**Application of rate change**

Brief set of examples concentrating on the rate of change application of derivatives is given in this section.

**Example** Find out all the points where the given function is not changing.

g ( x ) = 5 - 6 x -10 cos ( 2 x )

**Solution : **First we'll have to take the derivative of the function.

g′ ( x ) = -6 + 20 sin ( 2x )

Now, the function will not be altering if the rate of change is zero & thus to answer this question we have to determine where the derivative is zero. Thus, let's set this equivalent to zero & solve.

-6 + 20 sin ( 2 x ) = 0 ⇒ sin ( 2 x ) =6/20 = 0.3

Then the solution to this is,

2x = 0.3047 +2 ∏ n OR 2x = 2.8369 + 2 ∏ n n = ±1, ±2,....

x = 0.1524 + ∏ n OR x = 1.4185 + ∏ n n = 0, ±1, ±2,......