Question

Question #5baa9

Answer 1

Given
`y=f(x)=2sin(x-pi/4)`

To get x-intercepts we can put y=0 and solve for `x in [0,4pi]`

So `2sin(x-pi/4)=0`

`=>x-pi/4=npi" where " n in ZZ`

`=>x=npi+pi/4" where " n in ZZ`

Putting `n=0` we get

`x=pi/4`

Putting `n=1` we get

`x=(5pi)/4`

Putting `n=2` we get

`x=(9pi)/4`

Putting `n=3` we get

`x=(13pi)/4`

(a) So points of x-intercepts over `[0,4pi]` are

`(pi/4,0);((5pi)/4,0);((9pi)/4,0);((13pi)/4,0)`

b) For maximum value of y the value of `sin(x-pi/4)=1`

So `x-pi/4=pi/2`
`=>x= pi/4+pi/2=(3pi)/4`

For `x=(3pi)/4" " y=2`

The point(in (0,2pi) where the graph of this function reaches
maximum is `((3pi)/4,2)`