SOLUTION: solve for max/min f(x)=2sinx-1 on the interval [0,2 pi]

Question 1087366: solve for max/min f(x)=2sinx-1 on the interval [0,2 pi]
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

The largest that sin(x) can get is 1. So the max of f(x)=2*sin(x)-1 is f(x)=2*1-1=1
The smallest that sin(x) can get is -1. So the min of f(x)=2*sin(x)-1 is f(x)=2*(-1)-1=-3
---------------------------------
In summary:
Max = 1
Min = -3
Which is confirmed by the graph below

Take note that each "mountain" peaks as high as y = 1, and each "valley" bottoms out as low as y = -3. The up and down pattern you see is repeated forever in both directions along the x axis.

Graph generated by GeoGebra (free graphing software).

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