0

I've been posed with the question "Why is $f$ not invertible?"

I have learned that $x^2$ is not bijective unless I restrict it to only use non-negative Reals. However when I look at the curve of $2^x$ it looks to me that it passes the 1 unique x point throughout the whole chart. However, I need to prove why $2^x$ is not invertible and rewrite it to make it invertible. Do I need to restrict it from $R \rightarrow R^+$ ? What else am I not understanding? Thank you.

amWhy
  • 209,954
RyBoneCoder
  • 105

2 Answers2

2

Depend from where to where. But indeed, it's at least injective on $\mathbb R$, so on some co-domain, it will obviouly be bijective. Notice that whenever $a>0$, then $a^x$ is nothing else than $e^{x\ln(a)}$.

Walace
  • 710
  • 4
  • 12
  • Your response is a little over my head. Math is my weakness, I'm a cyber engineer and always struggle with math. :) – RyBoneCoder Apr 06 '20 at 20:30
  • @RyBoneCoder: If $f:X\to Y$ is injective, then $f:X\to f(X)$ is bijective. Here, since $2^x>0$ for all $x\in\mathbb R$, that $\lim_{x\to -\infty }2^x=0$ and $\lim_{x\to \infty }2^x=\infty $, then $x\mapsto 2^x$ is bijective $\mathbb R\to (0,\infty )$. – Walace Apr 06 '20 at 21:10
1

Note that the function $f(x)=2^x$ is not surjective because: $$\Im(f(x))=R^+$$ Being not surjective, $f(x)$ is not bijective. In this case we can't find the inverse because the inverse exists only when $f(x)$ is bijective.

If you restrict the codomain on $R^+$, the function is injective and surjective, so it's bijective. As noted before, the inverse is: $$y=g(x)=\log_2(x)$$

amWhy
  • 209,954
Matteo
  • 6,581