# Prove f(x) = x^3 + 3^x is a one-to-one function.

1. Aug 15, 2012

1. The problem statement, all variables and given/known data

Prove f(x) = x^3 + 3^x is a one-to-one function.

2. The attempt at a solution

Sum of one-to-one functions is a one-to-one function (I think/dont know how to prove).
x^3 is one-to-one, 3^x is one-to-one, thus f(x) is one-to-one. Surely there's a more rigorous proof.

2. Aug 15, 2012

### arildno

Your attempt is wrong. f(x)=x is one-to-one, so is g(x)=1-x. But the sum f+g is not one-to-one.

If a function is NOT one-to-one, what do you know occasionally occurs with the function that cannot happen with a one-to-one function?

3. Aug 15, 2012

Two x values give the same y value. Are you suggesting I use the horizontal line test?

4. Aug 15, 2012

### oli4

Hi quadreg, have you studied derivatives yet ?

Cheers...

5. Aug 15, 2012

Yes, I tried solving for critical points, but i get f'(x)=3x^2+(3^x)ln3 and can't solve for x where f'(x) = 0

6. Aug 15, 2012

### oli4

You only have to prove that the derivative is either always >=0 or <=0 can you do that ?
Cheers...

7. Aug 15, 2012

Ohh, that makes sense. 3x^2 is always >=0 and (3^x)ln3 is always >=0, so f'(x)>0.
Thankyou.

8. Aug 15, 2012

### oli4

Exactly, yo've got it :)

9. Aug 15, 2012

### Staff: Mentor

Mod note: Moving this thread to Calculus & Beyond

10. Aug 15, 2012

### LCKurtz

Almost. You have to be sure that if the derivative is allowed to be 0, it isn't zero on any interval. Of course, that isn't a problem in this example.

11. Aug 16, 2012

### oli4

Yes, of course, LCKurtz, :) it's a good idea to point that out, thanks :)