How do you decompose these functions?

Question

1. h(x) = 3x^2+1/x^6-8

 

2. h(x)= squareroot(4x^2-13)

 

please include steps on how you got your answer. Thank you.

 

Answer 1

1.  h(x) = (3u+1) / (u³ - 8), where u = x².

 

     Let f(x) = (3x+1)/(x³-8) and g(x) = x²

 

     Then f(g(x)) = f(x²) = (3x²+1) / ((x²))³ - 8) = (3x²+1) / (x⁶-8) = h(x)

 

 

2.  h(x) = √(4x² - 13)

 

     Let f(x) = √(4x - 13) and g(x) = x²

 

     Then f(g(x)) = f(x²) = √(4x²-13) = h(x)