graph also passes through $\left(1,1\right)$(1,1), which is our other stationary point.

Dot point $4$4, tells us that when $x=0$x=0, $y=-4$y=−4. That is the $y$y-intercept is $\left(0,-4\right)$(0,−4).

Dot point $5$5 tells us that between the two stationary points we have a decreasing function and elsewhere we have an increasing function.

Do: Plotting the points and indicating the known areas of increasing and decreasing, we obtain:

Joining the dots with a smooth curve and following the arrow trends we have a graph of a function with the given properties - this is not the only possibility, can you draw a few more?

Example 2

The graphs of $y=f'\left(x\right)$y=f′(x) and $y=f''\left(x\right)$y=f′′(x) of a function $y=f\left(x\right)$y=f(x) are shown below. The function passes through $\left(1,5\right)$(1,5), $\left(4,-2\right)$(4,−2) and $\left(7,3\right)$(7,3).

a) Determine the coordinates and nature of any stationary points of the graph $y=f\left(x\right)$y=f(x).

Think: From the graph of $f'\left(x\right)$f′(x) we can look for where the graph cuts the $x$x-axis. The $x$x-coordinates of zeros of the graph of $f'\left(x\right)$f′(x) will correspond to stationary points in the graph of $f\left(x\right)$f(x). To classify the stationary points we can look at the gradient function behaviour just either side of the points or look at the value and behaviour of the second derivative at these values.

Do: The graph of $f'\left(x\right)$f′(x) has zeros at $x=1$x=1, $4$4 and $7$7. So we have confirmed the given points are each stationary points.

At $x=1$x=1: Notice the derivative changes from positive to negative. Also the second derivative is negative, hence concave down. Thus we have a maximum turning point at $\left(1,5\right)$(1,5).

At $x=4$x=4: Notice the derivative changes from negative to positive. Also the second derivative is positive, hence concave up. Thus we have a minimum turning point at $\left(4,-2\right)$(4,−2).

At $x=7$x=7: Notice the derivative does not change sign. Also the second derivative is zero at this point and changes from negative to positive about this point, hence a change in concavity. Thus we have a stationary point of inflection at $\left(7,3\right)$(7,3).

b) The graph of $f''\left(x\right)$f′′(x) has an $x$x-intercept at $x=2.1$x=2.1, what feature will there be in the graph of $f\left(x\right)$f(x) at this coordinate?

Think: Does the graph change concavity? Does this coordinate match with a zero of $f'\left(x\right)$f′(x)?

Do: The graph of $f''\left(x\right)$f′′(x) changes from negative to positive - thus the graph of $f\left(x\right)$f(x) changes from concave down to concave up. So we have a point of inflection and since this does not align with a zero of $f'\left(x\right)$f′(x) we have a non-stationary point of inflection. From the minimum in the graph of $f'\left(x\right)$f′(x) we can see this is where the gradient of $f(x)$f(x) will be at its steepest negative gradient.

Being able to make connections between the graphs of $f\left(x\right)$f(x), $f'\left(x\right)$f′(x) and $f''\left(x\right)$f′′(x), and their key features is an important skill. Practice sketching different functions and their derivatives given one of the graphs as information. As well as the exercise for this lesson, you can try the following investigation to hone your skills.

Use the following applet to explore the sign of the derivative at any point along a function.

Created with Geogebra

Practice questions

Question 1

Consider the four functions sketched below.

Which of the sketches match the information for $f\left(x\right)$f(x) shown in the table? Select all the correct options.

Information about $f\left(x\right)$f(x):

$f'\left(-1\right)=0$f′(−1)=0

$f'\left(4\right)=0$f′(4)=0

$f'\left(x\right)>0$f′(x)>0 for $x>4$x>4

$f'\left(x\right)<0$f′(x)<0 elsewhere

A
B
C
D

Question 2

Consider the gradient function $f'\left(x\right)=\left(x-2\right)^2-4$f′(x)=(x−2)2−4 graphed below.

Loading Graph...

State the coordinates of the turning point of $f'\left(x\right)$f′(x).

Give your answer in the form $\left(x,y\right)$(x,y).
What kind of feature is at the point $\left(2,-5\right)$(2,−5) on the graph of $f\left(x\right)$f(x)?
Minimum turning point
A
$x$x-intercept
B
Maximum turning point
C
Point of inflection
D

Question 3

Given the following graph of $y=f\left(x\right)$y=f(x) select the graph that represents $y=f'\left(x\right)$y=f′(x).

A
B
C
D

Question 4

Consider the following graph of $y=f\left(x\right)$y=f(x).

Select a possible graph of the derivative.
A
B
C
D
For what intervals is the graph of $y=f\left(x\right)$y=f(x) increasing?

Give your answer as an inequality.
What are the $x$x-values of the stationary points?

Enter your answers on the same line, separated by commas.