Let p(x)=x+0.5sinx. On what intervals is p(x) increasing, and on what intervals is it decreasing?

Question

Let p(x)=x+0.5sinx. On what intervals is p(x) increasing, and on what intervals is it decreasing?

Find and classify the local exrema of p(x).

Answer 1

Take the derivative:

p'(x) = 1 + 0.5cos(x)

To find critical points, 1) find any places where the derivative DNE (none in this case) and 2) set the derivative equal to zero and solve:

1 + 0.5cos(x) = 0

0.5cos(x) = -1

cos(x) = -2

This has no solution therefore there are no critical points

In looking at the derivative, we see that since cos(x) oscillated between +1 and -1, that means 0.5cos(x) will oscillate between +1/2 and -1/2 and, since 1 plus those values is always positive, the function is increasing for all values of x. Increasing: (-∞, ∞) with no local extrema.

Answer 2

The derivative of x + .5sinx = 1+.5cosx

1+.5cosx is always positive: -1/2 <= 1 + .5cosx <= 1/2

The function is increasing from (-inf,inf)

It is never decreasing.

If you graph it, you'll see it's always increasing.