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3.3 How Derivatives Affect the
Shape of a Graph
What Does f ′ Say About f ?
3
What Does f ′ Say About f ?
To see how the derivative of f can tell us where a function is increasing or decreasing, look at Figure 1.
Between A and B and between C and D, the tangent lines have positive slope and so f′(x) > 0.
Figure 1
Between B and C the tangent lines have negative slope
and so f′(x) < 0. Thus it appears that f increases when f ′(x) is positive and decreases when f′(x) is negative.
To prove that this is always the case, we use the Mean Value Theorem.
What Does f ′ Say About f ?
5
Example 1
Find where the function f(x) = 3x4 – 4x3 – 12x2 + 5 is increasing and where it is decreasing.
Solution:
We start by differentiating f:
f′(x) = 12x3 – 12x2 – 24x = 12x(x – 2)(x + 1)
To use the I/D Test we have to know where f′(x) > 0 and where f′(x) < 0.
To solve these inequalities we first find where f′(x) = 0, namely at, x = 0, 2, and –1.
Example 1 – Solution
These are the critical numbers of f, and they divide the domain into four intervals (see the number line below).
Within each interval, f′(x) must be always positive or always negative.
We can determine which is the case for each interval from
cont’d
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Example 1 – Solution
A plus sign indicates that the given expression is positive, and a minus sign indicates that it is negative. The last
column of the chart gives the conclusion based on the I/D Test.
For instance, f′(x) < 0 for 0 < x < 2, so f is decreasing on (0, 2). (It would also be true to say that f is decreasing on the closed interval [0, 2].)
cont’d
Example 1 – Solution
The graph of f shown in Figure 2 confirms the information in the chart.
Figure 2
cont’d
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Local Extreme Values
Local Extreme Values
You can see from Figure 2 that f(0) = 5 is a local maximum value of f because f increases on (–1, 0) and decreases on (0, 2). Or, in terms of derivatives, f′(x) > 0 for –1 < x < 0
and f′(x) < 0 for 0 < x < 2.
In other words, the sign of f′(x) changes from positive to negative at 0. This observation is the basis of the following test.
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Local Extreme Values
The First Derivative Test is a consequence of the I/D Test.
In part (a), for instance, since the sign of f′(x) changes from positive to negative at c, f is increasing to the left of c and decreasing to the right of c. It follows that f has a local
maximum at c.
It is easy to remember the First Derivative Test by visualizing diagrams such as those in Figure 3.
Local maximum Local minimum
Figure 3(a) Figure 3(b)
Local Extreme Values
No maximum or minimum No maximum or minimum
Figure 3(c) Figure 3(d)
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Example 3
Find the local maximum and minimum values of the function
g(x) = x + 2 sin x 0 ≤ x ≤ 2π Solution:
We start by finding the critical numbers. The derivative is:
g′(x) = 1 + 2 cos x
so g′(x) = 0 when . The solutions of this equation are 2π/3 and 4π/3.
Example 3 – Solution
Because g is differentiable everywhere, the only critical numbers are 2π/3 and 4π/3. We split the domain into intervals according to the critical numbers. Within each interval, g′(x) is either always positive or always negative and so we analyze g in the following table.
cont’d
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Example 3 – Solution
Because g′(x) changes from positive to negative at 2π/3, the First Derivative Test tells us that there is a local
maximum at 2π/3 and the local maximum value is
≈ 3.83
cont’d
Example 3 – Solution
Likewise g′(x), changes from negative to positive at 4π/3 and so
cont’d
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Example 3 – Solution
The graph of g in Figure 4 supports our conclusion.
cont’d
g(x) = x + 2 sin x
Figure 4
What Does f ″ Say About f ?
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What Does f ″ Say About f ?
Figure 5 shows the graphs of two increasing functions on (a, b). Both graphs join point A to point B but they look different because they bend in different directions.
Figure 5(a) Figure 5(b)
What Does f ″ Say About f ?
In Figure 6 tangents to these curves have been drawn at several points. In (a) the curve lies above the tangents and f is called concave upward on (a, b). In (b) the curve lies below the tangents and g is called concave downward on (a, b).
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What Does f ″ Say About f ?
Figure 7 shows the graph of a function that is concave
upward (abbreviated CU) on the intervals (b, c), (d, e), and (e, p) and concave downward (CD) on the intervals (a, b), (c, d ), and (p, q).
Figure 7
What Does f ″ Say About f ?
Let’s see how the second derivative helps determine the intervals of concavity. Looking at Figure 6(a), you can see that, going from left to right, the slope of the tangent
increases.
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What Does f ″ Say About f ?
This means that the derivative f′ is an increasing function and therefore its derivative f″ is positive.
Likewise, in Figure 6(b) the slope of the tangent decreases from left to right, so f′ decreases and therefore f ″ is
negative.
Concave downward
Figure 6(b)
What Does f ″ Say About f ?
This reasoning can be reversed and suggests that the following theorem is true.
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Example 4
Figure 8 shows a population graph for Cyprian honeybees raised in an apiary. How does the rate of population
increase change over time? When is this rate highest?
Over what intervals is P concave upward or concave downward?
Figure 8
Example 4 – Solution
By looking at the slope of the curve as t increases, we see that the rate of increase of the population is initially very small, then gets larger until it reaches a maximum at about t =12 weeks, and decreases as the population begins to level off.
As the population approaches its maximum value of about 75,000 (called the carrying capacity), the rate of increase, P′(t), approaches 0.
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What Does f ″ Say About f ?
Example 6
Discuss the curve y = x4 – 4x3 with respect to concavity, points of inflection, and local maxima and minima. Use this information to sketch the curve.
Solution:
If f(x) = x4 – 4x3, then
f′(x) = 4x3 – 12x2 = 4x2(x – 3)
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Example 6 – Solution
To find the critical numbers we set f′(x) = 0 and obtain x = 0 and x = 3.
To use the Second Derivative Test we evaluate f″ at these critical numbers:
f″(0) = 0 f″(3) = 36 > 0
Since f′(3) = 0 and f ″(3) > 0, f(3) = –27 is a local minimum.
[In fact, the expression for f′(x) shows that f decreases to the left of 3 and increases to the right of 3.]
cont’d
Example 6 – Solution
Since f″(0) = 0, the Second Derivative Test gives no information about the critical number 0.
But since f′(x) < 0 for x < 0 and also for 0 < x < 3, the First Derivative Test tells us that f does not have a local
maximum or minimum at 0.
Since f″(x) = 0 when x = 0 or 2, we divide the real line into intervals with these numbers as endpoints and complete the following chart.
cont’d
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Example 6 – Solution
The point (0, 0) is an inflection point since the curve
changes from concave upward to concave downward there.
Also (2, –16) is an inflection point since the curve changes from concave downward to concave upward there.
Using the local minimum, the intervals of concavity, and the inflection points, we sketch the curve in Figure 11.
cont’d
Figure 11
What Does f ″ Say About f ?
Note
The Second Derivative Test is inconclusive when
f″(c) = 0. In other words, at such a point there might be a maximum, there might be a minimum, or there might be neither (as in Example 6).
This test also fails when f″(c) does not exist. In such cases the First Derivative Test must be used. In fact, even when both tests apply, the First Derivative Test is often the easier
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Example 7
Sketch the graph of the function f(x) = x2/3(6 – x)1/3. Solution:
Calculation of the first two derivatives gives
Since f′(x) = 0 when x = 4 and f ′(x) does not exist when x = 0 or x = 6, the critical numbers are 0, 4 and 6.
To find the local extreme values we use the First Derivative Test.
Since f′ changes from negative to positive at 0, f(0) = 0 is a local minimum.
Since f′ changes from positive to negative at 4, f(4) = 25/3 is a local maximum.
Example 7 – Solution
cont’d35
Example 7 – Solution
Looking at the expression for f″(x) and noting that x4/3 ≥ 0 for all x, we have f″(x) < 0 for x < 0 and for 0 < x < 6 and f″(x) > 0 for x > 6.
So f is concave downward on ( , 0) and (0, 6) and
concave upward on (6, ), and the only inflection point is (6, 0).
cont’d
Example 7 – Solution
The graph is sketched in Figure 12.
cont’d
Figure 12
