The function $g : [a, b]\to\mathbb{R}$ is continuous on $[a, b]$, differentiable on $(a, b)$ and satisfies $g'(x) > 0$. Explain $g$ is injective.

Question

The function $g : [a, b]\to\mathbb{R}$ is continuous on $[a, b]$, differentiable on $(a, b)$ and satisfies $g'(x) > 0$ for all $x \in (a,b)$.

Explain why $g$ is injective.

Does the Mean Value Theorem prove that a continuous and differentiable function is injective?

Answer 1

Let $x_1,x_2\in[a,b]$ be such that $x_1\neq x_2$. Suppose that $g(x_1)=g(x_2)$. Then by Rolle's theorem we would also have that $g'(\xi)=0$ for some $\xi\in(x_1,x_2)$, which is a contradiction.

Answer 2

You can prove this by assuming that there exists a y $\in$ (a,b) such that there is a $x_1$ and $x_2$ between a and b such that $g(x_1) = g(x_2) = y$, if this would be true the function is not injective. You can now proof that this assumption leads to a contradiction in the following way. So if $g(x_1) = g(x_2) = y$ then there exists a c between $x_1$ and $x_2$ sucht that g'(c) = 0 (This is Rolle's theorem).

But we know that for every x in between a and b, and thus also between $x_1$ and $x_2$, that g'(x) > 0 so we have found our contradiction. Our assumption that there exists a $x_1$ and $x_2$ such that g($x_1$) = g($x_2$) = y must be false and thus g needs to be a injective function.

Now this is proven using the mean value theorem because Rolle's theorem is just a special case of the mean value theorem.

Answer 3

Since $g'(x)>0$ the function $g$ is strictly increasing, therefore injective. Formally it follows from the Lagrange theorem $${g(x_2)-g(x_1)\over x_2-x_1}=g'(\xi), \qquad a\le x_1<\xi<x_2\le b$$ where $x_1,x_2$ are given.

Another explanation: assume $x$ denotes the time (in seconds) and $g(x)$ the distance (in metres) covered in the period $[a,x].$ Since the speed at every moment ($g'(x)$) is positive the distance is growing strictly.