# Homework Help: Calculus Analysis proof

1. Jul 16, 2010

### K29

1. The problem statement, all variables and given/known data
$$g$$ is a continuous and differentiable function on$$(a,b)$$. Prove that if for all $$x\in(a,b)$$, $$g'(x)\neq0\Rightarrow$$ g is an injective function.

Is there anything wrong about my proof? I'm not comfortable in making logical mathematical arguments: I usually make small inaccuracies. Or do a lot of work going down a bad path Any help appreciated.

3. The attempt at a solution

We select an arbitrary sub-interval on $$(a,b)$$. We call it $$(m,n)$$ We know that g is also continuous and differentiable on any sub-interval of the interval that it is continuous and differentiable on.

The "injective functoin" has the definition that $$f(a)=f(b)\Rightarrow a=b$$ That is, no two values on the domain can map to the same point on the co-domain.

So for now we shall assume that$$f(m)=f(n)$$. This allows us to make use of the Mean value theorem which states that if a function is continuous and, differentiable on an interval, in this case $$(m,n)$$ and $$f(m)=f(n)$$ then $$\exists c\in (m,n)$$ such that $$g'(c) = 0$$. This leads to a contradiction as $$g'(x)\neq0$$. Therefor no such interval $$(m,n)$$ can exist. Hence $$f(m)=f(n)\Rightarrow m=n$$ (The point m on the domain is equivalent to(the same point as) n) and $$g$$ is injective.

2. Jul 16, 2010

### Dick

The proof is ok. Your phrasing at the beginning is pretty awkward. You aren't selecting (m,n) to be an arbitrary sub-interval. You are doing a proof by contradiction. So you assume f is NOT injective. If f is NOT injective then there are two different points m and n such that f(m)=f(n). Now the rest of the proof is fine.

3. Jul 17, 2010

### K29

Thanks very much!