In this set of supplemental notes, I am going to walk you through the steps you will need to perform when investigating the characteristics of graphs. I will be working three examples simultaneously. I will first find the critical points for local maximums and minimums and for points of inflection. Then we will determine where the functions are increasing and decreasing. From that we will find the maximums and minimums. Next we will find out where the function is concave up or concave down and if there are any points of inflection. After these three examples, I will do several more examples completely.

The following will critical points of the function:

1.

Any value of x that makes f ' (x) = 0.

2.

Any value of x in which f ' (x) does not exist.

3.

If you are working on a closed interval, then the endpoints of the interval are also critical points.

4.

Any point that is either a vertical asymptote or a removable point of discontinuity.

EXAMPLE 1A:

Find the critical points for the function f (x) = x ³ - 2x ² + 3 on the interval [-5, 5].

SOLUTION:

f ' (x) = 3x ² - 4x = 0 ® x(3x - 4) = 0 ® x = 0 or x = 4/3

Since we are looking at this function over a closed interval, we must include the endpoints of the interval in the list of critical points. Therefore, the critical points for this function are x = -5, x = 0, x = 4/3, and x = 5.

EXAMPLE 2A:

Find the critical points for the following function.

SOLUTION:

First of all, let us find the domain for this function. Remember that the radicand must be greater than or equal to zero, so let us solve the following inequality.

2x - 6 ³ 0 ® 2x ³ 6 ® x ³ 3

Therefore, the domain for this function is the interval [3, ¥ ). Now, let us find the first derivative.

Notice that f ' (x) will never equal zero. Now let us look at the denominator. You will notice that x ¹ 3, so this will be the only critical point.

EXAMPLE 3A:

Find the critical points for the following function.

SOLUTION: First, let us see if we can simplify the function.

Notice that this function has a vertical asymptote at x = 2 and a removable point of discontinuity at x = -2. Those two points will automatically be considered as critical points. Now, let us find f ' (x).

Notice that f ' (x) will never equal zero, so let us look at the denominator. We should see that x ¹ 2, and that is what we concluded from above.

The following will also be critical points:

1.

Any value of x that makes f '' = 0.

2.

Any value of x in which f '' does not exist.

EXAMPLE 1B:

Find the second derivative and the critical points for the function f (x) = x ³ - 2x ² + 3 on the interval [-5, 5].

SOLUTION:

This is the only critical point for this function.

EXAMPLE 2B:

Find the second derivative and the critical points for the following function.

SOLUTION:

Notice that f '' will never equal zero, and x ¹ 3, so this will be the only critical point.

EXAMPLE 3B:

Find the second derivative and the critical points for the following function.

SOLUTION:

Notice that f '' will never equal zero, and x ¹ 2. Since x = -2 is a removable point of discontinuity, we will also include it in the list of critical points. So the critical points for this function is x = 2 and x = -2.

This completes the section on finding critical points.

Suppose that x = a, x = b, x = c, and x = d are the critical points for a function f (x). We would like to use these points to determine where the function is increasing or decreasing. To do this, we must set up the following intervals: (-¥ , a), (a, b), (b, c), (c, d), and (d, ¥ ), then we will evaluate f ' for a value contained in each interval.

Use the following criteria to determine if the function is increasing or decreasing on the given interval:

1.

If f ' > 0 on the interval, then the function is increasing on that interval.

2.

If f ' < 0 on the interval, then the function is decreasing on that interval.

A fringe benefit of determining if a function is increasing or decreasing on an interval is the ability to determine if the critical point is a local max or min. The following is what you have to look for to determine if the critical point is a max or a min.

1.

If the function goes from increasing to decreasing, then the critical point is a local max.

2.

If the function goes from decreasing to increasing, then the critical point is a local min.

EXAMPLE 1C:

Determine where the function f (x) = x ³ - 2x ² + 3 on the interval [-5, 5] is increasing and decreasing, and determine all local and absolute extrema.

SOLUTION:  Recall that the critical points of this function are x = 0, x = 4/3, x = -5, and x = 5. The intervals that we will be analyzing are (-5, 0), (0, 4/3), and (4/3, 5).

(-5, 0)

f '(-1) = 3 + 4 > 0

Increasing

(0, 4/3)

f '(1) = 3 - 4 < 0

Decreasing

(4/3, 5)

f '(2) = 12 - 8 > 0

Increasing

Since we are analyzing this function over a closed interval, we will not only have local extrema, we will have absolute extrema. To determine the local and absolute extrema, plug in the critical values back into the original function.

f (-5) = -125 - 50 + 3 = -172

f (0) = 3

f (4/3) = (64/27) - (32/9) + 3 = 49/27

f (5) = 125 - 50 + 3 = 78

Local max at (0, 3)

Absolute max at (5, 78)

Local min at (4/3, 49/27)

Absolute min at (-5, -172)

EXAMPLE 2C:

Determine where the following function is increasing and decreasing, and determine all local and absolute extrema.

SOLUTION: Recall that x ¹ 3 is the only critical point, therefore the intervals will be (-¥ , 3) and (3, ¥ ). (-¥ , 3) is not a part of the domain of this function, so disregard this interval.

(3, ¥ )

f ' (4) > 0

Increasing

This function is always increasing. f (3) = 0 is the absolute minimum for this function.

EXAMPLE 3C:

Determine where the following function is increasing and decreasing, and determine all local extrema.

SOLUTION: Recall that x ¹ 2 and x ¹ -2 and that these points are the only two critical points. Therefore, the intervals that we will be using are (-¥ , -2), (-2, 2), and (2, ¥ ).

(-¥ , -2)

f ' (-3) < 0

Decreasing

(-2, 2)

f ' (0) < 0

Decreasing

(2, ¥ )

f ' (3) < 0

Decreasing

This function is always decreasing. Therefore, it has no local maximum or minimum.

Suppose that x = a and x = b are critical points for f ''. Using these points, construct the following intervals (-¥ , a), (a, b), and (b, ¥ ). Picking a point from each interval, plug that value into f '' and evaluate the result using the following as a guide:

1.

If f '' > 0 on the interval, then the function is concave up (ccu).

2.

If f '' < 0 on the interval, then the function is concave down (ccd).

After determining the intervals where the function is concave up or concave down, we can determine the points of inflection. Any critical point (except where the function is not defined) in which the function changes concavity is a point of inflection.

EXAMPLE 1D:

Determine where the function f (x) = x ³ - 2x ² + 3 on the interval [-5, 5] is concave up and concave down, and determine all points of inflection.

SOLUTION: Recall that this function is defined over a closed interval, so the intervals that we will be considering are (-5, 2/3) and (2/3, 5).

(-5, 2/3)

f '' (0) < 0

ccd

(2/3, 5)

f '' (1) > 0

ccu

There is a point of inflection at the point x = 2/3. Does the graph of this function fit the data that we have gathered? Yes, it does.

EXAMPLE 2D:

Determine where the following function is concave up and concave down, and determine all points of inflection.

SOLUTION: Again, x = 3 is the only critical point for the second derivative. Therefore, there is only one interval we need to look at, and it is (3, ¥ ).

(3, ¥ )

f '' (4) < 0

ccd

This function does not have a point of inflection. Does the graph of this function fit the data that we have gathered? Yes, it does.

EXAMPLE 3D:

Determine where the following function is concave up and concave down, and determine all points of inflection.

SOLUTION: Recall that the two critical points for the second derivative are x ¹ -2 and x ¹ 2. Therefore, the intervals that we will be analyzing are (-¥ , -2), (-2, 2), and (2, ¥ ).

(-¥ , -2)

f ''(-3) < 0

ccd

(-2, 2)

f ''(0) < 0

ccd

(2, ¥ )

f ''(3) > 0

ccu

Since this function does not exist at x = 2, then that critical point can not be a point of inflection. Does the data that we have gathered fit the graph of the function? Yes it does.

Now here are a few more worked examples.

EXAMPLE 4:

For f (x) = 2x ³ - 18x, determine where it is increasing and decreasing, all extrema, where it is concave up and concave down, and points of inflection.

SOLUTION: First, let us find f ' (x) and the critical points.

Now to determine where the function is increasing and decreasing.

Now, find f '' (x) and the related critical points.

f '' = 12x = 0 ® x = 0

Now to determine where the function is concave up and concave down.

(-¥ , 0)

f '' (-1) = 12(-1) < 0

ccd

(0, ¥ )

f '' (1) = 12(1) > 0

ccu

Therefore, the point of inflection for this function is when x = 0 and f (0) = 0.

Here is the graph of this function. Does it fit the data that we have found? Yes, it does.

EXAMPLE 5:

For f (x) = x ⁴ - 4x ³ + 4x ², determine where it is increasing and decreasing, all extrema, where it is concave up and concave down, and points of inflection.

SOLUTION: First, find f ' (x) and the critical points.

The critical points are x = 0, x = 1, and x = 2. Now to determine where the function is increasing and decreasing.

(-¥ , 0)

f ' (-1) = 4(-1) ³ - 12(-1) ² + 8(-1) < 0

Decreasing

(0, 1)

f ' (0.5) = 4(0.5) ³ - 12(0.5) ² + 8(0.5) > 0

Increasing

(1, 2)

f ' (1.5) = 4(1.5) ³ - 12(1.5) ² + 8(1.5) < 0

Decreasing

(2, ¥ )

f ' (3) = 4(3) ³ - 12(3) ² + 8(3) > 0

Increasing

Local minimum

x = 0

f (0) = 0

Local maximum

x = 1

f (1) = 1 - 4 + 4 = 1

Local minimum

x = 2

f (2) = 2 ⁴ - 4(2) ³ + 4(2) ² = 0

Now find f '' (x) and the critical points.

Now to determine where the function is concave up and concave down.

f '' (-1) = 12 + 24 + 8 >0

ccu

f '' (1) = 12 - 24 + 8 < 0

ccd

f '' (2) = 48 - 48 + 8 > 0

ccu

Here is the graph of this function. Determine if the information that we have found fits the graph.

EXAMPLE 6:

For the following function, determine where it is increasing and decreasing, all extrema, where it is concave up and concave down, and points of inflection.

SOLUTION: Since the denominator will never equal zero, there will not be any critical points based on discontinuities of the function. So, let us find the first derivative and the critical points.

x = 0 is the only critical point. Now let us determine where the function is increasing and decreasing.

(-¥ , 0)

f '(-1) > 0

Increasing

(0, ¥ )

f '(1) > 0

Increasing

Therefore, this function has no local extreme values.

Now to find the second derivative and its critical points.

The critical points for the second derivative are x = 0, x = -1, and x = 1. Now to determine the concavity of this function.

(-¥ , -1)

f '' (-2) > 0

ccu

(-1, 0)

f '' (-0.5) < 0

ccd

(0, 1)

f '' (0.5) > 0

ccu

(1, ¥ )

f '' (2) < 0

ccd

Compare the information that we have collected with the graph of the function. Does the graph fit the data?

EXAMPLE 7:

For f (x) = x + sin x, determine where it is increasing and decreasing, all extrema, where it is concave up and concave down, and points of inflection on the interval [0, 2p ].

SOLUTION: Let us find the first derivative and the critical points. Since this function is defined over a closed interval, x = 0 and x = 2p are considered to be critical points.

f ' (x) = 1 + cos x = 0 ® cos x = -1 ® x = p

Therefore, the critical points are x = 0, x = p , and x = 2p .

Now to determine where the function is increasing and decreasing.

[0, p )

f ' (p /2) > 0

Increasing

(p , 2p ]

f ' (3p /2) > 0

Increasing

f (0) = 0

Absolute minimum

f (p ) = p

f (2p ) = 2p

Absolute maximum

Now to find the second derivative and its related critical points.

f ''(x) = -sin x = 0 ® x = p

[0, p )

f ''(p /2) < 0

ccd

(p , 2p ]

f ''(3p /2) > 0

ccu

Therefore, this function has a point of inflection at (p ,p ).

Look at the graph of this function, and determine if the data that we have found fits the graph.