(0, 2π). f(x) = x + 2 sin x

Question

 Apply the First Derivative Test to identify the relative extreme. relative maximum(x,y) relative minimum(x, y)?

Answer 1

First derivative test

f'(x) = 1+2cos(x) on (0, 2pi) 

set to 0 

cos(x) = -1/2

 

on the interval (0, 2pi) this only happens when X= 2pi/3 and X = 4pi/3

 

to determine whether relative max or min look at sign of derivative on 

(0, 2pi/3), (2pi/3, 4pi/3), (4pi/3, 2pi)

 

1.) try X = pi/2

f' = 1> 0

2) try pi

f' = -1 < 0

3.) try 3pi/2

f' = 1 > 0

 

since at X = 2pi/3 slope goes from positive to negative that is a relative max

since at X = 4pi/3 slope goes from negative to positive that is a relative min.

 

plug into f(X)

relative max (2pi/3, 2pi/3+sqr(3))

relative min (4pi/3, 4pi/3-sqr(3))

 

best - david