What is the x-coordinate for the minimum point in the function f(x) = 4 cos(2x − π) from x = 0 to x = 2π?

x = pi over 2
x = pi over 4
x = 2π
x = π

Question

What is the x-coordinate for the minimum point in the function f(x) = 4 cos(2x − π) from x = 0 to x = 2π?

 x = pi over 2
 x = pi over 4
 x = 2π
 x = π

anonymous · Answer

@aloud @phi @mathmate

phi · Answer

do you know what cos looks like from 0 to 2pi? https://www.google.com/search?q=cos%28x%29&gbv=2&sei=FSSEVeKzLIevsAWqqoKYAQ

anonymous · Answer

i thought it intercepts at -4

phi · Answer

the smallest value will be y= -4
for a "normal cosine"  from 0 to 2pi, the minimum value is at pi (exactly half way from 0 to 2pi)

anonymous · Answer

no I dont know how to graph them that well

phi · Answer

so you want to find where the "angle"  of your problem
 f(x) = 4 cos(2x − π) from x = 0 to x = 2π
equals pi

in other words, what is "x" so that
2x - pi = pi

anonymous · Answer

would you plug in the x=0 for the x in 2x-pi=pi

anonymous · Answer

like substitution

phi · Answer

The easiest way is to look at a graph: https://www.google.com/search?q=cos%28x%29&gbv=2&sei=FSSEVeKzLIevsAWqqoKYAQ#q=4+cos%282x+%E2%88%92+%CF%80%29

phi · Answer

but using algebra
2x - pi = pi
add +pi to both sides
2x - pi + pi = pi+pi

on the left  -pi+pi add up to zero.  on the right you get 2pi
2x = 2pi
now divide both sides by 2

anonymous · Answer

it would be x=pi because the 2's would cancel right?