3

In the first few lessons of an introductory Calculus class, there is this exercise:

Show that the polynomial $f(x) = x^n$ is injective whenever $n$ is odd, and is not injective whenever $n$ is even.

I understand why this statement holds intuitively -- when $n$ is odd, $f(x)$ will keep whatever sign $x$ has; but when $n$ is even, $f(x)$ will always be positive regardless of $x$'s sign. However, I have trouble writing a proof for it.

Claire
  • 431
  • 1
    What is the domain of $f(x)$? – peterwhy Aug 02 '22 at 21:23
  • For $n$ even just one counter-example is enough. For $n$ odd, remember that strict monotonicity implies injectivity. – zwim Aug 02 '22 at 21:25
  • @peterwhy It's not specified, but I am sure it's within the realm of real numbers since it's an introductory Calc course. – Claire Aug 02 '22 at 21:25
  • 1
    For even $n$, "$f(x)$ will always be positive" is not the right explanation. $e^x$ is always positive, yet is injective for $x\in\mathbb R$. – peterwhy Aug 02 '22 at 21:26
  • 1
    If $f(x)=x^{2n+1}$ then $f'(x)=(2n+1)x^{2n}$ is $≥0$ for all $x$, so it is strictly increasing. – lulu Aug 02 '22 at 21:31
  • 1
    @lulu : Note that the condition "$f'(x)\geq 0$ for all $x$" is not sufficient to imply that $f$ is strictly increasing. It is possible that $f\equiv 0$ on an interval. – MPW Aug 02 '22 at 21:37
  • @MPW. Sure, though in this case it is only $0$ at a single point. – lulu Aug 02 '22 at 21:37
  • @lulu : Indeed. But more is needed nonetheless, if a proof is to be made. – MPW Aug 02 '22 at 21:39

1 Answers1

0

To prove that $f$ is injective for odd $n$, you need to show that $x^n=y^n$ implies that $x=y$ for all real values of $x$ and $y$. To prove that $f$ is not injective for even $n$, you need to show that there are real numbers $x$ and $y$ such that $x^n=y^n$ and $x\neq y$.

The first proof is not at all straightforward. First, note that it is only possible for $x^n$ to equal $y^n$ when $x$ and $y$ have the same sign, so we can disregard the case where $x$ is positive and $y$ is negative (and vice versa). Let's consider the case where $x$ and $y$ are both positive. Assume that $x^n=y^n$. Then, $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dots+y^{n-1})=0$, and since the second bracketed term is positive, we must have $x=y$. I'll leave the case where $x$ and $y$ are both negative to you. Hint: if $x$ and $y$ are both negative, then $-x$ and $-y$ are both positive, so you can use the previous case to help you.

The second proof is much easier. Consider what happens when $x=1$ and $y=-1.$

Joe
  • 19,636