1
$\begingroup$

Question

How to determine the interval for the domain of an inverse function of a quadratic function

For an example the graph of $y=x^2$ 's inverse has the domain of x is larger or equal to 0 using algebraic methods. (if we were to use graphic it would be obvious)

I was wondering how we determine such interval of the inverse of a quadratic

3 Answers 3

3

A function is invertable on an interval for which it is injective. A function is injective when $\forall a,b,f(a)=f(b)\Rightarrow a=b$ (think about why this would guarantee an inverse!).

It's not too hard to show that a quadratic equation is injective on an interval if and only if the interval is entirely on one side of its local min/max. The $x$ coordinate of the min/max is given by $x=\frac{-b}{2a}$, so the quadratic is invertable on the intervals $(-\infty, \frac{-b}{2a}]$ and $[\frac{-b}{2a},\infty)$, as well as any smaller interval that is properly contained in one of those intervals.

The domain of the inverse function is obtained by just applying $f$ to these intervals. That gives $(-\infty, c-\frac{b^2}{4a}]$ and $[c-\frac{b^2}{4a},\infty)$ as the corresponding domains of the inverse function.

  • 1
    I find these downvotes slightly suspicious... – 2017-01-22
  • 0
    @SimplyBeautifulArt I guess you probably thought that I downvoted you and Stella's posts in order to put mine to the top. I do not engage in this type of targeted downvoting. Instead, it is because it does not answer this question, which clearly asks for the domain of the inverse function. – 2017-01-23
  • 1
    @Displayname You sure about that? (and not pointing you out at all, I just dislike downvotes without leaving a comment) IMO, showing the quadratic is not invert-able is a perfectly reasonable approach to the problem – 2017-01-23
  • 0
    @Displayname happy now? – 2017-01-23
  • 0
    @SimplyBeautifulArt While it is a reasonable approach, your answer makes it sound like the domain of the inverse function would be $[-b/2a,\infty)$ which would confuse the O.P. And everytime I see someone commenting about "suspicious downvotes" I think of an implication of targeted downvoting, sorry I misunderstood you. – 2017-01-23
  • 0
    @Displayname I'm sorry, but I'm confused. I thought **my** answer was pretty clear on the thought that there are two different cases. And don't worry about the misunderstanding, we should probably just drop it :P – 2017-01-23
2

Hint: If we have a quadratic function with image $X$ then the inverses' domain would be $X$.

Example: $f(x) = x^2+100x-271$. Clearly, this function's end behavior is going to infinity, and its lowest point is at its vertex. So its image is $[-2771,\infty)$, as $-2771$ is the y value of its vertex (Verification left to the O. P. as an exercise). As such, the domain of its inverse is $[-2771,\infty)$

1

A quadratic of the form $y=ax^2+bx+c$ with $a\ne0$, then it has a vertex at $x=-\frac b{2a}$, hence, the domain of the inverse will fall into two cases

If $a>0$, then $x\ge-\frac b{2a}$.

If $a<0$, then $x\le-\frac b{2a}$