Finding $h$ such that $h(U)$ has pdf $f(x) = \frac{3}{x^4} \mathbb{1}_{[1, +\infty]}$ when $U$ is uniform

Question

Let $U$ be a random variable distributed uniformly in $[0,1]$.

My task is to find function $h$ such that $Y = h(U)$ has density function denoted by $$f(x) = \frac{3}{x^4} \mathbb{1}_{[1, +\infty]}.$$

This is my attempt.
I know that $$f_Y(y) = \left| \frac{d}{dy} h^{-1} \right|$$ That observation leads to $$h^{-1}(y) = \frac{-1}{x^3} + C \text{ or } h^{-1}(y) = \frac{1}{x^3} + C$$ Is my attempt correct? How can I find the constant $C$?

Answer 1

Your attempt is not correct, and I think it is better here to switch from PDF to CDF.

The CDF $F$ that corresponds with PDF $f$ satisfies: $$F\left(x\right)=\int_{-\infty}^{x}f\left(u\right)du$$ leading here to: $$F\left(x\right)=1-x^{-3}\text{ if }x\geq1\text{ and }0\text{ otherwise.}$$

Then for $u\in\left(0,1\right)$ we find: $$h\left(u\right)=\inf\left\{ x\in\mathbb{R}\mid1-x^{-3}\geq u\right\} =\left(1-u\right)^{-\frac{1}{3}}$$

If it is unclear to you how I arrive here then have a look at point 3 in this answer where $h$ is interchanged with $\Phi$.