# Applications of Differentiation – Critical Numbers, Local Extrema, and the First Derivative Test

In this lesson, we will show you how to use the First Derivative Test to find critical numbers, local extrema, and intervals of increasing and decreasing of a function.

Before we explore the first derivative test, we need to know the definitions of critical numbers, local extrema, and intervals of increasing and decreasing.

Critical Numbers and Critical Points

Critical numbers are the x values where the derivative is equal to zero or undefined. The critical points are the points (which includes the y coordinate in addition to the x coordinate) where the graph is zero or undefined. Graphically, there are 3 cases for identifying critical numbers and critical points:

1. The places on the graph where the graph ‘switches’ from increasing to decreasing (and vice versa). These can be seen as ‘hill tops’ and ‘valleys’.
2. The places on the graph where there are abrupt changes in direction can be seen as sharp edges. At these points, the graph is not differentiable.
3. The places on the graph where the tangent line is vertical (therefore the slope is undefined).

Let’s examine each of the three cases:

Case 1: Examine the graph of F(x) = 5x5+3x2 Notice that the points in yellow are the points where the graph changes from increasing to decreasing and vice versa. Looking at our first critical number, x = -1.063, we notice that the graph switches from increasing to decreasing. Looking at our second critical number, x = 0, we see that the graph switches from decreasing to increasing.

Notice too that the tangent line at the highlighted points are completely horizontal, meaning the derivative is zero.

Our critical points are therefore (-1.063, 2.003), the ‘hill top’, and (0,0), the ‘valley’.

Case 2: Consider the Function F(x) = |x| Notice that the point highlighted in yellow is the place where there is an abrupt change in direction. On the graph, this is seen as a sharp edge. Because the function cannot be differentiated at that point, F’(0) = undefined. Therefore, x = 0 is a critical number, and our critical point is (0,0).

Case 3: Examine the circle 25 = x2+y2 If we look at point (-5,0) and (5,0), taking the slope of the tangent line at that point will give us a vertical line. A vertical line does not have a slope. Therefore, the slopes at the highlighted points are undefined.

On a related note, if we look at x = 0, the points (0,5) and (0,-5) are also critical points. If we take the slope of the tangent line at those points, the slope of the tangent line is completely horizontal, or equal to zero.

The critical numbers are therefore x = 5, x = -5 and x = 0, and our critical points are (0,5), (0,-5), (5,0), and (-5, 0)

Local Extrema

Local extrema are the local maxima and minima of a function. In other words, they are the largest and smallest value of the function over a specified interval.

The local extrema can be located at the critical points, where the derivative is zero or undefined. Graphically, local maxima can be seen as the ‘hill tops’ and the local minima can be seen as the ‘valleys’ of the function. For certain functions, the critical points are not always ‘hill tops’ or ‘valleys’ but rather are points where the function levels off. These points are called saddle points, or ‘plateaus’. Think of a saddle on a horse – in particular, the flat region on the back of a horse. Similarly, the saddle point of a function is the flat region of the graph. Consider the following figure: As you can see, the local minima (valleys), local maxima (hill tops), and saddle points (plateaus) are located at the critical points, where the derivative is zero.

Intervals of Increase and Decrease

As mentioned above, the critical points are the points where the function switches from increasing to decreasing and vice versa.

Consider the figure below: The orange point is the critical point. Notice that to the left of the critical point, the slope of the tangent line, or the derivative, is positive. Coincidentally, the graph is increasing when the derivative is positive. To the right of the critical point, the slope of the tangent line, or the derivative, is negative. Coincidentally, the graph is decreasing when the derivative is negative.

The First Derivative Test

Now that we have covered critical numbers, critical points, local extrema, and intervals of increase and decrease, now we can discuss how the first derivative test connects all these points together.

 Definition: The First Derivative Test If F’(x) changes from positive to negative at a critical number xc, then F(x) has a local maximum at xc.  ( xc, F(xc) ) is a local maximum If F’(x) changes from negative to positive at a critical number xc, then F(x) has a local minimum at xc.  ( xc, F(xc) ) is a local minimum. F’(x) does not change sign at xc (on both sides of xc, F’(x) is positive or on both sides of xc, F’(x) is negative), then F(x) is neither a local maximum or minimum at xc. ( xc, F(xc) ) is a saddle point. Guidelines for Performing the First Derivative Test Take the first derivative of the function. Set the first derivative equal to zero and solve for the variable. These will be the critical numbers. If F’(x) is a rational function , you also need to set the denominator Q(x) equal to zero to find the values that make F’(x) undefined. Plot the critical numbers on a number line. Find values from the number line to plug into F’(x) to determine intervals of increase and decrease. Plot the critical numbers into the original functions of F(x) to obtain the y coordinates of the critical points.

Example 1:

For the function below, find the critical numbers, intervals of increasing and decreasing, and the points that are either local minima, local maxima, or saddle points.

F(x) = 3x2 + 2x -1

Solution:

First, take the first derivative of the function:

Using the power rule, = 6x+2

Second, set the derivative equal to zero and solve for the variable:

0 = 6x+2

6x = -2

x = ( -1/3)  is the only critical number of the function.

Third, plot the critical number on the number line: Fourth, find values on the number line to plug into F’(x) to determine the intervals of increasing and decreasing:

Let’s find a value between – and -⅓ to plug into F’(x), such as -1:

F’(-1) = 6(-1)+2

F’(-1) = -6 + 2

F’(-1) = -4

Since the first derivative is negative, the function is decreasing on the interval of (- ∞, – ⅓).

Let’s find a value between -⅓ and ∞  to plug into F’(x), such as 1:

F’(1) = 6(1)+2

F’(1) = 6 + 2

F’(1) = 8

Since the first derivative is positive, the function is increasing on the interval of (- ⅓, ∞). From the number line, it is clear that at the critical number xc= -⅓, F(x) has a local minimum because the graph is decreasing on the left of xc and then increasing on the right of xc.

Fifth, plug in the critical numbers into the original function of F(x) to obtain the y coordinates of the critical points.

Since xc= -⅓, substitute -⅓ into the function:

F(x) = 3x2 + 2x -1

F(-⅓) = 3 (-1/3) 2 + 2 (-1/3) -1

F(-⅓) = 1/3 – 2/3 -1

F(-⅓) =  -4/3

Our critical point is (-⅓, -2) and is the local minimum.

With the first derivative test, we have discovered the following:

• The critical number of the function is xc= -⅓
• The critical point is (-⅓, -4/3) and is the local minimum.
• The interval of increase is (- ⅓, ∞)
• The interval of decrease is (- ∞, – ⅓)

Let’s graph the function F(x) to verify our findings: Notice that the derivative is negative on the interval (- ∞, – ⅓), and therefore, the function is decreasing on said interval. Notice that the derivative is positive on the interval of (- ⅓, ∞), and therefore, the function is increasing on said interval. Lastly, at the critical number, x = -⅓, the derivative is zero, and the point is a local minimum.

Example 2:

Let we have a function F(x) = 2x2 + 5x, find the following:

• Critical numbers
• Critical Points
• Intervals of Increasing and Decreasing
• Determine which critical points are local maximum, minimum, or saddle points.

Solution:

First, take the first derivative of the function:

Taking the derivative of the given function F(x):

d/dx F(x) =  d/dx (2x2+ 5x)

F’(x) = 4x + 5

Second, set the derivative equal to zero and solve for the variable:

Now put this derivative equal to zero,

0 = 4x + 5

Solving for x,

x = -5/4 is our only critical number.

Third, plot the critical number on the number line: Fourth, find values on the number line to plug into F’(x) to determine the intervals of increasing and decreasing:

Let’s find a value between -∞ and -5/4 to plug into F’(x), such as -2:

F’(x) = 4x + 5

F’(-2) = 4(-2)+5

F’(-2) = 4(-2) + 5

F’(-2) = -8 + 5

F’(-2) = -3

Since the first derivative is negative, the function is decreasing on the interval of (- ∞, – 5/4).

Let’s find a value between -5/4 and ∞  to plug into F’(x), such as 2:

F’(x) = 4x + 5

F’(2) = 4(2)+5

F’(2) = 4(2) + 5

F’(2) = 8 + 5

F’(2) = 13

Since the first derivative is positive, the function is increasing on the interval of (-5/4, ∞). From the number line, it is clear that at the critical number xc= -5/4, F(x) has a local minimum because the graph is decreasing on the left of xc and then increasing on the right of xc.

Fifth, plug in the critical numbers into the original function of F(x) to obtain the y coordinates of the critical points.

Since xc= -5/4, substitute -5/4 into the function:

F(x) = 2x2 + 5x,

F(-5/4) =2(-5/4)2 + 5(-5/4)

F(-5/4) =2(-5/4)2 + 5(-5/4)

F(-5/4) = -3.125

Our critical point is (-5/4, -3.125) and is the local minimum.

With the first derivative test, we have discovered the following:

• The critical number of the function is xc= -5/4
• The critical point is (-5/4, -3.125) and is the local minimum.
• The interval of increase is (- 5/4, ∞)
• The interval of decrease is (- ∞, – 5/4)

Let’s graph the function F(x) to verify our findings: Notice that the derivative is negative on the interval (- ∞, -5/4), and therefore, the function is decreasing on said interval. Notice that the derivative is positive on the interval of (- 5/4, ∞), and therefore, the function is increasing on the said interval. Lastly, at the critical number, x = -5/4, the derivative is zero, and the point is a local minimum.

Example 3:

Suppose we have given a function F(x) = 2x3 + x2  + 1. Find the following:

• Critical numbers
• Critical Points
• Intervals of Increase and Decrease
• Determine which critical points are local maximum, minimum, or saddle points.

Solution:

First, take the first derivative of the function.

Using the power rule,

d/dx F(x) =   d/dx(2x3 + x2 + 1)

F’(x) = 6x2 + 2x

Second, set the derivative equal to zero and solve for x to find the critical numbers:

0 = 6x2 + 2x

Since each term has a greatest common factor of 2x, factor a 2x from each term on the right side of the equation:

0 = (2x )(3x+1)

Set each factor equal to zero:

2x = 0

3x+1 = 0

Solving for each factor,

x = 0

x = -⅓

Our critical numbers are x = 0 and x = -0.33.

Third, plot the critical numbers on the number line: Fourth, find values on the number line to plug into F’(x) to determine the intervals of increasing and decreasing:

Let’s find a value between -∞ and -1/3 to plug into F’(x), such as -1:

F’(x) = 6x2 + 2x

F’(-1) = 6(-1)2 + 2(-1)

F’(-1) =  4

Since the first derivative is positive, the function is increasing on the interval of (- ∞, – ⅓).

Let’s find a value between -1/3 and 0 to plug into F’(x), such as -1/4:

F’(x) = 6x2 + 2x

F’(-¼ ) = 6(-1/4)2 + 2(-1/4)

F’(-1/4) =  -0.125

Since the first derivative is negative, the function is decreasing on the interval of (- ⅓ , 0).

Let’s find a value between 0 and   to plug into F’(x), such as 1:

F’(x) = 6x2 + 2x

F’(1) = 6(1)2 + 2(1)

F’(1) =  8

Since the first derivative is positive, the function is increasing on the interval of (0,  ∞). Looking at the number line, we see that at xc= -1/3, F(x) has a local maximum. At xc= 0 , F(x) has a local minimum.

Fifth, plug in the critical numbers into the original function of F(x) to obtain the y coordinates of the critical points.

Since xc= -1/3 and xc= 0, substitute them into the function:

F(x) = 2x3 + x2  + 1

F (0) = 0 + 0 +1

F(0) = 1

F (-1/3) = 2(-⅓)3 + (-⅓)2  + 1

F (-1/3 ) = 2(-⅓)3 + (-⅓)2  + 1

F(-⅓) = 1.037

Our two critical points are (0,1), which is the local minimum, and (-⅓, 1.037), which is the local maximum.

With the first derivative test, we have discovered the following:

• The critical numbers of the function are xc= -1/3 and xc= 0
• Our two critical points are (0,1), which is the local minimum, and (-⅓, 1.037), which is the local maximum.
• The intervals of increase is (- ∞, – ⅓) and (0, ∞)
• The interval of decrease is (- ⅓ , 0)

Let’s graph the function F(x) to verify our findings: Notice that the derivative is positive on the intervals (- ∞, -1/3) and (0, ∞), and therefore, the function is increasing in those intervals. Notice that the derivative is negative on the interval of (- ⅓, 0), and therefore, the function is decreasing on said interval. Lastly, at the critical numbers, x = -⅓ and x = 0, the derivative is zero, and our two critical points are (0,1), which is the local minimum, and (-⅓, 1.037), which is the local maximum.

Example 4:

Suppose we have given a function Find the following:

• Critical numbers
• Critical Points
• Intervals of Increasing and Decreasing
• Determine which critical points are local maximum, minimum, or saddle points.

Solution:

First, take the first derivative of the function.

Since F(x) is a rational function, we will use the quotient rule.

f(x) = -2

f'(x) = 0

g(x) = x2-4

g’(x) = 2x

Using the quotient rule of differentiation, Second, set the derivative equal to zero and solve for x to find the critical numbers:

Remember, since F’(x) is a rational function, we have to find the x values that make F’(x) equal to zero and undefined (rational functions are undefined when the denominator is equal to zero).

First, let’s find the x values that make F’(x) equal to zero: 0 = 4x

0 = x

Second, let’s find the x values that makes F’(x) undefined, which means that we have to set the denominator equal to zero:

0 = (x2-4)2

01/2 = [(x2-4)2]1/2

0 = x2-4

4 = x2

x = +/- 2

Therefore, our critical numbers (from increasing order) are -2, 0, and 2.

Third, plot the critical numbers on the number line: Fourth, find values on the number line to plug into F’(x) to determine the intervals of increasing and decreasing:

Let’s find a number in between -and -2, such as -3: Since F’(3) is positive, the function F(x) is increasing on the interval of (2, ∞).

Now we know the intervals of increasing and decreasing, we can create the sign chart: Fifth, plug in the critical numbers into the original function of F(x) to obtain the y coordinates of the critical points.

Our critical points were x = -2, x = 0, and x = 2. At xc = -2 and xc = 2, F(x) is undefined. Therefore, there are vertical asymptotes at x = -2 and x =2. The only critical point is (0, ½), which is a local minimum.

With the first derivative test, we have discovered the following:

• The critical numbers of the function are xc= -2, xc= 0, and xc= 2
• The critical point is (0, -½ ) and is the local minimum.
• The interval of increase is (0,2) and (2, ∞).
• The interval of decrease is  (- ∞, -2) and (-2, 0).

Below is a graph to confirm our results: Summary of Section Critical Numbers: x values where the derivative of a function is equal to zero or in undefined (x | F’(x) = 0 or undefined) Critical Points (Local Extrema): Points on the graph where the function is equal to zero or undefined. These are the locations of local extrema. Substitute the critical numbers back into F(x) to obtain the y coordinates and subsequently the critical points. Intervals of Increasing and Decreasing: The graph is increasing when F’’(x) is positive. The graph is decreasing when F’’(x) is negative. First Derivative Test: Using the first derivative of a function to find critical numbers, critical points (points of local extrema), and intervals of increasing and decreasing. Find F’(x). Set F’(x) = 0 and solve to find critical numbers. Create a sign chart to find intervals of increasing and decreasing. Substitute critical numbers into F(x) to obtain the critical point. Check your results by graphing.