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Determining Intervals on Which a Function is Increasing or Decreasing

Hannah Hill

Hannah Hill

7 min read

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Study Guide Overview

This study guide covers how to determine the intervals where a function is increasing or decreasing using the derivative. It explains that a positive derivative indicates an increasing function, while a negative derivative indicates a decreasing function. The guide emphasizes the importance of critical points (where the derivative is zero or undefined) in identifying these intervals. It provides a walkthrough example, step-by-step instructions, and practice problems to solidify understanding of this concept.

#5.3 Determining Intervals on Which a Function is Increasing or Decreasing

What can the derivative of a function show us about the function itself? Can it tell us when the function increases or decreases? Yes, it can! In this lesson, we’ll delve into how we can use derivatives to determine when a function increases or decreases. 📈

#🕑 When Does a Function Increase or Decrease?

In order to determine the intervals on which a function is increasing or decreasing, we first need to understand the concept of the derivative. The derivative of a function is the rate of change of the function at a given point.

Thus, we know the following:

  1. ➕=📈 If the derivative is positive at a certain point (which means the rate of change is positive at that point), the function is increasing at that point.
  2. ➖=📉 If the derivative is negative at a certain point, the function is decreasing at that point.

Take a look at this graph to see these trends in action. The gray line represents the function, fff, and the black line represents its derivative, f′f'f′.

!Untitled

Graph of a function and its derivative to demonstrate the trend

Image Courtesy of Informal Calculus

Now, you may be thinking that finding the interval where a function is increasing or decreasing is as simple as finding the interval where the function’s derivative is positive or negative, respectively. Well, it is! 🪄

Where can a function change from increasing to decreasing and vice versa? It can only change its direction from increasing to decreasing and vice versa at its critical points, points where the function’s derivative equals 000 or is undefined, and the points where the function itself is undefined.

So, for each of the intervals defined by the points where the function can change behavior, we can determine whether the function is increasing or decreasing on the interval by just plugging a point on that interval into the function’s derivative and seeing if the result is positive or negative. If it’s positive, then the function is increasing on that interval; if it’s negative, then the function is decreasing on that interval. Let’s give it a try!

#✏️ Function Behavior: Walkthrough

➡️ Let hhh be a function defined for all real numbers except 0.0.0. Also let h′h'h′ be defined as h′(x)=(x+7)x2h'(x)=\frac{(x+7)}{x^2}h′(x)=x2(x+7)​. On which intervals is hhh increasing?

From what we learned above, we can analyze the intervals where hhh is increasing by looking for the intervals where its derivative hhh is positive.

A function can only change its direction from increasing to decreasing and vice versa at its critical points and the points where the function itself is undefined. Based on the problem statement, we determine that in this case, the only points where hhh can change direction are x=−7x=-7x=−7 and x=0x=0x=0. We determine this by doing the following and solving:

h′(x)=(x+7)x2h'(x)=\frac{(x+7)}{x^2}h′(x)=x2(x+7)​

0=(x+7)x20=\frac{(x+7)}{x^2}0=x2(x+7)​

h′=0h'=0h′=0 at x=−7x=-7x=−7 so it is a critical point and hhh is undefined at 000 as stated in the problem.

These two points divide the number line into three intervals:

(−∞,−7),(−7,0),(0,∞)(-\infty,-7),(-7,0),(0,\infty)(−∞,−7),(−7,0),(0,∞)

Let’s evaluate h′h'h′ at each interval to see if it’s positive or negative on that interval and therefore see if hhh is increasing or decreasing on the interval. To evaluate h′h'h′ on an interval, we can choose any xxx-value within the interval to substitute xxx with and calculate.

In the table below, you’ll notice that we chose x=−8x=-8x=−8 for the first interval, x=−1x=-1x=−1 for the second interval, and x=1x=1x=1 for the third! Plug in, solve, and determine the behavior of the function.

Intervalxxxh′(x)h'(x)h′(x)Verdict
(−∞,−7)(-\infty,-7)(−∞,−7)x=−8x=-8x=−8h′(−8)=−164<0h'(-8)=-\frac{1}{64} <0h′(−8)=−641​<0hhh is decreasing
(−7,0)(-7,0)(−7,0)x=−1x=-1x=−1h′(−1)=6>0h'(-1)=6 > 0h′(−1)=6>0hhh is increasing
(0,∞)(0,\infty)(0,∞)x=1x=1x=1h′(1)=8>0h'(1)=8>0h′(1)=8>0hhh is increasing

In conclusion, hhh is increasing on the intervals (−7,0)(-7,0)(−7,0) and (0,∞)(0,\infty)(0,∞).

#🪜 Steps to Determining Function Behavior

Here are the steps we used if you prefer seeing them in list form:

  1. 🎯 Determine the critical points of the function, where f′(x)f'(x)f′(x) equals zero or is undefined.
  2. 🔢 Divide the function's domain into intervals based on the critical points determined. This is where you break the number line into pieces to see its behavior!
  3. 🔨 For each interval, choose a test point within that interval.
  4. ✏️ Evaluate the function's derivative at the chosen test point.
  5. 🏁 Interpret the result.
  6. If the result is positive, the function is increasing on that interval.
  7. If the result is negative, the function is decreasing on that interval.

Now you can put these steps into practice!


#📝 Function Behavior: Practice Problems

Time to solve some problems! 🔍

#❓ Function Behavior: Problems

#Function Behavior: Question 1

Let f(x)=x3−27xf(x)=x^3-27xf(x)=x3−27x.

On which interval(s) is fff decreasing?

#Function Behavior: Question 2

Let f(x)=x4−2x2f(x)=x^{4}-2x^{2}f(x)=x4−2x2.

On which interval(s) is fff increasing?

#✅ Function Behavior: Answers and Solutions

#Function Behavior: Question 1

The answer to this problem is (−3,3)(-3,3)(−3,3). Let’s get into why. ⬇️

We can analyze the intervals where fff is decreasing by looking for the intervals where its derivative fff is negative.

A function can only change its direction from increasing to decreasing and vice versa at its critical points and the points where the function itself is undefined. So first, take the derivative and find the critical points.

Since the derivative of fff is f′(x)=3x2−27f'(x)=3x^2-27f′(x)=3x2−27, the only points where fff can change direction are x=−3x=-3x=−3 and x=3x=3x=3 as f′(x)=0f'(x)=0f′(x)=0 at only these points.

These two points divide the number line into three intervals: (−∞,−3),(−3,3),(3,∞)(-\infty,-3),(-3,3),(3,\infty)(−∞,−3),(−3,3),(3,∞).

Let’s evaluate f′f'f′ at each interval to see if it’s positive or negative on that interval and therefore if fff is increasing or decreasing on the interval.

Intervalxxxf′(x)f'(x)f′(x)Verdict
(−∞,−3)(-\infty,-3)(−∞,−3)x=−4x=-4x=−4f(−4)=21>0f(-4)=21>0f(−4)=21>0fff is increasing
(−3,3)(-3,3)(−3,3)x=0x=0x=0f(0)=−27<0f(0)=-27<0f(0)=−27<0fff is decreasing
(3,∞)(3,\infty)(3,∞)x=4x=4x=4f(4)=21>0f(4)=21>0f(4)=21>0fff is increasing

In conclusion, fff is decreasing on the interval (−3,3)(-3,3)(−3,3).

#Function Behavior: Question 2

The answer to this problem is (−1,0)(-1,0)(−1,0) and (1,∞)(1,\infty)(1,∞). Here’s why:

We can analyze the intervals where fff is increasing by looking for the intervals where its derivative fff is positive.

A function can only change its direction from increasing to decreasing and vice versa at its critical points and the points where the function itself is undefined.

Since the derivative of fff is f′(x)=4x3−4xf'(x)=4x^3-4xf′(x)=4x3−4x, which we can rewrite as 4x(x+1)(x−1)4x(x+1)(x-1)4x(x+1)(x−1), the only points where fff can change direction are x=−1x=-1x=−1, x=0x=0x=0, and x=1x=1x=1 as f′(x)=0f'(x)=0f′(x)=0 at only these points.

These three points divide the number line into four intervals: (−∞,−1),(−1,0),(0,1),(1,∞)(-\infty,-1),(-1,0),(0,1),(1,\infty)(−∞,−1),(−1,0),(0,1),(1,∞).

Let’s evaluate f′f'f′ at each interval to see if it’s positive or negative on that interval and therefore if fff is increasing or decreasing on the interval.

Intervalxxxf′(x)f'(x)f′(x)Verdict
(−∞,−1)(-\infty,-1)(−∞,−1)x=−2x=-2x=−2f(−2)=−24<0f(-2)=-24<0f(−2)=−24<0fff is decreasing
(−1,0)(-1,0)(−1,0)x=−0.5x=-0.5x=−0.5f(−0.5)=1.5>0f(-0.5)=1.5>0f(−0.5)=1.5>0fff is increasing
(0,1)(0,1)(0,1)x=0.5x=0.5x=0.5f(0.5)=−1.5<0f(0.5)=-1.5<0f(0.5)=−1.5<0fff is decreasing
(1,∞)(1,\infty)(1,∞)x=2x=2x=2f(2)=24>0f(2)=24>0f(2)=24>0fff is increasing

In conclusion, fff is decreasing on the intervals (−1,0)(-1,0)(−1,0) and (1,∞)(1,\infty)(1,∞).


#⭐ Closing

Great work! You now have the steps and practice to determine where a function changes from increasing to decreasing.

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Question 1 of 9

If the derivative of a function f(x)f(x)f(x) is positive at a point, what does this indicate about the behavior of the function at that point? 🚀

The function is decreasing

The function is increasing

The function has a critical point

The function is constant