17 Stationary Points of Functions

A stationary point of a curve is a point on the curve at which the gradient is zero. In this section we will focus on the definitions of different types of stationary points of functions and how to find them. By the end of this section, you should have the following skills:

• An understanding of the deﬁnition of types of stationary points, including: local maxima, local minima and points of inﬂection.

• Find stationary points of a function using diﬀerentiation, including: local maxima, local minima and points of inﬂection.

• Test points to see if they are a stationary point.

• Test stationary points to determine their type of stationary point, using ﬁrst and second derivatives.

17.1 Local Maximum and Local Minimum Let f(x) be a function. It is useful to be able to ﬁnd and classify local maxima and minima. These are points at which the function switches from increasing to decreasing and vice-versa.

17.1.1 Local Minimum Here the graph is decreasing and then increasing as x increases through the local minimum. The point x = b in the diagram is a local minimum.

17.1.2 Local Maximum In contrast, the graph is increasing and then decreasing as x increases through the local maximum. The point x = a in the diagram is a local maximum.

1 Local minimum and local maximum.

Note that such a point as a is only a local maximum as it is possible that the function takes values elsewhere which are greater than the value at a. In our example, there are clearly values for the function which are greater than the value at x = a. Similarly for a local minimum.

Example 1 1. f(x) = 1 has a local maximum and a local minimum at every point!

2. Any linear function f(x) = ax+b, a 6= 0, b constants has neither local maxima nor local minima.

3. f(x) = x2 has a local minimum at x = 0.

4. f(x) = −x2 has a local maximum at x = 0.

5. a quadratic has either one local minimum or one local maximum (but not both).

6. f(x) = x3 − 3x has a local maximum at x = −1 and a local minimum at x = 1.

2 7. f(x) = sin(x) has local maxima at x = π/2, 5π/2 9π/2,... and has local minima at x = 3π/2, 7π/2 11π/2,....

8. f(x) = tan(x) has no local maxima or minima.

9. f(x) = ex has no local maxima or minima.

10. f(x) = ln(x) has no local maxima or minima.

11. f(x) = 1/x has no local maxima or minima.

12. f(x) = ex − x has a local minimum at x = 0 where f(0) = 1.

Note that this shows that ex − x ≥ 1 ⇒ ex ≥ 1 + x for all x.

3 13. 1 f(x) = (x − 1)(x − 2) has a local maximum at x = 3/2 where f(3/2) = −4.

Note carefully that a local maximum is not a global maximum in general. For example f(x) = x3 −3x has a local maximum of f(−1) = 2 at x = −1 but this is not a maximum for the function as clearly there are points at which this function is greater than 2. What is important here is that all other points x suﬃciently near to −1 have f(x) ≤ 2. Similarly for local minima,

4 x = 1 is a local minimum for f(x) = x3 − 3x but not a global minimum.

Exercise 1 Each of the following functions f(x) has a local maximum or minimum as indicated. Which of these is a global maximum or minimum?

(a) f(x) = x2 has a local minimum at x = 0.

(b) f(x) = −x2 has a local maximum at x = 0.

(d) f(x) = sin(x) has a local minimum at x = 3π/2.

(e) f(x) = 1/(x − 1)(x − 2) has a local maximum at x = 3/2.

(f) f(x) = ex − x has a local minimum at x = 0.

Solutions to exercise 1

(a) f(x) = x2 has a local minimum at x = 0.

(b) f(x) = −x2 has a local maximum at x = 0.

5 Graph of x2 and −x2.

Clearly y = x2 has a global maximum at x = 0 whereas y = −x2 has a local minimum at x = 0.

(d) f(x) = sin(x) has a local minimum at x = 3π/2.

Since sin(π/2) = 1 and sin(3π/2) = −1 and for all x, −1 ≤ sin(x) ≤ 1 we see that x = π/2 is a global maximum and x = 3π/2 is a global minimum.

(e) f(x) = 1/((x − 1)(x − 2)) has a local maximum at x = 3/2. The vertical lines at x = 1 and x = 2 are asymptotes.

6 We see that x = 3/2 is not a global maximum as for example f(0) = 1/2 > f(3/2).

(f) f(x) = ex − x has a local minimum at x = 0.

Note that if y = ex−x then dy/dx = ex−1 and so for x > 0, dy/dx > 0 and for x < 0, dy/dx < 0, hence the function is decreasing for x < 0 and increasing for x > 0, so x = 0 must be a global minimum.

7 17.2 Finding Local Maxima and Minima 17.2.1 Reminder about increasing and decreasing functions The following results from Increasing and Decreasing Functions are used in this section. Note that these results are true if f(x) is diﬀerentiable and f 0(x) is continuous, which we assume.

1. If f(x) is increasing for a range of values of x then df ≥ 0 dx over that range of values of x. 2. If f(x) is decreasing for a range of values of x then df ≤ 0 dx over that range of values of x.

17.2.2 Local Maxima We deal ﬁrst with local maxima - the details for local minima are similar. First recall that a local maximum at x = a occurs when a function changes from increasing to decreasing i.e.

• in an interval before b ≤ x ≤ a, f(x) is increasing ⇒ df/dx ≥ 0, b ≤ x ≤ a and • in an interval after a ≤ x ≤ c, f(x) is decreasing ⇒ df/dx ≤ 0, a ≤ x ≤ c.

8 Local Maximum.

Since df/dx ≥ 0, df/dx ≤ 0, x = a we must have:

If x = a is a local maximum then df = 0, x = a dx and df/dx decreases from positive to negative as x goes through a.

17.2.3 Local Minima

Local Minimum.

In a similar way we can show that if x = a is a local minimum then df = 0, x = a dx and df/dx increases from negative to positive as x goes through a.

17.2.4 Summary We have shown that if x = a is a local maximum or a local minimum then df = 0, x = a. dx This tells us that the tangent at x = a is horizontal i.e. parallel to the x-axis.

9 17.3 Stationary Points Any point x = a such that df = 0, x = a dx is called a stationary point. This is because the rate of change at x = a is 0 i.e. instantaneously stationary. So a local maximum or minimum is a stationary point. Note that for a local maximum or a local minimum the sign of df/dx changes as x goes through the stationary point.

17.3.1 Points of inflection There is another type of stationary point called a point of inflection. At these points we still have df/dx = 0 but df/dx does not change sign as x goes through a. For example y = x3 has a stationary point at x = 0, but dy/dx = 3x2 ≥ 0 for all x. So dy/dx does not change sign. Similarly, y = −x3 has a point of inflection at x = 0 as dy/dx ≤ 0 for all x.

Figure 1: y = x3.

10 17.4 Finding Stationary Points If f(x) is a function we solve the equation

df = 0 dx for x. The values of x we ﬁnd are the stationary points of f(x).

Example 2 Find the stationary points for the following functions f(x):

(a) f(x) = x2 − 5x + 6.

(b) f(x) = x3 − 6x2 + 7.

(d) f(x) = sin(x).

(e) f(x) = ex − 2x.

(f) f(x) = (x − 1)/(x2 + 5x + 3).

(g) f(x) = ln(x3 − 6x2 − 15x + 1).

(h) f(x) = 3x5 − 5x3 + 4. √ (i) f(x) = 7 + (2x2 − 10x) x.

Solution.

(a) df = 2x − 5. dx The stationary points are given by x such that df 5 = 0 ⇒ x = . dx 2 So one stationary point x = 5/2.

11 Graph of x2 − 5x + 6.

(b) df = 3x2 − 12x. dx The stationary points are given by x such that df = 0 ⇒ 3x2 − 12x = 3x(x − 4) = 0 ⇒ x = 0, x = 4. dx So the stationary points are x = 0, 4.

Graph of x3 − 6x2 + 7.

12 (c) df = 3x2 + 3. dx The stationary points are given by x such that df = 0 ⇒ 3x2 + 3 = 0 ⇒ x2 = −1. dx But there are no real number solutions of this equation. So no stationary points.

(d) df = cos(x). dx The stationary points are given by x such that df = 0 ⇒ cos(x) = 0 ⇒ x = ±π/2, ±3π/2, ±5π/2,.... dx There are an inﬁnite number of stationary points.

(e) df = ex − 2. dx The stationary points are given by x such that df = 0 ⇒ ex = 2 ⇒ x = ln(2.) dx So one stationary point x = ln(2).

13 Graph of ex − 2x.

(f)

df x2 + 5x + 3 − (x − 1)(2x + 5) = dx (x2 + 5x + 3)2 −x2 + 2x + 8 = (x2 + 5x + 3)2 −(x − 4)(x + 2) = . (x2 + 5x + 3)2

The stationary points are given by x such that df = 0 ⇒ x = 4, x = −2. dx So two stationary points.

14 Graph of (x − 1)/(x2 + 5x + 3).

(g) df 3x2 − 12x − 15 3(x − 5)(x + 1) = = . dx x3 − 6x2 − 15x + 1 x3 − 6x2 − 15x + 1 The stationary points are (in theory) given by x such that

df = 0 ⇒ x = 5, x = −1. dx WARNING. Does the domain of the function include these stationary points? x = −1 is OK as f(−1) = ln(7) but f(5) = ln(125−150−75+1) = ln(−99) which does not exist!! Hence only one possible stationary point x = −1. The graph of y = x3 − 6x2 − 15x + 1 between x = −4 and x = 9 is as follows and we see that it is negative between x = 0.0649952 and x = 7.88602, and for x < −2, hence y = ln(x3 − 6x2 − 15x + 1) is not deﬁned for these values.

15 Graph of x3 − 6x2 − 15x + 1.

(h) df = 15x4 − 15x2 = 15x2(x2 − 1). dx The stationary points are given by 15x2(x2 − 1) = 0 ⇒ x = 0, x = 1, x = −1

Graph of 3x5 − 5x3 + 4.

16 (i) df √ = 5x3/2 − 15x1/2 = 5(x − 3) x. dx √ The stationary points are given by 5(x − 3) x = 0 ⇒ x = 0, x = 3 Note that the domain of f(x) is all x ≥ 0.

√ Graph of 7 + (2x2 − 10x) x.

17.5 Finding local maxima and minima for a function f(x) Step 1: We ﬁnd all stationary points of f(x) i.e. solve

df = 0 dx for x.

Step 2: We then test all of these points to see if they are local maxima, local minima or perhaps points of inﬂection.

17 17.6 Testing Stationary Points Let a be a stationary point for the function f(x). If we have a local minimum at x = a then df/dx goes from negative to positive as x increases through the stationary point, and for a local maximum df/dx decreases from positive to negative.

In order to test for these changes in sign we deﬁne:

17.6.1 Test Points for a stationary point. Let a be a stationary point for f(x) and let b, c be any points such that:

1. b < a < c i.e. a lies between b and c.

2. No other stationary point other than a lies between b and c.

3. f(x) is deﬁned for all points in the interval [b, c].

We call b and c test points. Such test points give rise to a simple test for a local maximum :

17.6.2 Test for local maximum: Examining signs of the ﬁrst derivative Let b < c be test points for the stationary point a. If df/dx is positive for x = b and is negative for x = c then a is a local maximum.

Local maximum.

18 17.6.3 Test for local minimum: Examining signs of the ﬁrst derivative Let b < c be test points for the stationary point a. If df/dx is negative for x = b and is positive for x = c then a is a local minimum.

Local minimum.

17.6.4 Test for a point of inflection: Examining signs of the first derivative Remember that a stationary point need not be a local maximum or minimum, it could be a point of inflection as we have seen with f(x) = x3. This has one stationary point 0 but df/dx = 3x2 is always positive and never changes sign as is required for a local maximum or minimum. Thus there are two possibilities for a point of inflection: 1. df/dx > 0 for test points either side of the stationary point or

2. df/dx < 0 for test points either side of the stationary point.

Example 3 You are given the function f(x) and its stationary points. For each stationary point determine if it is a local maximum or local minimum (or neither).

(a) f(x) = x2 − 5x + 6. Stationary point is x = 5/2.

19 (b) f(x) = x3 − 6x2 + 7. Stationary points are x = 0, 4.

(d) f(x) = ex − 2x. Stationary point is x = ln(2).

(e) f(x) = (x − 1)/(x2 + 5x + 3). Stationary points are x = 4, −2.

(f) f(x) = 3x5 − 5x3 + 4. Stationary points are x = −1, 0, 1. √ (g) f(x) = 7 + (2x2 − 10x) x. Stationary points are x = 0, 3.

Solution.

(a) f 0(x) = 2x − 5

Taking test points 2, 3 either side of the stationary point 5/2 we have

f 0(2) = 4 − 5 = −1 < 0 f 0(3) = 6 − 5 > 0.

Since f 0(x) is going from negative to positive through this stationary point we have a local minimum.

(b) f 0(x) = 3x2 − 12x

First consider the stationary x = 0. Taking test points x = −1, 1 either side of this stationary point we have

f 0(−1) = 3 + 12 = 15 > 0 f 0(1) = 3 − 12 < 0.

Since f 0(x) is going from positive to negative through this stationary point we have a local maximum.

20 (Note that the values −1, 1 we chose did not include any stationary point other than 0). Next consider the stationary x = 4. Taking test points x = 3, 5 either side of this stationary point we have

f 0(3) = 27 − 36 = −9 < 0 f 0(5) = 75 − 60 = 15 > 0.

Since f 0(x) is going from negative to positive through this stationary point we have a local minimum. (Note that the values 3, 5 we chose did not include any stationary point other than 4.)

This has only one stationary point x = 0. Taking test points x = −1, x = 1 either side of this stationary point we have

f 0(−1) = 8 > 0 f 0(1) = 8 > 0.

Since f 0(x) is not changing sign through the stationary point there is no local maximum or minimum. This is a point of inﬂexion i.e. the ﬁrst derivative does not change sign as it passes through the stationary point.

(d) f 0(x) = ex − 2

Taking test points 1/2, 1 either side of the stationary point x = ln(2). (check using your calculator that 1/2 < ln(2) < 1) we have

f 0(1/2) = e1/2 − 2 < 0 f 0(1) = e − 2 > 0.

21 Since f 0(x) is going from negative to positive through this stationary point we have a local minimum.

(e) −(x − 4)(x + 2) f 0(x) = (x2 + 5x + 3)2 with stationary points x = 4, −2. First consider the stationary point x = −2. Taking test points x = −3, 0 either side of this stationary point we have −7. − 1 7 f 0(−3) = − = − < 0 (9 − 15 + 3)2 9 f 0(0) = 8/9 > 0.

Since f 0(x) is going from negative to positive through this stationary point we have a local minimum. (Note that the values −3, 0 we chose did not include the other stationary point 4.) Next consider the stationary point x = 4. Taking test points x = 0, 5 either side of this stationary point we have

f 0(0) = 8/9 > 0 7 f 0(5) = − < 0. 532 Since f 0(x) is going from positive to negative through this stationary point we have a local maximum. (Note that the values 0, 5 we chose did not include the other stationary point −2.)

(f) f 0(x) = 15x4 − 15x2 = 15x2(x2 − 1)

with stationary points at x = −1, 0, 1. At x = −1 we see that df/dx > 0 for a test point < −1 and

22 df/dx < 0 for a test point > −1. Hence x = −1 is a local maximum. At x = 0 we see that df/dx < 0 for a test point < 0 and df/dx < 0 for a test point > 0. Hence x = 0 is a point of inﬂection. At x = 1 we see that df/dx < 0 for a test point < 1 and df/dx > 0 for a test point > 1. Hence x = 1 is a local minimum.

Graph of 3x5 − 5x3 + 4.

(g) √ f 0(x) = 5x3/2 − 15x1/2 = 5(x − 3) x with stationary points x = 0, 3. Note that f(x) is not deﬁned for x < 0 and so we cannot test the point x = 0 to see what type of stationary point we have as we cannot take a test point < 0. At x = 3 we see that df/dx < 0 for a test point < 3 and df/dx > 0 for a test point > 3. Hence x = 3 is a local minimum.

23 √ Graph of 7 + (2x2 − 10x) x.

Exercise 2 You are given the function f(x) and you have to ﬁnd its stationary points. For each stationary point determine if it is a local maximum or local minimum (or neither).

(a) f(x) = x2 + 6x − 3.

(b) f(x) = 2x3 − 3x2 − 12x + 5.

(d) f(x) = ln(x) + 1/x.

(e) f(x) = x/(x2 + 5x + 6).

(f) f(x) = x − sin(x)

Solutions to exercise 2

24 (a) f(x) = x2 + 6x − 3. df = 2x + 6. dx The stationary points are given by x such that df = 0 ⇒ x = −3. dx So one stationary point x = −3. Take the test points x = −4, x = −2 and we ﬁnd that the ﬁrst derivative is < 0 at x = −4 and is > 0 at x = −2. So x = −3 is a local minimum.

(b) f(x) = 2x3 − 3x2 − 12x + 5.

df = 6x2 − 6x − 12. dx The stationary points are given by x such that df = 0 ⇒ 6x2 − 6x − 12 = 0 ⇒ x2 − x − 2 = 0. dx The solutions of this quadratic equation are x = 2 and x = −1. So there are two stationary points. For x = −1 take the test points x = −2, x = 0 and we find that the first derivative is > 0 at x = −2 and is < 0 at x = 0 so a local maximum. For x = 2 take the test points x = 1, x = 3 and we find that the first derivative is < 0 at x = 1 and is > 0 at x = 3 so a local minimum.

df = 5x4 + 3x2 = x2(5x2 + 3). dx The stationary points are given by x such that df = 0 ⇒ x2(5x2 + 3) = 0 ⇒ x = 0 dx

25 as 5x2 + 3 ≥ 0. So there is one stationary point x = 0. Take the test points x = −1, x = 1 and we find that the first derivative is > 0 at x = −1 and is > 0 at x = 1 so a point of inflection.

(d) f(x) = ln(x) + 1/x. df 1 1 = − . dx x x2 The stationary points are given by x such that df x − 1 = 0 ⇒ = 0 ⇒ x = 1. dx x2 So there is one stationary point. Take the test points x = 0.5, x = 2 and we ﬁnd that the ﬁrst derivative is < 0 at x = 0.5 and is > 0 at x = 2 so a minimum. Note that we could not take x = 0 as a test point as it does not lie in the domain of f(x).

(e) f(x) = x/(x2 + 5x + 6).

df x2 + 5x + 6 − x(2x + 5) = dx (x2 + 5x + 6)2 −x2 + 6 = (x2 + 5x + 6)2 √ √ (x − 6)(x + 6) = − . (x2 + 5x + 6)2

The stationary points are given by x such that df √ √ = 0 ⇒ x = 6, x = − 6. dx

So two stationary√ points. For x = − 6, take test points, x = −2.5, −2.4 and we see that df/dx < 0 at x = −2.5 and df/dx > 0 at x = −2.4 . So df/dx√ is increasing as it goes through the stationary point, so x = − 6 is a minimum.

26 Note that we could not use x = −3, −2 as test points as the function is not deﬁned at these√ points. Similarly, for x = 6, using the test points x = 2, 3, df/dx√ is decreasing as it goes through this stationary point, hence x = 6 is a maximum.

(f) f(x) = x − sin(x). df = 1 − cos(x). dx The stationary points are given by x such that df = 0 ⇒ cos(x) = 1 ⇒ x = 2nπ, n = 0, ±1,.... dx So an inﬁnite number of stationary points. These are all points of inﬂection as we have df/dx = 1 − cos(x) > 0 for any test points either side of 2nπ. This is because cos(x) ≤ 1 for all x and cos(x) = 1 only when x = 2nπ, n = 0, ±1,... .

Graph of x − sin(x).

27 17.6.5 Choosing Test Points - a warning The following example demonstrates that you have to be careful in choosing your test points. You must make sure that the function is deﬁned at all points in the interval between the test points.

Example 4 Let f(x) = ln(x3 − 6x2 − 15x + 1). Find the stationary points and for each stationary point determine whether or not it is a local maximum, local minimum or neither. Solution.

df 3x2 − 12x − 15 = dx x3 − 6x2 − 15x + 1 3(x − 5)(x + 1) = . x3 − 6x2 − 15x + 1 The stationary points are (in theory) given by x such that

df = 0 ⇒ x = 5, x = −1. dx WARNING. Does the domain of the function include these stationary points? x = −1 is OK as f(−1) = ln(7) but f(5) = ln(125 − 150 − 75 + 1) = ln(−99) which does not exist! Hence only one possible stationary point x = −1. To determine the type of this stationary point consider the test points x = −2, x = 0. We have x = −2 ⇒ df/dx = −21 < 0 and x = 0 ⇒ df/dx = −15 < 0. Hence at first sight it seems that there is no change in sign, so neither a local max. or local min. BUT there is a problem here. In fact x = −2 is not in the domain of f(x) i.e. f(−2) is not defined as f(−2) = ln(−1) which does not exist. So we have to choose our test points carefully to make sure that f(x) is defined at all points between the test points. Instead of x = −2 choose x = −1.5 and now f(−1.5) = ln(6.625) and it is true that f(x) is defined at all values in between −1.5 and 0. So these are our test points.

28 We have x = −1.5 ⇒ df/dx = 0.7924528 > 0 and x = 0 ⇒ df/dx = −15 < 0. Hence we have a local maximum at x = −1.

Exercise 3 Let f(x) = ln(x3 − 5x2 + 7x − 2). Find the stationary points of f(x). For each stationary point determine whether it is a local max. or local min. or neither. Solutions to exercise 3

df 3x2 − 10x + 7 (3x − 7)(x − 1) = = . dx x3 − 5x2 + 7x − 2 x3 − 5x2 + 7x − 2 The stationary points are (in theory) given by x such that

df = 0 ⇒ x = 7/3, x = 1. dx WARNING. Does the domain of the function include these stationary points? x = 1 is OK as f(1) = ln(1) = 0 but f(7/3) = ln(−0.185) which does not exist! Hence only one possible stationary point x = 1. To determine the type of this stationary point consider the test points x = 0.9, x = 1.1. These points are in the domain of f(x) so they can be used. We have x = 0.9 ⇒ df/dx > 0 and x = 1.1 ⇒ df/dx < 0. Hence we have a local maximum at x = 1.

17.7 Second Derivative Test There is another way of testing a stationary point. This is based on the value of the second derivative at the stationary point. Let f(x) have the stationary point x = a.

29 17.7.1 Second Derivative Test. Let d2f v = evaluated at x = a. dx2 Then

1. v < 0 ⇒ x = a is a LOCAL MAXIMUM.

2. v > 0 ⇒ x = a is a LOCAL MINIMUM.

3. v = 0 ⇒ no further information. There may be a point of inflection. Note that the second derivative test cannot determine whether or not a stationary point is a point of inflection. We have to use the first derivative test and look at the signs at test points either side of the stationary point.

Example 5 Let f(x) = x3 − 6x2 + 9x − 2. Find all the stationary points and for each stationary point determine whether it is a local max. or local min. or neither. Solution.

We have df/dx = 3x2 − 12x + 9 = 3(x2 − 4x + 3) = 3(x − 3)(x − 1). The stationary points are given by df/dx = 0 i.e. x = 3, x = 1. Now d2f = 6x − 12. dx2

At d2f x = 1, = 6 − 12 = −6 < 0. dx2 So x = 1 is a local maximum. At d2f x = 3, = 18 − 12 = 6 > 0. dx2 So x = 3 is a local minimum.

30 Graph of x3 − 6x2 + 9x − 2.

Exercise 4 You are given the function f(x) and you have to ﬁnd its stationary points. For each stationary point determine if it is a local maximum or local minimum (or neither) using the Second Derivative Test. (Note that these are the same functions as Exercise 2 and if you have found their stationary points already you can check that the Second Derivative Test gives the same local max or min (or neither).)

(a) f(x) = x2 + 6x − 3.

(b) f(x) = 2x3 − 3x2 − 12x + 5.

(d) f(x) = ln(x) + 1/x.

(e) f(x) = x/(x2 + 5x + 6).

(f) f(x) = x − sin(x)

31 Solutions to exercise 4

(a) df = 2x + 6. dx One stationary point x = −3 We have at this stationary point that

d2f = 2 > 0. dx2 So x = −3 is a local minimum.

(b) f(x) = 2x3 − 3x2 − 12x + 5. df = 6x2 − 6x − 12. dx The stationary points are x = 2 and x = −1.

d2f = 12x − 6x. dx2 At x = 2 the second derivative is > 0 so a local minimum. At x = −1 the second derivative is < 0 so a local maximum.

d2f = 20x3 + 6x. dx2 At x = 0 the second derivative is 0 so no information as to its type. So we have to use the ﬁrst derivative test and we see from the solutions to Exercise 2 that this is a point of inﬂection.

32 (d) f(x) = ln(x) + 1/x. df 1 1 = − . dx x x2 The stationary point is x = 1.

d2f 1 2 = − + . dx2 x2 x3 At x = 1 the second derivative is > 0 so a local minimum.

(e) f(x) = x/(x2 + 5x + 6).

df x2 + 5x + 6 − x(2x + 5) = dx (x2 + 5x + 6)2 −x2 + 6 = (x2 + 5x + 6)2 √ √ (x − 6)(x + 6) = − . (x2 + 5x + 6)2 √ √ The stationary points are x = 6, x = − 6 It is much easier to use the ﬁrst derivative test as diﬀerentiating again may lead to many mistakes! √ We have see from√ the solutions to Exercise 2 that x = − 6 is a minimum and x = 6 is a maximum.

(f) f(x) = x − sin(x). df = 1 − cos(x). dx The stationary points are given by x such that df = 0 ⇒ cos(x) = 1 ⇒ x = 2nπ, n = 0, ±1,.... dx

33 So an inﬁnite number of stationary points.

d2f = sin(x) = 0 dx2 at all these stationary points. So no information using this test. However, the solution in Exercise 2 using the ﬁrst derivative test shows that these are all points of inﬂexion.

17.8 Why does the second derivative test work? 17.8.1 Local Maximum Note that for a local maximum the first derivative is going from positive to negative as x goes through the stationary point. Hence the first derivative is decreasing. But we have seen previously that if a function is decreasing its derivative is ≤ 0. Hence if a is a local maximum, the derivative of the first derivative is ≤ 0 at x = a i.e. d2f ≤ 0 at x = a. dx2

17.8.2 Local Minimum Note that for a local minimum the first derivative is going from negative to positive as x goes through the stationary point. Hence the first derivative is increasing and so its derivative is ≥ 0. Hence if a is a local minimum, the derivative of the first derivative is ≥ 0 at x = a i.e. d2f ≥ 0 at x = a. dx2

17.8.3 Which test should I use to determine the type of a stationary point? You have two tests. One is to find the signs of the first derivative before and after the stationary point. The other is to find the sign of the value of the second derivative at the

34 stationary point as detailed above. It is easier to use the second derivative test if the second derivative is easy to ﬁnd, for example in the last example where we had a polynomial. However in an example such as

f(x) = ln(x3 − 6x2 − 15x + 1) the second derivative is significantly harder to find - it is easy to make al- gebraic mistakes and it is easier to examine the changing signs of the first derivative. But you have to be careful about the test points. The next example uses both methods. Note that in this method we say classify the stationary points to mean find out whether or not they are local max, local min or neither.

Example 6 Find and classify the stationary points of x − 1 f(x) = . x2 + 8 Solution.

Using the quotient rule we have

(x2 + 8) × 1 − 2x × (x − 1) f 0(x) = (x2 + 8)2 −x2 + 2x + 8 = (x2 + 8)2 −(x − 4)(x + 2) = . (x2 + 8)2

Hence the stationary points given by f 0(x) = 0 are x = 4, x = −2.

35 Graph of (x − 1)/(x2 + 8).

Now we classify them using the diﬀerent methods.

• Method 1: Examining signs of the ﬁrst derivative.

Stationary point x = 4. The test points we choose are x = 3, x = 5 as there are no stationary points other than 4 in between them and f(x) is deﬁned at all points in the interval [3, 5]. Now f 0(3) > 0 and f 0(5) < 0 so we have a local maximum as f 0(x) is decreasing through the stationary point.

Stationary point x = −2. The test points we choose are x = −3, x = 0 as there are no stationary points other than −2 in between them and f(x) is deﬁned at all points in the interval [−3, 0]. Now f 0(−3) < 0 and f 0(0) > 0 so we have a local minimum as f 0(x) is increasing through the stationary point.

• Method 2: Sign of second derivative. Remember that if a is a stationary point then

36 (a) f 00(a) < 0 ⇒ a is a local maximum (b) f 00(a) > 0 ⇒ a is a local minimum.

First we ﬁnd f 00(x)

−x2 + 2x + 8 f 0(x) = ⇒ (x2 + 8)2 (x2 + 8)2.(−2x + 2) − (−x2 + 2x + 8).4x(x2 + 8) f 00(x) = (x2 + 8)4 (x2 + 8)[(x2 + 8)(−2x + 2) − (−x2 + 2x + 8).4x] = (x2 + 8)4 2x3 − 6x2 − 48x + 16 = . (x2 + 8)3

Stationary point x = 4. We have f 00(4) = (128 − 96 − 192 + 16)/243 < 0. So x = 4 is a local maximum, agreeing with the ﬁrst method. Stationary point x = −2. We have f 00(−2) = (−16 − 24 + 96 + 16)/123 > 0. So x = −2 is a local minimum, agreeing with the ﬁrst method.

You can judge which of the above methods is easier for this example - but note that a lot of mistakes can be made in ﬁnding the second derivative and its value at the stationary points.

37 17.9 Videos

Stationary Points 1

This video ﬁnds the stationary points of f(x) = x3 − 3x2 − 1 and classi- ﬁes them.

Stationary Points 2

The stationary points of f(x) = 3x4 + 4x3 − 5 are found and shown to be a local maximum, a local minimum and a point of inﬂection.

Stationary Points 3 √ This video ﬁnds the stationary points of f(x) = 40 + (6x3 − 56x) x and shows that you have to take into account the domain of the function.

Diﬀerentiating: Maximising a Function

In this video a formula for drug absorption is given and a maximum value is found.