1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

Notice that

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $f^{-1}(y) = -2 \sqrt y$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.