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 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 $1
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 graph of $y=f'\left(x\right)$y=f′(x) of a function $y=f\left(x\right)$y=f(x) is shown below. Consider the coordinates $\left(1,5\right)$(1,5), and $\left(4,-2\right)$(4,−2)
 |
a) Determine the coordinates and nature of any local minimum/maximum 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. As the curve to the left of $x=1$x=1 is above the $x$x-axis and below it to the left, the nature of $$ 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 taking into account the gradient of the curve either side of the coordinate, it is concave up. Thus we have a minimum turning point at $\left(4,-2\right)$(4,−2).
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 $f'\left(x\right)=\left(x-1\right)^2-5$f′(x)=(x−1)2−5 drawn below.
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).
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).
.svg)
A.svg)
B.svg)
C.svg)
D
Question 4
Consider the following graph of $y=f\left(x\right)$y=f(x).
Select a possible graph of the derivative.
.svg)
A.svg)
B.svg)
C.svg)
DFor 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.