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)?

1 Answer
Nov 18, 2017

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]}