The functions f and g are such that f(x) = 3x³ + 1 for x > 0 and g(x) = \frac{4}{x²} for x > 0 (a) Work out gf(1) The function h is such that h = (fg)^{-1} (b) Find h(y) - Edexcel - GCSE Maths - Question 22 - 2021 - Paper 1

For part (b), we need to find h(y) where h is the inverse of the function fg.

1. Start by finding the expression for fg:

   $fg(x) = f(g(x))$

   Substitute for g(x) first:

   $g(x) = \frac{4}{x^2}$

   Then calculate f(g(x)):

   $f(g(x)) = f\left(\frac{4}{x^2}\right)$

   Now, substitute ( g(x) ) into f:

   $= 3\left(\frac{4}{x^2}\right)^3 + 1$

   Simplifying further:

   $= 3 \cdot \frac{64}{x^6} + 1 = \frac{192}{x^6} + 1$

Thus, we have:

$fg(x) = \frac{192}{x^6} + 1$

2. To find h(y), we solve the equation: ( y = fg(x) ),

   $y = \frac{192}{x^6} + 1$

   Rearranging gives:

   $y - 1 = \frac{192}{x^6}$

   Solving for x gives:

   $x^6 = \frac{192}{y - 1}$

   Taking the sixth root,

   $x = \left(\frac{192}{y - 1}\right)^{\frac{1}{6}}$

Thus, we find:

$h(y) = \left(\frac{192}{y - 1}\right)^{\frac{1}{6}}$