22.17 - The Second Derivative Test

In the previous section, we saw the concavity test. In this section, we will see the second derivative test for finding local extrema.

1. Consider the graph in fig.22.55 of the previous section. With some modifications, it is shown again below as fig.22.57.
• Two modifications are made to the earlier fig.22.55
(i) The point (0,0) is specially marked because, it is a local maximum.
(ii) The vertical white dashed line is drawn through the local minimum. This is for getting a better understanding about the change of sign of f ' on either sides of the local minimum. Such a line should have to be drawn through the local maximum also. But there, we already have the y-axis.

Fig.22.57

• The vertical magenta dashed line divides the graph into two portions: concave down portion and concave up portion.

2. Consider the concave down portion. Any concave down portion will contain a peak point. This peak point will be a local maximum. We want a relation between this local maximum and the second derivative f ''. The relation can be established in 4 steps:
(i) We see that, in the concave down portion:
   ♦ The green line is below the x-axis.
   ♦ This happens because, in this portion, all f '' values are −ve. We have seen the reason in the previous section 22.16.
(ii) We see that, in the concave down portion:
• The yellow curve is above x-axis up to the local maximum.
   ♦ This happens because, in this portion, all f ' values are +ve upto the local maximum. We have seen the reason in a previous section. See fig.22.46(b) of section 22.15.
• The yellow curve is below x-axis after the local maximum.
   ♦ This happens because, in this portion, all f ' values are −ve after the local maximum. We have seen the reason in a previous section. See fig.22.46(b) of section 22.15.
(iii) We see that, in the concave down portion:
• The yellow curve changes sign at the local maximum. That means, f ' changes sign at the local maximum.
(We know that, at the point where the "change in sign" occurs, f ' will be zero)
(iv) Based on the above three steps, we can establish the relation:
• At the local maximum:
   ♦ f ' will be zero.
   ♦ f '' will be −ve.

3. Consider the concave up region. Any concave up region will contain a valley point. This valley point will be a local minimum . We want a relation between this local minimum and the second derivative f ''. The relation can be established in 4 steps:
(i) We see that, in the concave up portion:
   ♦ The green line is above the x-axis.
   ♦ This is because, in this portion, all f '' values are +ve. We have seen the reason in the previous section 22.16.
(ii) We see that, in the concave up portion:
• The yellow curve is below x-axis up to the local minimum.
   ♦ This happens because, in this portion, all f ' values are −ve up to the local minimum. We have seen the reason in a previous section. See fig.22.46(a) of section 22.15.
• The yellow curve is above x-axis after the local maximum.
   ♦ This happens because, in this portion, all f ' values are +ve after the local maximum. We have seen the reason in a previous section. See fig.22.46(a) of section 22.15.
(iii) We see that, in the concave up portion:
• The yellow curve changes sign at the local minimum. That means, f ' changes sign at the local minimum.
(We know that, at the point where the "change in sign" occurs, f ' will be zero)
(iv) Based on the above three steps, we can establish the relation:
• At the local minimum:
   ♦ f ' will be zero.
   ♦ f '' will be +ve.
4. So now we have an effective test to determine whether a critical point is a local maximum or a local minimum. The test can be done in 2 steps:
Step I: Determine all critical points
Step II: Determine the sign of f '' at each of those points.
   ♦ If f '' is −ve at a critical point, then that point is a local maximum.
   ♦ If f '' is +ve at a critical point, then that point is a local minimum.
◼ This test is known as the second derivative test.

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• Note that, the second derivative test does not tell us anything about the case when f '' = 0
• That means, the test is not applicable at points where f '' = 0. At such a point, we must use the first derivative test to find whether that point is a local maximum or local minimum. It is also possible that, such a point is neither a local maximum nor a local minimum.

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Solved example 22.56
Find local extrema of the function f given by f(x) = 3x⁴ + 4x³ − 12x² + 12
Solution:
Step I: Finding the critical points
1. Write the first two derivatives:
f '(x) = 12x³ + 12x² − 24x
f ''(x) = 36x² + 24x − 24
2. Equate the first derivative to zero and solve for x:
12x³ + 12x² − 24x = 0
⇒ 12x(x² + x − 2) = 0
⇒ x(x² + 2x − x − 2) = 0
⇒ x[x(x + 2) − (x + 2)] = 0
⇒ x[(x + 2) (x − 1)] = 0
⇒ x = 0, x = −2 and x = 1
3. So the points in category I are: x = 0, x = −2 and x = 1
4. We obtained f '(x) = 12x³ + 12x² − 24x
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only three critical points are: x = 0, x = −2 and x = 1 

Step II: Find f '' at the critical points
1. f ''(0) =  36(0)² + 24(0) − 24 = −24
2. f ''(−2) =  36(−2)² + 24(−2) − 24 = 36(4) − 48 −24 = 144 − 72 = 72
3. f ''(1) =  36(1)² + 24(1) − 24 = 36(1) + 24 −24 = 36

Step III: Applying the second derivative test
1. Since f ''(0) is −ve, there is a local maximum at x = 0.
2. Since f ''(−2) is +ve, there is a local minimum at x = −2.
3. Since f ''(1) is +ve, there is a local minimum at x = 1.

Step IV: Finding the actual extrema values:
1. The local maximum value at x = 0 is given by:
f(0) = 3(0)⁴ + 4(0)³ − 12(0)² + 12 = 12
2. The local minimum value at x = −2 is given by:
f(−2) = 3(−2)⁴ + 4(−2)³ − 12(−2)² + 12 = −20
3. The local minimum value at x = 1 is given by:
f(1) = 3(1)⁴ + 4(1)³ − 12(1)² + 12 = 7

Step V (optional): Drawing the graph

• The graph is shown in fig.22.58 below:

Fig.22.56

   ♦ f is drawn in red color.
   ♦ (−1,7) is a local minimum.
   ♦ (0,12) is a local maximum.
   ♦ (2,20) is a local minimum.

Solved example 22.57
Find all points of local extrema of the function f given by f(x) = 2x³ − 6x² + 6x + 5
Solution:
Step I: Finding the critical points
1. Write the first two derivatives:
f '(x) = 6x² − 12x + 6
f ''(x) = 12x − 12
2. Equate the first derivative to zero and solve for x:
6x² − 12x + 6 = 0
⇒ 6(x² −2x + 1) = 0
⇒ x² −2x + 1 = 0
⇒ (x − 1)² = 0
⇒ x = 1
3. So the only one point in category I is: x = 1
4. We obtained f '(x) = 6x² − 12x + 6
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 1 

Step II: Find f '' at the critical points
• f ''(1) = 12(1) − 12 = 0

Step III: Applying the second derivative test
1. The second derivative test is not applicable at the point where f '' = 0.
2. So we have to apply the first derivative test for this problem
(We have already solved this problem by applying the first derivative test. See solved example 22.52 in section 22.15. So we can use those steps to complete this problem)

Step IV: Dividing the number line
• The critical point is x = 1. Therefore, the number line can be divided into two intervals:
(−∞, 1) and (1,∞)

Step V: Finding the sign of f '(x) in each interval
First interval:
• A convenient number in the first interval is −1.
• f '(−1) = 6(−1)2 − 12(−1) + 6 = 6 + 12 + 6 = 24
• Therefore, the sign of f '(x) in the first interval is +ve.

Second interval:
• A convenient number in the second interval is 2.
• f '(2) = 6(2)2 − 12(2) + 6 = 24 − 24 + 6 = 6
• Therefore, the sign of f '(x) in the second interval is +ve.

Step VI: Analyzing the change of signs at critical point
• At the critical point 1, there is no change of sign for f '(x). So this critical point is neither a point of local maximum nor a point of local minimum.

Step VII (optional): Drawing the graph

• The graph is drawn for the solved example 22.52 in section 22.15.

Solved example 22.58
Find all points of local extrema of the function f given by f(x) =x⁵− 5x³
Solution:
Step I: Finding the critical points
1. Write the first two derivatives:
f '(x) = 5x⁴ − 15x²
f ''(x) = 20x³ − 30x
2. Equate the first derivative to zero and solve for x:
5x⁴ − 15x² = 0
⇒ 5x²(x² − 3) = 0
⇒ 5x² = 0 and (x² − 3) = 0
⇒ x = 0, x = √3 and x = −√3
3. So the three points in category I are:
x = 0, x = √3 and x = −√3
4. We obtained f '(x) = 5x⁴ − 15x²
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only three critical points are:
x = 0, x = √3 and x = −√3

Step II: Find f '' at the critical points
• f ''(0) = 20(0)³ − 30(0) = 0
• f ''(√3) = 20(√3)³ − 30(√3) = √3[20(√3)² − 30]
= √3[20(3) − 30] = √3[60 − 30] = 30√3
• f ''(−√3) = 20(−√3)³ − 30(−√3) = −√3[20(−√3)² − 30]
= −√3[20(3) − 30] = −√3[60 − 30] = −30√3

Step III: Applying the second derivative test
1. Since f ''(0) is zero, the second derivative test is not applicable at x = 0.
2. Since f ''(√3) is +ve, there is a local minimum at x = √3.
3. Since f ''(−√3) is −ve, there is a local maximum at x = −√3.

Step IV: Since the second derivative is zero at x = 0, we have to apply the first derivative test at that point. For that, we divide the number line
• The critical points are x = 0, x = √3 and x = −√3. Therefore, the number line can be divided into four intervals:
(−∞, −√3), (−√3,0), (0,√3) and (√3,∞)
• Since we are checking x = 0, we need to consider only two intervals: (−√3,0) and (0,√3)

Step V: Finding the sign of f '(x) in each interval
First interval: (−√3,0)
• A convenient number in the first interval is −1.
• f '(−1) = 5(−1)⁴ − 15(−1)² = 5 − 15 = −10
• Therefore, the sign of f '(x) in the first interval is −ve.

Second interval: (0,√3)
• A convenient number in the second interval is 1.
• f '(1) = 5(1)⁴ − 15(1)² = 5 − 15 = −10
• Therefore, the sign of f '(x) in the second interval is −ve.

Step VI: Analyzing the change of signs at critical point
• At the critical point 0, there is no change of sign for f '(x). So this critical point is neither a point of local maximum nor a point of local minimum.

Step VII (optional): Drawing the graph

Fig.22.58

   ♦ f is drawn in red color.
   ♦ (−1.73,10.39) is a local maximum.
   ♦ (0,0) is neither a local maximum nor a local minimum.
         ✰ This is a point of inflection.
   ♦ (1.73,−10.39) is a local minimum.

Solved example 22.59
Find two positive numbers whose sum is 15 and the sum of whose squares is minimum.
Solution:
Step I: Writing the problem as a function
1. Sum of two positive numbers is to be 15
So if one number is x, then the other number is (15−x)
2. If S is the sum of the squares, then we can write:
S = x² + (15 − x)2
3. The value of S varies depending on the value of x. So S is a function of x. We can write:
S = f(x) = x² + (15 − x)²
4. We want S to have the minimum possible value. That means, we want the least of all the local extrema of f. That means, we need to find all local extrema and compare them.

Step II: Finding the critical points
1. Write the first two derivatives:
• f '(x) = 2x + 2(15 − x)(−1)
= 2x − 2(15 − x)
= 2x −30 + 2x
= 4x −30
• f ''(x) = 4
2. Equate the first derivative to zero and solve for x:
4x −30 = 0
⇒ 4x = 30
⇒ x = 7.5
3. So the only one point in category I is: x = 7.5
4. We obtained f '(x) = 4x −30
• This function is defined for all real numbers. That means, there is no input x at which f '(x) is not defined.
• Therefore, there will be no points in category II.
5. So the only one critical point is: x = 7.5

Step III: Find f '' at the critical point
• f ''(7.5) =  4. This is because, f ''(x) is a constant 

Step IV: Applying the second derivative test
• Since f ''(7.5) is +ve, there is a local minimum at x = 7.5
• Since there is only one critical point, there are no other points to compare. We can write:
The minimum value of the function occurs at x = 7.5

Step V: Finding the actual minimum value:
• The local maximum value at x = 7.5 is given by:
f(7.5) = (7.5)² + (15 − 7.5)² = 2(7.5)² = 112.5

Step VI: Final result:
• Out of the two numbers, the first number x = 7.5
• So second number (15 − x) = 7.5
• The minimum sum = (7.5)² + (7.5)² = 2(7.5)² = 112.5 

Check:
• Let us put x = 8. Then (x − 8) = 7
• Then S = 8² + 7² = 113. This is greater than 112.5

Step VI (optional): Drawing the graph

• The graph is shown in fig.22.59 below:

Fig.22.59

   ♦ f is drawn in red color.
   ♦ (7.5,112.5) is the local minimum.

◼ In general, if sum of two positive numbers is k and sum of their squares is to be minimum, then there is only one option:
Both numbers must be equal to k/2.

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Based on the above solved examples, we can write a comparison between first derivative test and second derivative test. It can be written in 2 steps:
1. In the first derivative test, we check the sign of f ' on either sides of each critical point. Based on the signs, we decide whether that critical point is a local maximum or a local minimum.
2. In the second derivative test, we check the sign of f '' at each critical point. Based on the sign, we decide whether that critical point is a local maximum or a local minimum.
3. So in many cases, the second derivative test will be easier to apply.

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In the next section, we will see a few more solved examples.

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