f(x) = cos(x - 4) - 8 what is the maximum value that this function takes?
The negative 4 shifts it to the right 4 and the -8 shifts it down 8. So you would have the maximum value(since cos(x) normally goes to 1). Of 1-8=-7
f'(x)=-sin(x-4) -sin(x-4)=0 sin(x-4)=0 remember sin0=0, sinpi=0, and the rest is repeats x-4=0 => x=4 x-4=pi => x=pi+4 but x=4 and x=pi+4 is for time around a circle (we can keep going around the circle over and over and till foreverness ends) x=4+2npi and x=pi+4+2npi for n=0,1,2,3,4,5,... now f(4)=cos(4-4)-8=cos(0)-8=1-8=-7 f(pi+4)=cos(pi+4-4)-8=cos(pi)-8=-1-8=-9 so since -9<-7 then the minimum value would be -9 and the maximum value would be -7 the minumums occur at x=pi+4+2npi maximums occur at x=4+2npi
the options that i have, are: a) 2 b)4 c)8 d)12 so, i dont know ... :/
the word value means the y value the function is below the x-axis it is impossible for f to take on any of those y-values
was what i thought.. thank you (:
if they said maximum, then it would be x=4 but they said value
but like i said above there are infinity many maximums