1.05 Critical points and average rate of change
Slopes of tangents
We can find the rate of change of a linear function by finding the slope. How can we find the rate of change of a non-linear function? Consider these scenarios:
- The vertical speed of a projectile
- The reproduction rate of a colony of bacteria
- The revenue growth of a business over a year
- The flow in and out of a body of water
Exploration
We looked at the horizontal velocity of a projectile earlier. Can we use the same technique to find this rate of change? Suppose that the vertical height of the projectile ($y$y, in meters) is given by $y=40t-5t^2$y=40t−5t2. Let’s plot this relationship first:
Notice that the function is non-linear. Does it have a constant rate of change?
We can test this by calculating the average rate of change between various points.
For two points on a curve, $\left(x_1,y_1\right)$(x1,y1) and $\left(x_2,y_2\right)$(x2,y2), the average rate of change is the slope between those two points.
$\text{Average rate of change }=\frac{y_2-y_1}{x_2-x_1}$Average rate of change =y2−y1x2−x1
Let’s choose $\left(1,35\right)$(1,35), $\left(3,75\right)$(3,75), $\left(5,75\right)$(5,75), and $\left(7,35\right)$(7,35):
Calculating the rate of change using the previous technique gives us:
| $\frac{\text{Change in height from }t=1\text{ to }t=3}{\text{Change in time from }t=1\text{ to }t=3}$Change in height from t=1 to t=3Change in time from t=1 to t=3 | $=$= | $\frac{75-35}{3-1}$75−353−1 |
| $=$= | $\frac{40}{2}$402 | |
| $=$= | $20$20 | |
| $\frac{\text{Change in height from }t=3\text{ to }t=5}{\text{Change in time from }t=3\text{ to }t=5}$Change in height from t=3 to t=5Change in time from t=3 to t=5 | $=$= | $\frac{75-75}{5-3}$75−755−3 |
| $=$= | $\frac{0}{2}$02 | |
| $=$= | $0$0 | |
| $\frac{\text{Change in height from }t=5\text{ to }t=7}{\text{Change in time from }t=5\text{ to }t=7}$Change in height from t=5 to t=7Change in time from t=5 to t=7 | $=$= | $\frac{35-75}{7-5}$35−757−5 |
| $=$= | $\frac{-40}{2}$−402 | |
| $=$= | $-20$−20 |
When we choose three different intervals to find the rate of change, we get three different results. The issue is that since the function is non-linear the rate of change is variable.
Since the rate of change is variable, we are looking for a way to find the rate of change at a particular time. We still want to use the slope of a line, but even if we fix one point on the graph, we cannot guarantee that any second point will give us the correct rate of change. And if we choose the same point both times we will get $\frac{0}{0}$00 which is undefined.
A line is a tangent (or tangent line) to a curve at a specific point if it touches and travels in the same direction as the curve at that point. Since it travels in the same direction it has the same rate of change as the function at that point. So the slope of the tangent will give us the rate of change of the function.
In this topic we will only use tangent to refer to lines. However, in other contexts any kind of curve can meet any other kind of curve at a point and travel in the same direction at that point. These two curves are both tangents to each other.
Exploration
Returning to the previous example, tangents to the curve at three points have been drawn in:
The tangent to the curve at $t=1$t=1 is $y=30t-5$y=30t−5. The slope of this tangent is $30$30 and so the rate of change of the height with respect to time after $1$1 second is $30$30 m/s. We can use the other two tangents to find the rate of change at $t=4$t=4 and $t=5$t=5:
| Time | Equation of tangent | Rate of change |
|---|---|---|
| $t=1$t=1 | $y=30t+5$y=30t+5 | $30$30 m/s |
| $t=4$t=4 | $y=80$y=80 | $0$0 m/s |
| $t=5$t=5 | $y=-10t+125$y=−10t+125 | $-10$−10 m/s |
The rate of change of a non-linear function does vary with respect to the independent variable. However, we can find the rate of change for any particular value of the independent variable if we know the tangent at this value.
It’s worth considering the situations where we cannot draw a tangent on a graph. The first requirement is that the point is on the domain of the function. If the function is undefined at a horizontal value then the tangent will also be undefined.
Secondly, the function must not suddenly change direction or position at the point. Since the function has two or more directions at these points, they do not have tangents. Examples of these points are given in the graph below:
A line is a tangent (or tangent line) to a curve at a specific point if it touches and travels in the same direction as the curve at that point.
The rate of change of a non-linear function at a specific point is the slope of the tangent to the function at that point.
Practice questions
Question 1
Consider the function $f\left(x\right)=2x-4$f(x)=2x−4 shown here.
If we were to draw a tangent to the function for any $x$x value, what is the slope of that tangent?
Question 2
What is the slope of the tangent at the given point?
Question 3
What is the slope of the tangent at the given point?
Identify increasing, decreasing, or constant sections of function domains
Using the slope of the tangent, we can find the rate of change of most functions. For non-linear functions, the rate of change will vary with respect to the independent variable. When it comes to interpreting the rate of change, there are four situations.
First, if the rate of change is positive ($m>0$m>0), then the function is increasing. Second, if the rate of change is negative ($m<0$m<0), then the function is decreasing. We can see this on a graph:
Here the slope of the tangent is positive if the vertical values increase as the graph goes to the right, and the slope of the tangent is negative if the vertical values decrease as the graph goes to the right.
We can also interpret this in terms of the rate of change of the variables. For example, acceleration is the rate of change of velocity with respect to time. If the acceleration is positive (so that the slope of the velocity with respect to time is positive) then the velocity is increasing, and if the acceleration is negative then the velocity is decreasing.
The third situation comes when the rate of change is zero ($m=0$m=0). In this case, the function is constant. Since the rate of change of non-linear functions is variable, it will usually only be zero at specific points on the domain. We call these points critical points.
In this graph, the tangent has slope $0$0 so it is a horizontal line. The rate of change is $0$0 at this specific point, but nowhere else on the function, so this is a critical point. Notice that to the left of the critical point the function is decreasing and to the right it is increasing. We call this critical point a turning point.
In this graph, the tangent also has slope $0$0, so this is a critical point. However, on both sides of the critical point, the function is increasing. We call this critical point a point of inflection.
The fourth situation is when there is no tangent at that point. This could be because the function is not defined at that point, or it could be that the function suddenly changes position or direction. In either case, the function is not increasing, decreasing or stationary. The function could be increasing on both sides, decreasing on both sides or increasing on one side and decreasing on the other.
Worked examples
Question 4
A company's daily revenue, $R$R, in thousand dollars over $12$12 months has been plotted below. We want to classify the entire domain of the function graphed below into regions which are increasing, decreasing or constant:
Think: While we do not know exactly what this function is, we have enough information in the graph to work this out. Since a function cannot change from increasing to decreasing without passing through a critical point, the first step is to find all of the critical points and the second step is to classify each region between the critical points.
Do: First, notice the critical points at $t=3$t=3, $5$5, and $10$10. The function is constant at these three points. To classify the rest of the function, divide the domain into the intervals $\left(0,3\right)$(0,3), $\left(3,5\right)$(3,5), $\left(5,10\right)$(5,10), and $\left(10,12\right)$(10,12).
The function value is higher when $t=3$t=3 than when $t=0$t=0. So between $t=0$t=0 and $t=3$t=3 the function is increasing. The function value is lower when $t=5$t=5 than when $t=5$t=5. So between $t=3$t=3 and $t=5$t=5 the function is decreasing.
Following the same process, the function is increasing from $t=5$t=5 to $t=10$t=10 and increasing from $t=10$t=10 to $t=12$t=12. However, we cannot say that the function is increasing from $t=5$t=5 to $t=12$t=12 because there is a critical point at $t=10$t=10.
In conclusion, the function is constant when $t$t is in the set $\left\{3,5,10\right\}${3,5,10}, increasing when $t$t is in the set $\left(0,3\right)\cup\left(5,10\right)\cup\left(10,12\right)$(0,3)∪(5,10)∪(10,12) and decreasing when $t$t is in the set $\left(3,5\right)$(3,5).
Reflect: Knowing the intervals when the function is increasing and decreasing lets us interpret the rate of change of the revenue. Since the function is always positive, the company is always taking in money.
This is true even when the function is decreasing. The rate of change is negative on the interval $\left(3,5\right)$(3,5), so the company is taking in less money during this period than they did at the start of month $3$3 but the rate of change is slow enough that the revenue never completely stops.
Also, as the time gets close to $10$10 months, the rate of change reaches $0$0. Since this is a critical point of inflection, the revenue is growing on either side of that day, but not on the day itself.
A function can be described as either increasing, decreasing or constant. Some functions will be increasing, decreasing or constant only in specific intervals of the domain.
- A function is increasing (on an interval) if the rate of change is positive ($m>0$m>0)
- A function is decreasing (on an interval) if the rate of change is negative ($m<0$m<0)
- A function is constant (on an interval) if the rate of change is zero ($m=0$m=0)
A critical point is a point on the function where the function is constant.
- An extremum is a critical point where the function is increasing on one side and decreasing on the other. At these points, the function has a maximum or a minimum value.
- A point of inflection is a critical point where the function is increasing on both sides or decreasing on both sides
Practice questions
Question 5
Which of the following describes the rate of change of the function $f\left(x\right)=-x+10$f(x)=−x+10?
The rate of change of the $y$y values is:
Positive for all values of $x$x
ANegative for all values of $x$x
BZero for all values of $x$x
CPositive for negative values of $x$x and negative for positive values of $x$x
D
Question 6
Consider the function $f\left(x\right)=\left(x+2\right)\left(x-4\right)$f(x)=(x+2)(x−4) drawn below.
What is the $x$x-value of the critical point of $f\left(x\right)$f(x)?
What is the region of the domain where $f\left(x\right)$f(x) is increasing?
Write the answer in interval notation.
What is the region of the domain where $f\left(x\right)$f(x) is decreasing?
Write the answer in interval notation.
Question 7
Consider the function $f\left(x\right)=-2\left(x+4\right)^3-2$f(x)=−2(x+4)3−2 drawn below.
What is the $x$x-value of the critical point of $f\left(x\right)$f(x)?
What are the regions of the domain where $f\left(x\right)$f(x) is decreasing?
Write the regions in interval notation separated by commas.
Average rates of change
We can find the rate of change of a non-linear function at a point by looking at the slope of the tangent. However, this will only tell us about the rate of change at one point and it requires knowing the tangent first.
Exploration
Suppose that the relationship between the population ($n$n) of a colony of bacteria and the time ($t$t, in days) is described by the relationship $n=2^t$n=2t. We want to find the growth rate, which is the rate of change of the population. We can plot this relationship:
Instead of a tangent, we will use a secant. A secant is a line which touches a curve at two specific points. We can choose two points on this curve, $\left(1,2\right)$(1,2) and $\left(5,32\right)$(5,32), and draw the line connecting them, $y=7.5x-5.5$y=7.5x−5.5:
This secant has a slope of $7.5$7.5 so we say that the average rate of change from day $1$1 to day $5$5 is $8$8 per day.
We can draw secants through any two points of a graph. Here we have drawn the secants connecting $\left(1,2\right)$(1,2) to $\left(3,8\right)$(3,8), $\left(2,4\right)$(2,4) to $\left(4,16\right)$(4,16), and $\left(3,8\right)$(3,8) to $\left(5,32\right)$(5,32). These secants have slopes of $3$3, $6$6, and $12$12 respectively.
Notice that the average rate of change is variable since this is a non-linear function. This is true even when we take secants around the same point. $\left[1,5\right]$[1,5] and $\left[2,4\right]$[2,4] are both intervals around $3$3, but in the first case the average rate of change is $7.5$7.5 and in the second it is $6$6.
Worked example
Question 8
The volume of a lake over five weeks has been recorded below:
| Week | $0$0 | $1$1 | $2$2 | $3$3 | $4$4 | $5$5 |
|---|---|---|---|---|---|---|
| Volume (m3) | $123000$123000 | $142000$142000 | $135000$135000 | $111000$111000 | $104000$104000 | $123000$123000 |
(a) Find the average rate of change of volume in the first week
(b) Find the average rate of change of volume over the whole five weeks
(c) Find the average rate of change of volume in the last three weeks
Think: We don't know the function mapping the week to the volume. However, the average rate of change only requires two points on the graph. So we can find the average rate of change from just the data points.
Do: For each period, the average rate of change will be the change in volume divided by the number of weeks:
| (a) | average rate of change of volume in the first week | $=$= | $\frac{\text{change in volume}}{\text{number of weeks}}$change in volumenumber of weeks |
| $=$= | $\frac{142000-123000}{1-0}$142000−1230001−0 | ||
| $=$= | $\frac{19000}{1}$190001 | ||
| $=$= | $19000$19000 | ||
| (b) | average rate of change of volume over the whole five weeks | $=$= | $\frac{\text{change in volume}}{\text{number of weeks}}$change in volumenumber of weeks |
| $=$= | $\frac{123000-123000}{5-0}$123000−1230005−0 | ||
| $=$= | $\frac{0}{1}$01 | ||
| $=$= | $0$0 | ||
| (c) | average rate of change of volume in the last three weeks | $=$= | $\frac{\text{change in volume}}{\text{number of weeks}}$change in volumenumber of weeks |
| $=$= | $\frac{135000-123000}{5-2}$135000−1230005−2 | ||
| $=$= | $\frac{-12000}{3}$−120003 | ||
| $=$= | $-4000$−4000 |
Reflect: We can tell that the function is non-linear because the average rate of change is variable. Also notice that the average rate of change can be positive, negative or zero depending on the interval we choose. This is a significant limitation of average rates of change.
A secant is a line which intersects with a curve at two points
The average rate of change of a function over an interval is the slope of the secant on the function between the endpoints of the interval
Practice questions
Question 9
Does the graphed function have a constant or a variable rate of change?
Constant
AVariable
B
Question 10
Consider a function which takes certain values, as shown in the table below.
| $x$x | $3$3 | $6$6 | $8$8 | $13$13 |
|---|---|---|---|---|
| $y$y | $-12$−12 | $-15$−15 | $-17$−17 | $-22$−22 |
Find the average rate of change between $x=3$x=3 and $x=6$x=6.
Find the average rate of change between $x=6$x=6 and $x=8$x=8.
Find the average rate of change between $x=8$x=8 and $x=13$x=13.
Do the set of points satisfy a linear or non-linear function?
Linear
ANon-linear
B
Question 11
Does the function that is satisfied by the ordered pairs: $\left\{\left(-2,-5\right),\left(1,-20\right),\left(2,-25\right),\left(7,-50\right),\left(9,-60\right)\right\}${(−2,−5),(1,−20),(2,−25),(7,−50),(9,−60)} have a constant or a variable rate of change?
Constant
AVariable
B