The functions f and g are such that f(x) = 3x^3 + 1 for x > 0 and g(x) = \frac{4}{x^2} 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

To find h(y), we first express h in terms of f and g:

1. Given that h = (fg)^{-1}, we need to find fg(x): $fg(x) = f(g(x))$

2. First, compute g(x): $g(x) = \frac{4}{x^2}$

3. Next, substitute g(x) into f: $f(g(x)) = f\left(\frac{4}{x^2}\right) = 3\left(\frac{4}{x^2}\right)^3 + 1$ $= 3\left(\frac{64}{x^6}\right) + 1 = \frac{192}{x^6} + 1$

4. Now we have fg(x): $fg(x) = \frac{192}{x^6} + 1$

5. To find the inverse, we let y = fg(x): $y = \frac{192}{x^6} + 1$

6. Rearranging for x gives: $y - 1 = \frac{192}{x^6} \Rightarrow x^6 = \frac{192}{y - 1} \Rightarrow x = \left(\frac{192}{y - 1}\right)^{1/6}$

Thus, h(y) is: $h(y) = \left(\frac{192}{y - 1}\right)^{1/6}$