25 Increasing and Decreasing Functions

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1 - 25 Increasing and Decreasing Functions It is useful in mathematics to define whether a function is increasing or decreasing. In this section we will use the differential of a function to determine this for a range of given values. By the end of this section, you should have the following skills: An understanding of the definition of the Newton Quotient and its use in determining properties of functions. An understanding of the definition of increasing, decreasing and monotonic functions. Use the differential of a function to determine whether it is increasing or decreasing Increasing or Decreasing We can use the derivative to find out whether or not a function is increasing or decreasing over a range of values The Newton Quotient Let f(x) be a function. Let x and x + h belong to the domain of the function then the Newton Quotient is: f(x + h) f(x) N(h) =. h We used the Newton Quotient previously to define the derivative by taking the limit as h 0, and we use it here to show how the derivative can give information on whether or not the function is increasing or decreasing Increasing Functions We say that a function f(x) is increasing in the interval (a, b) if x y f(x) f(y) for all such x, y in (a, b). 1

2 Decreasing Functions We say that a function f(x) is decreasing in the interval (a, b) if x y f(x) f(y) for all such x, y in (a, b) Monotonic Functions A function is called monotonic on an interval if it is either increasing or decreasing on the interval The Newton Quotient and Monotonic Functions The Newton Quotient, given by the equation: N(h) = satisfies the properties below. f(x + h) f(x) h 1. f(x) increasing in the interval (a, b) N(h) 0. Proof. If x is in (a, b) and h is such that x + h is also in (a, b) then: Case 1: h > 0 f(x + h) f(x) 0 N(h) 0. Case 2: h < 0 f(x + h) f(x) 0 N(h) f(x) decreasing in the interval (a, b) N(h) 0. Proof. If x is in (a, b) and h is such that x + h is also in (a, b) then: Case 1: h > 0 f(x + h) f(x) 0 N(h) 0. Case 2: h < 0 f(x + h) f(x) 0 N(h) 0. Hence as lim h 0 N(h) = f (x) and: 1. N(h) 0 for all h lim h 0 N(h) 0 f (x) N(h) 0 for all h lim h 0 N(h) 0 f (x) 0. 2

3 The Differential and Monotonic Functions We obtain the result: 1. f (x) 0 over a range of values of x f(x) increasing over that range. 2. f (x) 0 over a range of values of x f(x) decreasing over that range. Example 1 Over what range of values is f(x) = x 2 2x+3 increasing? Solution. We have f (x) = 2x 2 and f (x) 0 2x 2 0 x 1. Hence f(x) is increasing for x 1. Similarly you can show that f(x) is decreasing for x 1. Of course this is a quadratic and we already know that by completing the square we can find the minimum value which is 2 at x = 1 i.e. x 2 2x + 3 = (x 1) Graph of x 2 2x + 3. Example 2 Over what range of values is f(x) decreasing? f(x) = x 3 x 2 5x

4 Solution. We have f (x) = 3x 2 2x 5 = (x + 1)(3x 5). f(x) is decreasing over a range of values if f (x) 0 over that range. Solving the quadratic inequality we have f (x) 0 (x + 1)(3x 5) 0 1 x 5/3. Hence f(x) is decreasing for 1 x 5/3. Graph of x 3 x 2 5x + 3. Example 3 Over what range of values is f(x) decreasing? Solution. Using the Quotient Rule we have f(x) = x 1 x f (x) = x2 + 2x + 3 (x 2 + 3) 2 = (1 + x)(3 x) (x 2 + 3) 2. 4

5 f(x) is decreasing over a range of values if f (x) 0 over that range. Since (x 2 + 3) 2 > 0 for all x we have that f (x) 0 (1 + x)(3 x) 0 (x + 1)(x 3) 0 on changing signs. This is another quadratic inequality and we have f (x) 0 x 1 or x 3. Hence f(x) is decreasing for x 1 or x 3. Graph of (x 1)/(x 2 + 3). Exercise 1 (a) Over what range of values is f(x) = 3x 2 + x 100 increasing? (b) Over what range of values is f(x) = x 3 + 2x 2 x + 6 decreasing? 5

6 (c) Over what range of values is f(x) increasing? f(x) = 2x 1 x + 3. (d) Over what range of values is f(x) increasing? f(x) = x 2 x 2 3. (e) Over what range of values is f(x) decreasing? f(x) = x 1 x Solutions to exercise 1 (a) We have f (x) = 6x + 1 and f (x) 0 6x x 1/6. Hence f(x) is increasing for x 1/6. Similarly you can show that f(x) is decreasing for x 1/6. Of course this is a quadratic and we already know that by completing the square we can find the maximum value which is 1199/12 at x = 1/6 i.e. 3x 2 + x 100 = 3(x 1/6) /12. 6

7 Graph of 3x 2 + x 100 7

8 (b) We have f (x) = 3x 2 + 4x 1 = (x 1)(3x 1). f(x) is decreasing over a range of values if f (x) 0 over that range. Using our earlier work on inequalities we have f (x) 0 (x 1)(3x 1) 0 (x 1)(3x 1) 0 x 1 or x 1/3. Hence f(x) is decreasing for x 1 or x 1/3. Graph of x 3 + 2x 2 x

9 (c) Differentiating we obtain df dx = 2(x + 3) (2x 1) (x + 3) 2 = 7 (x + 3) 2. This is always positive and hence the function is always increasing on its domain R { 3}. (d) Using the Quotient Rule we have Graph of (2x 1)/(x + 3). f (x) = (x2 3) (x 2)(2x) (x 2 3) 2 = (1 x)(x 3) (x 2 3) 2. f(x) is increasing over a range of values if f (x) 0 over that range. Since (x 2 3) 2 > 0 for all x ± 3 we have that f (x) 0 (1 x)(x 3) 0 (x 1)(x 3) 0 9

10 on changing signs. We then have f (x) 0 1 x 3. Hence f(x) is increasing for 1 x 3. (e) We have Graph of (x 2)/(x 2 3)). x2 + 2 (x 1) f (x) = x = (x2 + 2) (x 1)x (x 2 + 2) 3/2 = x + 2 (x 2 + 2). 3/2 x x 2 +2 As (x 2 +2) 3/2 > 0 for all x we have that f(x) is decreasing if f (x) 0 i.e. if x x 2. So we have shown that f(x) is decreasing for x 2. 10

11 Graph of (x 1)/ x