Find $h(x)=[1,4[$

Question

Let $f: \mathbb R \longrightarrow \mathbb R$ and $g: \mathbb R \longrightarrow \mathbb R$ given by $$f(x)=x^2 \quad \text{and} \quad g(x)=x+5$$ Consider the function $$h: \mathbb R \longrightarrow \mathbb R$$ given by $h(x)=f \circ g$. What is the ur $h^{-1}([1,4[)$ ?

The equation I have to solve is $h(x)=[1,4[$. I do not know how to solve it analytically so I tried to solve the problem by drawing.

Q1: $h(x)$ is a number but $[1,4[$ is an interval (set). How can I make sense of this equation? A number and a set can never be the same.

Q2: What theorems, definitions etc. should I use to solve it analytically?

Answer 1

For each $x\in\mathbb R$,\begin{align}x\in h^{-1}\bigl([1,4)\bigr)&\iff h(x)\in[1,4)\\&\iff(x+5)^2\in[1,4)\\&\iff x+5\in[1,2)\text{ or }x+5\in(-2,-1]\\&\iff x\in[-4,-3)\text{ or }x\in(-7,-6].\end{align}

Answer 2

$h = f \circ g$, so $h^{-1} = g^{-1} \circ f^{-1}$. Then \begin{align*} h^{-1}([1,4[) &= g^{-1} \circ f^{-1}([1,4[) \\ &= g^{-1}(]-2,-1] \cup [1,2[) \\ &= ]-7,-6] \cup [-4,-3[ \text{.} \end{align*}