I will test b), you do the others
interval: π/4 <= x <= π/2
function: f(x) = y = 3cos (4πx + π) - 3 <---- I have a feeling you have a typo and meant:
y = 3cos (4x + π) - 3
I will assume as such
f(π/2) = 3cos (4(π/2) + π) - 3 = -3-3 = -6
f(π/4) = 3cos(4(π/4) - 3 = - 6
avg rate of change = (-6 - (-6))/(π/2 - π/4) = 0
looks like I got lucky and picked the right one
I suggest to do the others to see if more than one answer was possible
Which interval gives an average change of 0 for the function y = 3cos (4πx + π) - 3?
a) π/3 <= x <= π/2
b) π/4 <= x <= π/2
c) π/2 <= x <= 3π/2
d) π/3 <= x <= π/2
2 answers
I have the same question as Craig, however with no typo