for functions like y = a sin(kx) and y = a cos(kx)
the amplitude is a, that is, y lies between -a and +a
and the period is 2π/k
so for yours 2π/k = 60
k = 2π/60 = π/30
we are also supposed to have f(0) = 8, so it could be a cosine function
f(x) = 8cos(πx/30)
for g(0)=0, let's use a sine curve, since it starts at (0,0)
g(x) = 8sin(πx/30)
for f(x) = 5
8cos (πx/30) = 5
cos (πx/30) = .625
set your calculator to radians (RAD),
and I get πx/30 = .89566... , stored it in memory
x = (30/π)(.89566...) = 8.553 correct to 3 decimals
but the cosine is positive in I or IV
so x = 2π - 8.553 = -2.2398.. , a negative, we don't want that
but if we add a period to it, we should get another
x = -2.2398 + 60 = 57.730
another solution would occur one period later, so
x = 60 + 8.553 = 68.553
so the three smallest are: x = 8.553, 57.730, and 68.553
Proof:
http://www.wolframalpha.com/input/?i=plot+f(x)+%3D+8cos(%CF%80x%2F30)+,+0+%3C+x+%3C+100
There are many examples of graphs that have period 60 and whose y-values vary between -8 and 8. First, find a formula for such a function f, given that f(0)=8. Next, find a formula for such a function g, given that g(0)=0.
For your f, also find the three smallest positive x that makes f(x)=5
I don't know where to start for this problem...
2 answers
Okay, that makes so much more sense. Thank you so much!