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

[quadreg]

Homework Statement

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.

 

[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.

Try to answer the following:
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?

 

[quadreg]

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.

Try to answer the following:
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?

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

 

[oli4]

Hi quadreg, have you studied derivatives yet ?

Cheers...

 

[quadreg]

Hi quadreg, have you studied derivatives yet ?

Cheers...

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

 

[oli4]

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

 

[quadreg]

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

 

[oli4]

Exactly, yo've got it :)

 

[Mark44]

Mod note: Moving this thread to Calculus & Beyond

 

[LCKurtz]

Exactly, yo've got it :)

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.

 

[oli4]

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