Okay, so I just want to confirm something that's been bugging me with regards to inverse functions.

We know that a function $f^{-1}$ is the inverse of $f$ if

  • $ \forall x \in \operatorname{Dom}(f), f^{-1}[f(x)]=x $
  • $ \forall x \in \operatorname{Dom}(f^{-1}), f[f^{-1}(x)]=x$

Also, the domain of $f$ is the range of $f^{-1}$ and the range of $f$ is the domain of $f^{-1}$.

So am I right in saying that $f(x)=\frac{1}{4}x^2$ where $ x \leq 0$ has no inverse?

Normally I'd say that $f^{-1}(x)= \pm 2\sqrt{x}$
But the domain of $f$ is $x\leq 0$ therefore the range of $f^{-1}$ is $\leq 0$
So $f^{-1}(x)= - 2\sqrt{x}$ will satisfy this condition
However we now have that $f[f^{-1}(x)]=f[-2\sqrt{x}]=\frac{1}{4}(-2\sqrt{x})^2=x$
BUT $f^{-1}[f(x)]=f^{-1}[\frac{1}{4}x^2]=-2\sqrt{\frac{1}{4}x^2}=-x \ne x$

Am I right in saying that this $f(x)$ indeed has no inverse?


Notice that

$$f^{-1} \left[ \frac14 x^2\right] = -2 \sqrt{\frac14x^2}=-2 \cdot \frac12 \color{blue}{|x|}=-\color{blue}{|x|}=x$$

  • $\begingroup$ The domain of $f^{-1}$ is $x \ge 0$ so $|x|=x$ and therefore $-|x|=-x$ $\endgroup$ – KennyB Oct 7 '17 at 15:32
  • $\begingroup$ The domain of $f^{-1}$ is $y \geq 0$. $y = \frac14 x^2 \geq 0$. The domain of $f$ is $x \leq 0$. $|x|=-x$. $\endgroup$ – Siong Thye Goh Oct 7 '17 at 15:36
  • $\begingroup$ Oh yes, of course! Thank you. $\endgroup$ – KennyB Oct 7 '17 at 15:41

There is a slight misunderstanding about invertible functions.

A function $f:A \to B$ can be thought of as relations $f \subseteq A \times B$ with the special property

$$\text{$(x,a) \in f$ and $(x, b) \in f$ implies $a=b$}$$

The inverse of $f$ is defined as $f^{-1} = \{(y,x) : (x,y) \in f \} \subseteq B \times A$, and it always exists; but it is not always a function. A function is invertible if its inverse is also a function.

For the function \begin{array}{cclc} f: & (-\infty, \infty) &\to &[0, \infty) \\ & x &\mapsto &\frac 14x^2 \end{array}

$f^{-1}$ is not a function since $(\frac 14x^2, x) \in f^{-1}$ and $(\frac 14x^2, -x) \in f^{-1}$.

For the function \begin{array}{cclc} f: & [0, \infty) &\to &[0, \infty) \\ & x &\mapsto &\frac 14x^2 \end{array}

$f^{-1}$ is a function and $f^{-1}(y) = 2 \sqrt y$.

For the function \begin{array}{cclc} f: & (-\infty, 0] &\to &[0, \infty) \\ & x &\mapsto &\frac 14x^2 \end{array}

$f^{-1}$ is a function and $f^{-1}(y) = -2 \sqrt y$.

  • $\begingroup$ They should make this definition more readily available in texts then. Also, in this example $f:(-\infty,0] \rightarrow [0,\infty)$. Surely that changes things as mentioned. $\endgroup$ – KennyB Oct 8 '17 at 13:27

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