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I have a problem from a textbook involving functions. The function $f(x)$ is defined by $f(x)=x^2-3$ for $[x\in\mathbb{R}, x\ge0]$. Here is the problem: Find the values of $x$ such that $f(x)={f^{-1}(x)}$. Obviously ${f^{-1}(x)}=\sqrt{x+3}$ for $[x\in\mathbb{R}, x\ge-3]$. Then, $f(x)={f^{-1}(x)}$, so $x^2-3=\sqrt{x+3}$, and I can solve this to find the solution.

However in the solution given by the textbook it is stated that $f(x)={f^{-1}(x)}$, then $f(x)=x$ and $x^2-3=x$, which gives the same solution. However, I do not understand why $f(x)=x$. How is this justified? I am sure that I am missing something obvious. To me it would seem that $ff(x)={ff^{-1}(x)}$ so ${f^{2}(x)}=x$.

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  • $\begingroup$ @anonymouse I would vote for that as an answer. $\endgroup$
    – David K
    Aug 10, 2016 at 18:01

2 Answers 2

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Migrated and expanded from a comment.

Geometrically, you can think of inverting a function as reflecting it about the line $y = x$ (I'll leave the fuzzy details as to why to you; as a starting point, remember that the algebraic algorithm for inverting a function is "replace all the $x$'s with $y$'s and all the $y$'s with $x$'s").

If a function and its inverse have an intersection point, it must occur on the line $y = x$. This is because the only points that are fixed under reflection about $y = x$ are exactly the points on the line $y = x$. (Draw a point in the plane and reflect it to see where it ends up.)

So, the following are all equivalent:

$$f(x) = f^{-1}(x), f(x) = x, f^{-1}(x) = x$$

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    $\begingroup$ "If a function and its inverse have an intersection point, it must occur on the line $y=x$" Is this always true? Suppose $f(x) = -x, f(x) = f^{-1} (x)$ for all $x.$ $\endgroup$
    – Doug M
    Aug 10, 2016 at 18:30
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    $\begingroup$ The correct statement is: if $f(x)=x$, then $f(x) = f^{-1}(x)$. The converse does not hold in general. $\endgroup$
    – Nex
    Aug 10, 2016 at 19:20
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$$f(x)=x^2-3$$ For clarity, let us denote, $$f(x)=y$$ Then, $$y=x^2-3$$ $$x=\sqrt{y+3}=f^{-1}(y)$$ So, when you write, $$f^{-1}(x)=\sqrt{x+3}$$ you are just interchanging the symbols of $y$ and $x$. Keeping the symbols as it is, it reduces to, $$f(x)=x$$

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