Find the period of the function y = 3/2 tan(1/3^x).
А) pi
B) pi/3
C) 3pi
D pi/6

Explanation:

I'm assuming you meant to say

y = (3/2)*tan( (1/3)x )

If so, then that equation is in the form

y = A*tan(Bx)

The B coefficient is B = 1/3 and it directly ties together to the period T.

T = pi/B

T = pi/(1/3)

T = pi*(3/1)

T = 3pi answer is choice C

Side note: This formula only works for tangent and cotangent functions.

can you post the graph?