Help with trig- determining period, domain, and range?

Question

Let y= -4cos (4x-pi) + 6

Determine the period, domain, and range of the function

Any help on this would be appreciated. Please explain completely so that I fully understand and can do similar problems on my own. Thanks so much and 10 points to a best answer.

Answer 1

y= -4cos (4x-pi) + 6

Factor 4 out of (4x - pi)

y = - 4 cos 4 ( x - pi/4) + 6

The period is (2pi) / 4 or pi/2

Domain is all reals because you can substitute any number in for x.

Range is 2 <= y <= 10 because the vertical shift is 6 and the amplitude is 4, so the graph will go up 4 and down 4 units from the 6.  The negative sign in front of the amplitude doesn't affect the range, it just flips the graph over so you will start at a minimum instead of a max when you graph.  I hope this helps !

Answer 2

y=-4cos(4x-pi)+6=>

y=-4cos(pi-4x)+6=>

y=4cos4x+6  (simplify the equation)

The domain of x is

D={x: -inf.<x<inf.}

The range of y=f(x) is

R=[y:2=<y<=10}

Let T be the period,then f(x)=f(x+T),

following this concept, we may write

4cos4(x+T)+6=4cos4x+6=>

sin(4x+2T)sin2T=0=>

T=pi/2