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17Calculus Derivatives - Increasing and Decreasing Intervals

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Increasing and Decreasing Intervals

On the page discussing the first derivative, we gave you an overview of increasing and decreasing intervals. Here we discuss how to use critical points to find intervals that are increasing and decreasing.

Topics You Need To Understand For This Page

derivatives graphing 1st derivative critical points

On the last page, you learned how to calculate critical points. One of the interesting properties of critical points is that, if the graph of a continuous function is going to change direction, we are guaranteed that the function will change direction at a critical point and no where else. So this gives us a fool-proof way of determining increasing and descreasing intervals. After determining the critical points, we use them to build a table to determine where the increasing and decreasing intervals exist. To help us organize our information, we can build a table like this.

Table Format For Increasing and Decreasing Intervals

Interval

\( -\infty < x < c_1 \)

\( c_1 < x < c_2 \)

\( c_2 < x < c_3 \)

\( c_3 < x < \infty \)

Test \(x\)-value

Sign of \(g'(x)\)

Conclusion

The x-values \(c_1, c_2 . . . \) are the critical numbers and discontinuities that we found. Your function may have fewer or more critical points than we have listed here. So you need to set up your table based on what you found. The point of this table is to demonstrate how the table is built. The first row of intervals should cover the entire domain and only the domain of the function. The breaks should be only at critical points and discontinuities. In this table, the domain is \((-\infty, +\infty)\). Your domain may be different.

Since we are guaranteed that a continuous function will not change direction at any other point than a critical point, we can choose any point in an interval as a test value. So, for example, if our interval is \(-1 < x < 2\), we can choose \(0\), \(1\), \(-0.5\), \(0.25\) or any other point in the interval. Be careful to notice that the interval is an open interval, so you cannot choose one of the endpoints.

The possible conclusions in the last row are increasing or decreasing based on the sign of the slope. This tells us what is going on in each interval.

Okay, now it's time for some practice problems. After that, the next step is the first derivative test, where you learn how to use critical numbers and the information about increasing and decreasing intervals to find maximums and minimums of a function.

Before we go on, let's work some practice problems.

You Can Have an Amazing Memory: Learn Life-Changing Techniques and Tips from the Memory Maestro

Practice

Unless otherwise instructed, determine the increasing and decreasing intervals for these functions.

\(f(x)=3x+2\)

Problem Statement

Determine the increasing and decreasing intervals for \(f(x)=3x+2\).

Solution

Krista King Math - 1336 video solution

video by Krista King Math

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\(f(x)=8-2x^2\)

Problem Statement

Determine the increasing and decreasing intervals for \(f(x)=8-2x^2\).

Solution

Krista King Math - 1337 video solution

video by Krista King Math

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\(f(x)=6x-2x^2\)

Problem Statement

Determine the increasing and decreasing intervals for \(f(x)=6x-2x^2\).

Solution

Krista King Math - 1338 video solution

video by Krista King Math

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\(f(x)=x^4-2x^2+1\)

Problem Statement

Determine the increasing and decreasing intervals for \(f(x)=x^4-2x^2+1\).

Solution

Krista King Math - 1339 video solution

video by Krista King Math

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\(f(x)=3x^4+4x^3-12x^2\)

Problem Statement

Determine the increasing and decreasing intervals for \(f(x)=3x^4+4x^3-12x^2\).

Solution

Krista King Math - 1340 video solution

video by Krista King Math

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\(\displaystyle{f(x)=xe^{-x/2}}\)

Problem Statement

\(\displaystyle{f(x)=xe^{-x/2}}\)

Solution

Although her final answer is correct in this video, she makes a mistake in the calculations. She sets up the equation \(e^{-x/2} = 0\) and tries to solve it, eventually saying that \(x=0\). However, if you plug in \(x=0\), you will find that \(e^{-0/2} = e^0 = 1\). Her mistake is that there is no value for x that will solve \(e^{-x/2} = 0\). And so, \(x=0\) does not solve \(e^{-x/2} = 0\) and should never show up in her list of possible critical numbers. Other than that, this is a good problem and shows some other good techniques.

Krista King Math - 1341 video solution

video by Krista King Math

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Practice Instructions

Unless otherwise instructed, determine the increasing and decreasing intervals for these functions.

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When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

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