SOLUTION: please help me: The range of F in F(x)=2sinx;-pi/2<x<pi is : HERE ARE THE POSSIBLE ANSWERS: a)-2<_F(x)<_2 b)-2<_F(x)<_0 c)F(x)<_2

Algebra ->  Trigonometry-basics -> SOLUTION: please help me: The range of F in F(x)=2sinx;-pi/2<x<pi is : HERE ARE THE POSSIBLE ANSWERS: a)-2<_F(x)<_2 b)-2<_F(x)<_0 c)F(x)<_2      Log On


   

Question 242933: please help me:
The range of F in F(x)=2sinx;-pi/2 HERE ARE THE POSSIBLE ANSWERS:
a)-2<_F(x)<_2
b)-2<_F(x)<_0
c)F(x)<_2
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
If you graph f%28x%29=2%2Asin%28x%29, you get



and you can see that the min y value is -2 while the max y value is 2. So the range is -2%3C=f%28x%29%3C=2 making the answer a)


Another way you can do it is look at a table of values for 2*sin(x) and determine the max/min values.