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,......