Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let the function $y=f(x)=(1+x^2)sgn(x)$. Prove that $y^{-1}$ is a continuous function.

I know that the $y$ is not continuous, but don't know how to treat the problem at first. Thank you for your help.

share|improve this question
    
What is $sgn$??? Is it the sign function? –  Babak S. Jan 30 '13 at 7:15
    
Yes, Babak. its –  Basil R Jan 30 '13 at 7:19

2 Answers 2

up vote 4 down vote accepted

Your notation is a bit confusing. I think you mean that $f^{-1}$ (i.e. the inverse function, not the reciprocal) is a continuous function. Thus you're writing $x = f^{-1}(y)$.

There are three cases to consider: $x > 0$, $x < 0$ and $x = 0$. When $x > 0$, $y = 1 + x^2 > 1$ and $x = \sqrt{y-1}$. This is a continuous function on $(1,\infty)$. Similarly, when $x < 0$, $y = -1 - x^2 < -1$ and $x = -\sqrt{-1 - y}$. This is a continuous function on $(-\infty, -1)$. Finally, when $x = 0$, $y = 0$. Thus $$ f^{-1}(y) = \cases{\sqrt{y-1} & for $x > 1$\cr 0 & for $x = 0$\cr -\sqrt{-1-y} & for $x < -1$\cr \text{undefined} & otherwise\cr}$$ Since the three sets $(-\infty, -1)$, $\{0\}$ and $(1,\infty)$ are separated, $f^{-1}$ is continuous on its domain which is their union.

share|improve this answer
    
+1. I think you want $y$'s instead of $x$'s in your expression for $f^{-1}$ (e.g. $\text{ for } y\geq 1$ instead of $\text{ for } x\geq 1$). –  Stefan Hansen Jan 30 '13 at 7:30
    
Thank you robert. –  Basil R Jan 30 '13 at 7:30
    
And you want strict inequalities: $f^{-1}$ is undefined at $\pm1$. –  Marc van Leeuwen Jan 30 '13 at 7:35
    
@MarcvanLeeuwen: yes, of course. I edited it. –  Robert Israel Jan 30 '13 at 18:33

The function $f$ is discontinuous (only) at $0$, and since $1+x^2\geq1$ for all $x\in\mathbf R$, the range of $f$ is contained in $(-\infty,-1)\cup\{0\}\cup(1,+\infty)$. As $f$ is strictly increasing and tends to $\pm\infty$ as $x\to\pm\infty$, the range of $f$ is in fact equal to that set, and $f$ is a bijection from $\mathbf R$ to that range.

The inverse of $f$ is therefore a well defined function $ (-\infty,-1)\cup\{0\}\cup(1,+\infty)\to\mathbf R$; continuity is a local property, and therefore determined separately on each of these three connected components of the domain of $f^{-1}$. The component $\{0\}$ is discrete (it is an isolated point), so continuity there is vacuously satisfied. For the other two components, we are dealing with the inverse of a restriction of $f$ to $(-\infty,0)$ respectively to $(0,+\infty)$, and these restrictions are continuous and strictly increasing. You should know how to prove that the inverse of such a function is continuous, and this settles the continuity of $f^{-1}$ everywhere on its domain.

share|improve this answer
    
Let $D \subset \mathbb{R}$ and let $x \in D$. Then continuity at $x$ is vacuous -- i.e., every $f: D \rightarrow \mathbb{R}$ is continuous at $x$ -- iff $x$ is an isolated point of $D$. This holds in the case at hand...but $D$ having empty interior is not enough. –  Pete L. Clark Jan 30 '13 at 7:41
    
@PeteL.Clark: Thanks, you're right. I'll correct. –  Marc van Leeuwen Jan 30 '13 at 7:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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