Proving a function is injective

[Cha0t1c]

Homework Statement

Let f: ]1, +inf[ → ]0, +inf[ be defined by f(x)=x^2 +2x +1. Prove f is injective.

Relevant Equations

f(a) = f(b) ==> a=b

Hello,
Let f: ]1, +inf[ → ]0, +inf[ be defined by f(x)=x^2 +2x +1.

I am trying to prove f is injective.

Let a,b be in ]1, +inf[ and suppose f(a) = f(b).

Then, a^2 + 2a + 1 = b^2 + 2b + 1.

How do I solve this equation such that I end up with a = b?

Solution:

(a + 1) ^2 = (b + 1)^2

sqrt[(a+1)^2] = sqrt[(b+1)^2]

abs(a + 1) = abs(b + 1)

since a>1 and b>1

a + 1 = b +1
thus a = b

hence f is injective.

[Moderator's note: Moved from a technical forum and thus no template.]

 

[PeroK]

Hello,
Let f: ]1, +inf[ → ]0, +inf[ be defined by f(x)=x^2 +2x +1.

I am trying to prove f is injective.

Let a,b be in ]1, +inf[ and suppose f(a) = f(b).

Then, a^2 + 2a + 1 = b^2 + 2b + 1.

How do I solve this equation such that I end up with a = b?

It's a quadratic equation.

 

[Cha0t1c]

It's a quadratic equation.

(a+1)^2 = (b+1)^2, How did I not see that?!

 

[Mark44]

(a+1)^2 = (b+1)^2, How did I not see that?!

But you're not done.
If $(a + 1)^2 = (b + 1)^2$ then $a + 1 = \pm (b + 1)$. What more do you need to do to conclude that a = b?

 

[Cha0t1c]

But you're not done.
If $(a + 1)^2 = (b + 1)^2$ then $a + 1 = \pm (b + 1)$. What more do you need to do to conclude that a = b?

But a and b are strictly greater than 1 for all a and b since the domain is ] 1, +inf [. Doesn't it free us from this situation since b + 1 will always be greater than 0?

 

[member 587159]

But a and b are strictly greater than 1 for all a and b since the domain is ] 1, +inf [. Doesn't it free us from this situation since b + 1 will always be greater than 0?

Yes, it does.

 

[WWGD]

If you know how to differentiate you can use that to see where the function is strictly increasing/decreasing and thus not taking the same value twice.

 

[Mark44]

If you know how to differentiate you can use that to see where the function is strictly increasing/decreasing and thus not taking the same value twice.

But the thread was posted in the Precalc section. If the thread was posted appropriately, we can't assume any prior knowledge of calculus.

 

[member 587159]

But the thread was posted in the Precalc section. If the thread was posted appropriately, we can't assume any prior knowledge of calculus.

Yes, but the argument still holds without calculus if you know about parabolas.

 

[Mark44]

Yes, but the argument still holds without calculus if you know about parabolas.

My comment was about the use of calculus techniques such as differentiation in a thread posted in the Precalc section, not about any properties of parabolas.

 

[mathwonk]

I like the idea of looking at the increasing nature of the function (x+1)^2 on the domain [-1,inf), which is obvious even without differentiating, for this problem.