Problem: If k is a positive number and f is a function such that, for every positive number x,
[f(x2+1)]x=k
then, for every positive number y,
[f(y29+y2)]y12
is equal to
Answer Choices:
A. k
B. 2k
C. kk
D. k2
E. yk
Solution:
[f(y29+y2)]y2=⎣⎢⎢⎡[f((y3)2+1)]y3⎦⎥⎥⎤2=k2.