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This page covers the two main topics we use the first derivative for to understand what is going on with the graph of a function, increasing and decreasing intervals and maximums and minimums. First, we discuss what we mean by increasing and decreasing functions and intervals, then introduce critical numbers and finally explain the first derivative test for determining maximums and minimums.

Increasing and Decreasing Intervals

In this section, we discuss what is meant by increasing and decreasing by looking at various functions. You need to know this when using critical numbers to determine increasing and decreasing intervals.

plot 1 - an increasing function

A function that is increasing is one that has a positive slope. Plot 1 is an example of an increasing function since it's slope is \(m=2\), which is positive, i.e. \( m > 0\). To keep this straight in my head, I think of myself as walking on the curve (in this case a straight line) from left to right (increasing x). If I am going up, the slope is positive and the function is increasing.

plot 2 - a decreasing function

A decreasing function is one that has a negative slope. In plot 2, the slope is \(m=-2\) and, since the slope is negative, \( m < 0\), this is a decreasing function. Using the same idea as above, I keep this straight by thinking about walking on the curve. If I am going downhill, the slope is negative and, therefore, is a decreasing function.

Some functions are neither increasing nor decreasing. In this case, we say the function is constant, the graph will have a slope of zero and be a horizontal line.

plot 3

Okay, so now that we have the idea of the behavior of entire functions, we are going to work with functions that are increasing in part of the graph and decreasing in others. The idea is break up the function into intervals (sections of the graph based on the x-value) so that only one thing is going on at a time. Let's look at the graph in plot 3.

Notice that there are two things going on. First, the graph is increasing from \(x=-2\) to \(x=-1\). Second, the graph is decreasing from \(x=-1\) to \(x=3\).
We would describe this graph as:
- increasing on the interval \( -2 < x < -1 \), and
- decreasing on the interval \( -1 < x < 3 \)

Notice that we specified open intervals, i.e. we did not include the endpoints. This is usually the way it is done. Also, we do not say anything for \( x < -2 \) and \( x > 3\) because those x-values are not in the domain. This is shown by the obviously filled circles at the points \((-2,0)\) and \((3,-1)\).

Okay, so now that you have a feel for what increasing and decreasing functions look like, let's discuss critical numbers. We use critical numbers to know where a function changes from increasing to decreasing or vis-versa.

Critical Points

In this section, we discuss critical numbers and critical points, what they are and how to find them. Then we discuss how to use critical numbers to determine increasing and decreasing intervals.

Critical numbers are x-values of functions where something special happens. These x-values must be in the domain. Two possible things can occur at critical number, \(x=c\).
1. If the derivative exists at \(x=c\), then the derivative at \(x=c\) is zero.
2. The derivative does not exist at \(x=c\).
If one of those two things happen, then \(x=c\) is called a critical value or critical number.
Note: If the term 'critical point' is used, then that refers to the point \((c,f(c))\), not just the number \(x=c\).

To find the critical numbers, we take the derivative, set it equal to zero and solve of the x-values. If those x-values are in the domain of the function, then they are critical values. Finally, we look at the derivative and determine where the derivative doesn't exist. If those points are in the domain, then those points are also considered critical points.

Important thing to remember

The value \(x=c\) must be in the domain of the function. If it is not in the domain, it cannot be a critical number.

The interesting thing that happens at critical points is that the function levels out at these points giving us three possible situations.
1. maximum
2. minimum
3. saddlepoint

plot 4

Maximums and minimums, which may be either relative or absolute, are points where the graph is either increasing or decreasing, it levels off and changes direction.
Saddlepoints occur where a graph is either increasing or decreasing, it levels off at a point then the graph continues on in the same direction. The graph in plot 4 shows these types of points.

Okay, before we go on, let's work some practice problems. For these problems, determine the critical numbers.

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new.

First Derivative - Practice Problems Conversion

[A01-1333] - [A02-1334] - [A03-1336] - [A04-1337] - [A05-1338] - [A06-1339] - [A07-1340] - [A08-1341] - [A09-1342]

[A10-1343] - [A11-1345] - [A12-1346] - [A13-1347] - [A14-1917] - [B01-1335] - [B02-1344]

Please update your notes to this new numbering system. The display of this conversion information is temporary.

GOT IT. THANKS!

critical numbers
\( f(x)=x^{1/3}-x^{-2/3} \)

Problem Statement

critical numbers
\( f(x)=x^{1/3}-x^{-2/3} \)

Final Answer

\(x=0\) and \(x=-2\)

Problem Statement

critical numbers
\( f(x)=x^{1/3}-x^{-2/3} \)

Solution

1333 solution video

video by Krista King Math

Final Answer

\(x=0\) and \(x=-2\)

close solution

critical numbers
\(\displaystyle{f(x)=\frac{x^2}{x^2-9}}\)

Problem Statement

critical numbers
\(\displaystyle{f(x)=\frac{x^2}{x^2-9}}\)

Solution

1334 solution video

video by PatrickJMT

close solution

critical numbers
\(\displaystyle{g(x)=x\sqrt{16-x^2}}\)

Problem Statement

critical numbers
\(\displaystyle{g(x)=x\sqrt{16-x^2}}\)

Solution

1335 solution video

video by PatrickJMT

close solution

Increasing and Decreasing Intervals

Using this procedure to determine the critical numbers, we are guaranteed that the function will not change direction at any other point. 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 the 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. Unless otherwise instructed, determine the increasing and decreasing intervals.

increasing/decreasing
\(f(x)=3x+2\)

Problem Statement

increasing/decreasing
\(f(x)=3x+2\)

Solution

1336 solution video

video by Krista King Math

close solution

increasing/decreasing
\(f(x)=8-2x^2\)

Problem Statement

increasing/decreasing
\(f(x)=8-2x^2\)

Solution

1337 solution video

video by Krista King Math

close solution

increasing/decreasing
\(f(x)=6x-2x^2\)

Problem Statement

increasing/decreasing
\(f(x)=6x-2x^2\)

Solution

1338 solution video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

1339 solution video

video by Krista King Math

close solution

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

Problem Statement

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

Solution

1340 solution video

video by Krista King Math

close solution

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

Problem Statement

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

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.

1341 solution video

video by Krista King Math

close solution

First Derivative Test

We use the First Derivative Test to determine if a critical point is a maximum or minimum or neither. The idea is to break up the function into separate sections and then to analyze each section. The function is broken at discontinuities and critical points. So the first step is to make sure we know what the domain is and then to find the critical points. Once we have a set of points that are either discontinuities or critical points, we test values in between the breaks. We plug each test point into the first derivative. If the value is greater than zero, then the function is increasing. If the value is less than zero, then the function is decreasing.

To keep track of what we are doing, we can build a table like this one. Let's say we have a function \(g(x)\) and we have determined break points (critical points and discontinuities) at \(x=c_1\) and \(x=c_2\) and the function is defined for \(x < c_1\) and for \(x > c_2 \). Also, the entire domain needs to be covered by the intervals in the first row.

Table Format For The First Derivative Test

Interval/Point

\( -\infty \) \( < x < \) \(c_1\)

\( x = c_1 \)

\( c_1 < x < c_2 \)

\( x = c_2 \)

\( c_2 < x < \infty \)

Test x-value

Sign of \(g'(x)\)

Conclusion

So, how to do you know what test x-value to use? Basically, you can use any value inside the interval in the first row. Notice that the intervals are open, meaning that they do not include the endpoints. So the point you choose cannot be an endpoint but can be any other point in the interval.

The possible conclusions for the interval columns are increasing or decreasing. The possible conclusions for the point columns are maximum or minimum or neither, if the point is in the domain of \(g(x)\). If the point is not in the domain, we just leave it blank since it has no meaning.

A critical point is a minimum if the function goes from decreasing to increasing, left to right.
A critical point is a maximum if the function goes from increasing to decreasing, left to right.
If the function does not change direction, the critical point is neither a maximum nor a minimum, which we call a saddle point.

A maximum or minimum can be referred to as an extremum ( plural extrema ) if our discussion can apply to either one.
If the extremum is the largest (for a maximum) or the smallest (for a minimum) value of the function everywhere, it is called an absolute extremum. Otherwise, it is called a relative extremum. Relative extremum are sometimes called local extremum.

Before going on to practice problems, here is a good video to watch that explains all of this with an example.

PatrickJMT - Increasing/Decreasing , Local Maximums/Minimums [min-secs]

video by PatrickJMT

Okay, time for some practice problems and then we start looking at the second derivative and what it can tell us about the graph.

next: second derivative →

Instructions - Unless otherwise instructed, determine the maximuma and minimums (extrema) for these problems using the first derivative test.

maximums/minimums
\(f(x)=x^2\)

Problem Statement

maximums/minimums
\(f(x)=x^2\)

Solution

1342 solution video

video by PatrickJMT

close solution

maximums/minimums
\(f(x)=(x^2-1)^3\)

Problem Statement

maximums/minimums
\(f(x)=(x^2-1)^3\)

Solution

1343 solution video

video by PatrickJMT

close solution

maximums/minimums
\(f(x)=x^4-2x^2\)

Problem Statement

maximums/minimums
\(f(x)=x^4-2x^2\)

Solution

1345 solution video

video by Krista King Math

close solution

maximums/minimums
\(f(x)=x^3-4x^2\)

Problem Statement

maximums/minimums
\(f(x)=x^3-4x^2\)

Solution

1346 solution video

close solution

maximums/minimums
\(y=e^x(x-2)\)

Problem Statement

maximums/minimums
\(y=e^x(x-2)\)

Solution

1347 solution video

close solution

Find the extrema of the function \(f(x)=\sin^2x\) over the interval \([0,3]\).

Problem Statement

Find the extrema of the function \(f(x)=\sin^2x\) over the interval \([0,3]\).

Solution

1917 solution video

video by Krista King Math

close solution

absolute maximums/minimums
\(\displaystyle{f(x)=\frac{x^2-4}{x^2+4}}\) on \([-4,4]\)

Problem Statement

absolute maximums/minimums
\(\displaystyle{f(x)=\frac{x^2-4}{x^2+4}}\) on \([-4,4]\)

Solution

1344 solution video

video by PatrickJMT

close solution
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