Finding the inverse of the function $f$ defined by $f(x) = x^2 - 4x$ for $x \geq 2$

Question

Let $f$ be the function defined by $$f(x)=x^2-4x$$ for $x\ge 2$. Find the inverse function indicating its domain and the range.

When I try to make $x$ the subject I get $+$and $-$ two answers for the inverse function. Is it the proper answer? The answer I got is $$ x=2 \pm \sqrt{y+4}.$$ Is it correct and how do I determine the domain and range of the function? Thanks

Answer 1

If You understand $f$ to be a function on $\mathbb{R}$ it is not invertible but the restrictions $f|_{\mathbb{R}^{\leq 2}}$ and $f|_{\mathbb{R}^{\geq 2}}$ are invertible with inverses $g_1:\mathbb{R}^{\geq-4}\rightarrow \mathbb{R}^{\leq 2},x\mapsto 2-\sqrt{x+4}$ and $g_2:\mathbb{R}^{\geq-4}\rightarrow \mathbb{R}^{\geq 2},x\mapsto 2+\sqrt{x+4} $ respectively.

Answer 2

yes we get $$x_{1,2}=2\pm \sqrt{4+y}$$ and it must be $$4+y\geq 0$$

Answer 3

The function $f$ defined by $f(x) = x^2 - 4x$ for $x \geq 2$ has domain $\text{Dom}_f = [2, \infty)$. We can find its range by completing the square. \begin{align*} f(x) & = x^2 - 4x\\ & = (x^2 - 4x + 4) - 4\\ & = (x - 2)^2 - 4 \end{align*} The graph of $y = (x - 2)^2 - 4$ is a parabola with vertex $(2, -4)$ that opens upwards. The restriction that $x \geq 2$ means that the graph consists only of the right half of the parabola. Since the graph is continuous and increases without bound as $x$ approaches $\infty$, the range of $f$ is $\text{Ran}_f = [-4, \infty)$.

Solving for the inverse yields \begin{align*} y & = (x - 2)^2 - 4\\ y + 4 & = (x - 2)^2\\ \sqrt{y + 4} & = |x - 2| \end{align*} Since $x \geq 2$, $|x - 2| = x - 2$. Thus, \begin{align*} \sqrt{y + 4} & = x - 2\\ 2 + \sqrt{y + 4} & = x \end{align*} Therefore, the inverse function is $$f^{-1}(x) = 2 + \sqrt{x + 4}$$ The domain of the inverse is $\text{Dom}_{f^{-1}} = [-4, \infty)$ and the range of the inverse is $\text{Ran}_{f^{-1}} = [2, \infty)$.

Notice that the domain of the function is the range of its inverse and the range of the function is the domain of its inverse.

Since the point $(a, b)$ is on the graph of $f$ if and only if the point $(b, a)$ is on the graph of $f^{-1}$, the graphs of $f$ and $f^{-1}$ are symmetric with respect to the line $y = x$, as shown in the figure below.