Question

A function f is defined by f(x) 5-2sin2x For 0<x<pie a)find the .range of f. b)sketch the graph of y=f (x)?

Answer 1

The range is `f(x) in [3,7]`. See the graph below

Explanation:

To find the range, proceed as follows :

`-1 <= sinx <=1`

`-1 <= sin2x <=1`

`1 >= -sin2x >=-1`

`2 >= -2sin2x >=-2`

`2+5 >= (5-2sin2x )>=-2+5`

`3<= f(x)=(5-2sin2x) <=7`

Therefore,

the range is `f(x) in [3,7]`

To sketch the graph in the domain `x in (0, pi)`

Calculate the following values

`color(white)(aaaa)` `x` `color(white)(aaaa)` `f(x)`

`color(white)(aaaa)` `0` `color(white)(aaaa)` `5`

`color(white)(aaaa)` `pi/4` `color(white)(aaaa)` `3`

`color(white)(aaaa)` `pi/2` `color(white)(aaaa)` `5`

`color(white)(aaaa)` `3/4pi` `color(white)(aaaa)` `7`

`color(white)(aaaa)` `pi` `color(white)(aaaa)` `5`

`5-2sin2x=6`

`2sin2x=-1`

`sin2x=-1/2`

`2x=7pi/6`, `=>`, `x=7/12pi`

`2x=11/6pi`, `=>`, `x=11/12pi`

See the graph below

graph{(y-5+2sin(2x))(y-6)=0 [-10, 10, -5, 5]}