# A Non-Inversible Function

If a function $f(x)$ from reals to reals satisfies that $2f(2x)-f(x^2)^2=1$ then prove that $f^{-1}(x)$ does not exist.

source: me Note by Daniel Liu
7 years ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

## Comments

Sort by:

Top Newest

We basically have to prove that there exists distinct$x_1,x_2$ such that $f(x_1)=f(x_2)$ because if $f^{-1}(x)$ does exist under this, then $f^{-1}(f(x_1))=x_1, f^{-1}(f(x_2))=x_2$, which violates the definition of a function that $f(x)$ corresponds to at most one value.

For this problem simply take $x=0,2$ and we have it.

To elaberate, we first find a value $x$ such that $2x=x^2=y$ so that we get an equation with one variable, it turns out that this would lead to only one solution for $f(y)$ since $(f(y)-1)^2=0\Rightarrow f(y)=1$. Since $x^2=2x$ gives $x=0,2$, we have $f(0)=1,f(2)=1$.

- 7 years ago

Log in to reply

Excellent Solution. Did not think that way!

- 7 years ago

Log in to reply

Yep, that's correct. Quite an interesting problem, until you realize that the solution is very quick and easy.

- 7 years ago

Log in to reply

for x(2-x) has two solutions 0 and 2 .. !! the values of the function is 1 for two elements in its domain i.e x=0 and 2 .. its not onto .. !

- 7 years ago

Log in to reply

For what domain?

- 7 years ago

Log in to reply

That's a good question to ask.

Staff - 7 years ago

Log in to reply

I think all reals works, but if you see any contradiction feel free to tell me and I can change it.

- 7 years ago

Log in to reply

Put x=2, it gives f(4)=1 . Now differentiating the given equality, 4f'(2x)-2f(x^2)f'(x^2)*2x = 0. put x=2, it gives ,f(4)=1/2 ..clearly, the function is non-invertible..

- 7 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...