Chapter 4.3, Problem 18E

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516

Chapter
Section

### Calculus: Early Transcendental Fun...

7th Edition
Ron Larson + 1 other
ISBN: 9781337552516
Textbook Problem

# Intervals on Which a Function Is Increasing or Decreasing In Exercises 11-22, find the open intervals on which the function is increasing or decreasing. f ( x ) = sin 2 x + sin x , 0 < x < 2 π

To determine

To calculate: The open intervals for which the provided function f(x)=sin2x+sinx,  0<x<2π is increasing or decreasing.

Explanation

Given:

The function f(x)=sin2x+sinx,Â Â 0<x<2Ï€

Formula Used:

For any function g(x) that is differentiable in the interval (a,b) and continuous in the interval [a,b], the function g(x) can be classified as an increasing or a decreasing function based on the following conditions:

If d(g(x))dx>0 for the interval (a,b), the function g(x) is said to be increasing in the interval [a,b].

If d(g(x))dx<0 for the interval (a,b), the function g(x) is said to be decreasing in the interval [a,b].

Calculation:

Differentiate the provided function and equate it to zero to obtain the critical points:

d(f(x))dx=sin2x+cosxsin2x+cosx=0cosx(2sinx+1)=0x=Ï€2,7Ï€6,3Ï€2,11Ï€6

These critical points give five open intervals (0,Ï€2),(Ï€2,7Ï€6)(7Ï€6,3Ï€2),(3Ï€2,11Ï€6),(11Ï€6,2Ï€).

Now take test points to see the behaviour of the function in these intervals.

Let Ï€4âˆˆ(0,Ï€2).

f'(Ï€4)=sin2(Ï€4)+cos(Ï€4)=1+12>0

Hence, the function is increasing in the interval (0,Ï€2)

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