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**

The same thing applies to curves also.

Fig 2a.Fig 2b.

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 – x ^{2}* a) decreasing b) increasing?

Solution

*f(x)=4 – 4x – x ^{2}*

*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)= x ^{3}+4x^{2}+5x* a) decreasing b) increasing?

### Solution

*f(x)= x ^{3}+4x^{2}+5x*

*f’(x)=3x ^{2}+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*= 4t ^{3 }– 3t.*

### Solution

V*= 4t ^{3 }– 3t*

*V’=12t ^{2}-3*

*V’=3(4t ^{2}-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*