NZ Level 7 (NZC) Level 2 (NCEA)
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Identifying Increasing/Decreasing or Constant sections of Function Domains
Lesson

Using the gradient 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 gradient of the tangent is positive if the vertical values increase as the graph goes to the right, and the gradient 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 gradient 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 stationary points.

In this graph, the tangent has gradient $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 stationary point. Notice that to the left of the stationary point the function is decreasing and to the right it is increasing. We call this stationary point a turning point.

In this graph, the tangent also has gradient $0$0, so this is a stationary point. However, on both sides of the stationary point, the function is increasing. We call this stationary point a stationary 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 example

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 stationary point, the first step is to find all of the stationary points and the second step is to classify each region between the stationary points.

Do: First, notice the stationary 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 stationary 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 stationary point of inflection, the revenue is growing on either side of that day, but not on the day itself.

Increasing, decreasing and constant functions

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 stationary point is a point on the function where the function is constant.

  • A turning point is a stationary point where the function is increasing on one side and decreasing on the other
  • A stationary point of inflection is a stationary point where the function is increasing on both sides or decreasing on both sides

 

Practice questions

Question 1

Which of the following describes the rate of change of the function $f\left(x\right)=-x+10$f(x)=x+10?

  1. The rate of change of the $y$y values is:

    Positive for all values of $x$x

    A

    Negative for all values of $x$x

    B

    Zero for all values of $x$x

    C

    Positive for all values of $x$x

    A

    Negative for all values of $x$x

    B

    Zero for all values of $x$x

    C

Question 2

Consider the function $f\left(x\right)=\left(x+4\right)\left(x-2\right)$f(x)=(x+4)(x2) drawn below.

Loading Graph...

  1. What is the $x$x-value of the stationary point of $f\left(x\right)$f(x)?

  2. What is the region of the domain where $f\left(x\right)$f(x) is increasing?

    Write the answer in interval notation.

  3. What is the region of the domain where $f\left(x\right)$f(x) is decreasing?

    Write the answer in interval notation.

Question 3

Consider the function $f\left(x\right)=-\left(x-4\right)^3+7$f(x)=(x4)3+7 drawn below.

Loading Graph...

  1. What is the $x$x-value of the stationary point of $f\left(x\right)$f(x)?

  2. 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.

Outcomes

M7-9

Sketch the graphs of functions and their gradient functions and describe the relationship between these graphs

91262

Apply calculus methods in solving problems

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