Decreasing and Increasing functions

Recall that the derivative of a function gives us the gradient of that function. Also notice that the function in fig 1a is increasing whilst the function in fig 1b is decreasing moving from left to right.

Fig 1a                                                                                                                                                                              Fig 1b

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The same thing applies to curves also.

Fig 2a.                                                                                                                                                                    Fig 2b.

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In fig 2a, the given function is decreasing on the interval, x<0, and has negative gradient. The same function is increasing on the interval x>0, and has positive gradient on this interval.

Similarly the function in fig 2b is increasing, and has a positive gradient on the interval x<0. The same function is decreasing and has a negative gradient on the interval x>0.

It can therefore be inferred from the above diagrams that , an increasing function has a positive gradient whilst a decreasing function has a negative gradient. This is always true.

We can therefore use differentiation to determine whether a function is increasing or decreasing, and the range of values for which it is increasing or decreasing.

Worked Example 1

For what values of x is the function f(x)=4 – 4x – x2  a) decreasing b) increasing?

Solution

f(x)=4 – 4x – x2 

f’(x)= – 4 – 2x

a) f(x) is decreasing if f’(x) < 0.

i.e   – 4 – 2x <0

-2x < 4

x > –2


b) f(x) is increasing if f’(x) > 0.

i.e   – 4 – 2x > 0

-2x < 4

x < –2

Worked Example 2

For what values of x is the function f(x)= x3+4x2+5x    a) decreasing b) increasing?

Solution

f(x)= x3+4x2+5x

f’(x)=3x2+8x+5 

f’(x)=(x+1)(3x+5)

a) (x+1)(3x+5)<0

-5/3<x<-1

b) (x+1)(3x+5)>0

x<-5/3 or x>-1


Worked Example 3

Find the range of  values of t for which V is increasing if  V= 4t3 – 3t.

Solution

V= 4t3 – 3t

V’=12t2-3

V’=3(4t2-1)

V’=3(2t – 1)(2t+1)

V is increasing if V’>0.

3(2t – 1)(2t+1)>0

(2t – 1)(2t+1)>0

t < -0.5 or t > 0.5