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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

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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.

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

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  • $\begingroup$ But isn't against the rule that a function must not be many to one. For same y value you are getting 2 values right? $\endgroup$ – emil Aug 19 '17 at 11:20
  • $\begingroup$ yes of Course, but your function $f(x)=x^2-4x$ is not monoton. so you will get two inverse $\endgroup$ – Dr. Sonnhard Graubner Aug 19 '17 at 11:24
  • $\begingroup$ it is a Parabel in the form $$f(x)=(x-2)^2-4$$ $\endgroup$ – Dr. Sonnhard Graubner Aug 19 '17 at 11:28
  • $\begingroup$ We are given that the domain of $f$ is $[2, \infty)$, so we must take the positive root. $\endgroup$ – N. F. Taussig Aug 19 '17 at 13:03
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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. Consequently, the graphs of $f$ and $f^{-1}$ are symmetric with respect to the line $y = x$, as shown in the figure below.

graph_of_quadratic_function_with_restricted_domain_and_its_inverse

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