22.15 - First Derivative Test For Local Extrema

In the previous section, we we saw the analytical method for finding absolute extrema. In this section, we will learn how to test whether, a critical point is a local extremum.

• We have seen the method for finding the critical points. But all those critical points, need not be local extrema points.
• We want an analytical method to test whether a critical point is a local extremum. The method can be explained in 12 steps:
1. In fig.22.44(a) below, the red curve is the graph of a function f in the interval (a,b).

Fig.22.44

• From the graph, we see that, f is decreasing in (a,b).
• Take any point from (a,b). Draw a tangent at that point.
    ♦ That tangent will be sloping downwards as we move from left to right.
    ♦ That means, that tangent has a negative slope.
(Two sample tangents are drawn in the fig.a)
    ♦ Negative slope indicates that, derivative at that point is negative.

2. In fig.22.44(b) above, the red curve is the graph of a function f in the interval (a,b).
• From the graph, we see that, f is decreasing in (a,b).
• Take any point from (a,b). Draw a tangent at that point.
    ♦ That tangent will be sloping downwards as we move from left to right.
    ♦ That means, that tangent has a negative slope.
(Two sample tangents are drawn in the fig.a)
    ♦ Negative slope indicates that, derivative at that point is negative.

3. In fig.22.44, we saw two functions. They have different shapes. But both are decreasing functions. They give the same result.
• The result can be summarized in three steps:
(i) Take any function.
(ii) Suppose that, the function is decreasing in (a,b).
(iii) Then all derivatives in (a,b) will be negative.

4. In fig.22.45(a) below, the red curve is the graph of a function f in the interval (a,b).

Fig.22.45

• From the graph, we see that, f is increasing in (a,b).
• Take any point from (a,b). Draw a tangent at that point.
    ♦ That tangent will be sloping upwards as we move from left to right.
    ♦ That means, that tangent has a positive slope.
(Two sample tangents are drawn in the fig.a)
    ♦ Positive slope indicates that, derivative at that point is positive.

5. In fig.22.45(b) above, the red curve is the graph of a function f in the interval (a,b).
• From the graph, we see that, f is increasing in (a,b).
• Take any point from (a,b). Draw a tangent at that point.
    ♦ That tangent will be sloping upwards as we move from left to right.
    ♦ That means, that tangent has a positive slope.
(Two sample tangents are drawn in the fig.b)
    ♦ Positive slope indicates that, derivative at that point is positive.

6. In fig.22.45, we saw two functions. They have different shapes. But both are increasing functions. They give the same result.
• The result can be summarized in three steps:
(i) Take any function.
(ii) Suppose that, the function is increasing in (a,b).
(iii) Then all derivatives in (a,b) will be positive.

7. In fig.22.46(a) below, the red curve is the graph of a function f in the interval (a,b).

Fig.22.46

• The point c in the interval (a,b), is a critical point.
• We can note five important facts:
(i) Point c is a local minimum.
(ii) To the left of c, all tangents will be having a negative slope.
(One sample tangent is drawn on the left side)
(iii) So all derivatives on the left are negative. That means, the portion between points a and c, is of decreasing nature.
(iv) To the right of c, all tangents will be having a positive slope.
(One sample tangent is drawn on the right side)
(v) So all derivatives on the right are positive. That means, the portion between points c and b, is of increasing nature.
• Based on the above five facts, we can write:
If c is a local minimum in (a,b), then:
    ♦ Sign of the derivative changes at c
    ♦ This change is from negative to positive.
    ♦ Portion between a and c is decreasing in nature.
    ♦ Portion between c and b is increasing in nature.

8. In fig.22.46(b) above, the red curve is the graph of a function f in the interval (a,b).
• The point c in the interval (a,b), is a critical point.
• We can note five important facts:
(i) Point c is a local maximum.
(ii) To the left of c, all tangents will be having a positive slope.
(One sample tangent is drawn on the left side)
(iii) So all derivatives on the left are positive. That means, the portion between points a and c, is of increasing nature.
(iv) To the right of c, all tangents will be having a negative slope.
(One sample tangent is drawn on the right side)
(v) So all derivatives on the right are negative. That means, the portion between points c and b, is of decreasing nature.
• Based on the above five facts, we can write:
If c is a local maximum in (a,b), then:
    ♦ Sign of the derivative changes at c
    ♦ This change is from positive to negative.
    ♦ Portion between a and c is increasing in nature.
    ♦ Portion between c and b is decreasing in nature.

9. In both figs.22.46(a) and 22.46(b), the derivative exists at the point c. Let us see the cases where the derivative does not exist.
• In fig.22.47(a) below, c is a cusp point. Every thing that we wrote in (7) is applicable for this fig.
• In fig.22.47(b) below also, c is a cusp point. Every thing that we wrote in (8) is applicable for this fig.

Fig.22.47

• So we can write:
It does not matter whether the derivative exists at c or not. We can find whether c is a local maximum or local minimum, just by finding the signs of the derivative on either sides of c.

10. Now let us see the case where c is neither a local maximum nor a local minimum. In fig.22.48(a) below, the red curve is the graph of a function f in the interval (a,b).

Fig.22.48

• The point c in the interval (a,b), is a critical point.
• We can note six important facts:
(i) Point c is a not a local minimum.
(ii) Point c is a not a local maximum.
(iii) To the left of c, all tangents will be having a negative slope.
(One sample tangent is drawn on the left side)
(iv) So all derivatives on the left are negative. That means, the portion between points a and c, is of decreasing nature.
(v) To the right of c also, all tangents will be having a negative slope.
(One sample tangent is drawn on the right side)
(vi) So all derivatives on the right are negative. That means, the portion between points c and b, is also of decreasing nature.
• Based on the above six facts, we can write:
If c is neither a local minimum nor a local maximum in (a,b), then:
    ♦ Sign of the derivative will not change at c
    ♦ If it is −ve on the left, it will be −ve on right also.
    ♦ If the portion between a and c is decreasing in nature, the portion between c and b will also be decreasing in nature.

11. In fig.22.48(b) above, the red curve is the graph of a function f in the interval (a,b).
• The point c in the interval (a,b), is a critical point.
• We can note six important facts:
(i) Point c is a not a local minimum.
(ii) Point c is a not a local maximum.
(iii) To the left of c, all tangents will be having a +ve slope.
(One sample tangent is drawn on the left side)
(iv) So all derivatives on the left are +ve. That means, the portion between points a and c, is of increasing nature.
(v) To the right of c also, all tangents will be having a +ve slope.
(One sample tangent is drawn on the right side)
(vi) So all derivatives on the right are +ve. That means, the portion between points c and b, is also of increasing nature.
• Based on the above six facts, we can write:
If c is neither a local minimum nor a local maximum in (a,b), then:
    ♦ Sign of the derivative will not change at c
    ♦ If it is +ve on the left, it will be +ve on right also.
    ♦ If the portion between a and c is increasing in nature, the portion between c and b will also be increasing in nature.

12. Based on the above 11 steps, we can write the test. It can be written in four steps:
(i) f is a continuous function in the interval (a,b). Point c falls with in (a,b).
(ii) f(c) is a local minimum if:
    ♦ The derivative is −ve between a and c.
    ♦ The derivative is +ve between c and b.
(iii) f(c) is a local maximum if:
    ♦ The derivative is +ve between a and c.
    ♦ The derivative is −ve between c and b.
(iv) f(c) is neither a local maximum nor a local minimum if:
    ♦ The derivative has the same sign between [a and c] and [c and b].

• The above four steps taken together, is known as the first derivative test for finding local extrema.

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Now we will see some solved examples.

Solved example 22.51
Find all points of local maxima and local minima of the function f given by f(x) = x³ − 3x + 3
Solution:
Step I: Finding the critical points and dividing the number line
1. First we write the derivative:
f '(x) = 3x² − 3

2. This derivative must be equated to zero.
3x² − 3 = 0
⇒ 3(x² − 1) = 0
⇒ x² = 1
⇒ x = +1 and x = −1

3. So the points in category I are: x = 1 and x = −1
4. We obtained f '(x) = 3x² − 3
• 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 two critical points are: x = 1 and x = −1
6. Therefore, the number line can be divided into three intervals:
(−∞,−1), (−1,1) and (1,∞)

Step II: Finding the sign of f '(x) in each interval
First interval:
• A convenient number in the first interval is −2.
• f '(−2) = 3(−2)² − 3 = 3(4) − 3 = 12 − 3 = 9
• Therefore, the sign of f '(x) in the first interval is +ve. 

Second interval:
• A convenient number in the second interval is 0.2.
• f '(0.2) = 3(0.2)² − 3 = 3(0.04) − 3 = 0.12 − 3 = −2.88
• Therefore, the sign of f '(x) in the second interval is −ve.

Third interval:
• A convenient number in the third interval is 2.
• f '(2) = 3(2)² − 3 = 3(4) − 3 = 12 − 3 = 9
• Therefore, the sign of f '(x) in the third interval is +ve.

Step III: Analyzing the change of signs at critical points
• At the first critical point −1, the sign of f '(x) changes from +ve to −ve. So this critical point is a point of local maximum.
• At the second critical point +1, the sign of f '(x) changes from −ve to +ve. So this critical point is a point of local minimum.

• Fig.22.49 below shows the graph

Fig.22.49

Solved example 22.52
Find all points of local maxima and local minima of the function f given by f(x) = 2x³ − 6x² + 6x + 5
Solution:
Step I: Finding the critical points and dividing the number line
1. First we write the derivative:
f '(x) = 6x² − 12x + 6

2. This derivative must be equated to zero.
6x² − 12x + 6 = 0
⇒ x² − 2x + 1 = 0
⇒ (x − 1)² = 0
⇒ (x − 1) = 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
6. Therefore, the number line can be divided into two intervals:
(−∞, 1) and (1,∞)

Step II: 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 III: 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.

• Fig.22.50 below shows the graph

Fig.22.50

• We see that:
There is only one critical point, and it is neither a point of local maximum nor a point of local minimum.

Solved example 22.53
Find all points of local maxima and local minima of the function f given by f(x) = x¹¹ − 6x¹⁰.
Solution:
Step I: Finding the critical points and dividing the number line
1. First we write the derivative:
f '(x) = 11x¹⁰ −60x⁹.

2. This derivative must be equated to zero.
11x¹⁰ −60x⁹ = 0
⇒ x⁹(11x − 60) = 0
⇒ x⁹ = 0 and 11x − 60 = 0
⇒ x = 0 and 11x  = 60
⇒ x = 0 and x = 60/11

3. So the points in category I are: x = 0 and x = 60/11
4. We obtained f '(x) = 11x¹⁰ −60x⁹.
• 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 two critical points are: x = 0 and x = 60/11
6. Therefore, the number line can be divided into three intervals:
(−∞,0), (0,60/11) and (60/11,∞)

Step II: Finding the sign of f '(x) in each interval
First interval:
• A convenient number in the first interval is −1.
• f '(−1) = 11(−1)¹⁰ −60(−1)⁹ = 11+60 = 71
• Therefore, the sign of f '(x) in the first interval is +ve. 

Second interval:
• A convenient number in the second interval is 1.
• f '(1) = 11(1)¹⁰ −60(1)⁹ = 11−60 = −49
• Therefore, the sign of f '(x) in the second interval is −ve.

Third interval:
• A convenient number in the third interval is 10.
• f '(10) = 11(10)¹⁰ −60(10)⁹ = (10)⁹ [11(10) −60]
= (10)⁹ [110 −60] = (10)⁹ [50]
• Therefore, the sign of f '(x) in the third interval is +ve.

Step III: Analyzing the change of signs at critical points
• At the first critical point 0, the sign of f '(x) changes from +ve to −ve. So this critical point is a point of local maximum.
• At the second critical point 60/11 (=5.455), the sign of f '(x) changes from −ve to +ve. So this critical point is a point of local minimum.

• Fig.22.51 below shows the graph

Fig.22.51

• We see that:
Some portion near the critical point x = 0, is horizontal. In fact it is not horizontal. It appears to be horizontal because of the large scale of the graph. This can be proved in two steps:
(i) f(−0.1) = (−0.1)¹¹ − 6(−0.1)¹⁰
= (−1)(0.1)¹¹ − 6(0.1)¹⁰
= (0.1)¹⁰ [(−1)(0.1) − 6]
= (0.1)¹⁰ [−0.1 − 6]
= (0.1)¹⁰ [−6.1]
= (−1)(0.1)¹⁰ [6.1]
• This is a −ve quantity. That means, just to the left of the critical point x=0, the point on the graph is below the x-axis.
(ii) f(0.1) = (0.1)¹¹ − 6(0.1)¹⁰
= (0.1)¹⁰ [0.1 − 6]
= (0.1)¹⁰ [−5.9]
= (−1)(0.1)¹⁰ [5.9]
• This is a −ve quantity. That means, just to the right of the critical point x=0, the point on the graph is below the x-axis.

• We also see that, point x = 5.455 appears to be a cusp point. But in reality, there is a smooth transition at that point. If we enlarge the surrounding region of that point, we will be able to see the smooth transition. 

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In the next section, we will see the concavity test.

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