# HW 7: Challenge 3

Problem: Let $f$ be a differentiable function on an interval $A$. Show that if $f'(x) \neq 0$ on $A$, then $f$ is one-to-one. Provide an example to show that the converse of the statement need not be true.

Proof:  Suppose $f$ is a differentiable function on some interval $A \in \mathbb{R}$. Let $f(x) = f(y)$ for some arbitrary $x, y \in A$. By Theorem 5.2.3, $f$ is also continuous on $A$ and so by the Mean Value Theorem, $\exists c \in (a,b)$ such that $f'(c) = \frac{f(x)-f(y)}{x-y}$. By some algebra, $f(x)-f(y) = f'(c)(x-y)$. Since $f(x) = f(y)$$f'(c)(x-y) = 0$. So either $f'(c) =0$ or $(x-y)=0$. However, by assumption, $f'(x) \neq 0$ so $(x-y)=0$, so $x=y$ and thus $f$ is one-to-one.

Corrected Proof: (Proof by Contradicition) Suppose $f$ is a differentiable function on some interval $A \in \mathbb{R}$ with the condition that $f'(c) \neq 0$. Let $f(x) = f(y)$ for some arbitrary $x, y \in A$. By Theorem 5.2.3, $f$ is also continuous on $A$ and so by the Mean Value Theorem, $\exists c \in (x,y)$ such that (WLOG) for $y>x$, $f'(c) = \frac{f(x)-f(y)}{x-y}$. By some algebra, $f(x)-f(y) = f'(c)(x-y)$. Since $f(x) = f(y)$, $f'(c)(x-y) = 0$. Since $y>x$ by contruction of the mean value thereorem, $f'(c)=0$ However, this is a contradiction to the assumption that $f'(c) \neq 0$, thus $x=y$ and thus $f$ is one-to-one.

Example: Let $f(x) = x^5$ on the interval $(-3,3)$. We see that $f'(x) = 5x^4$. If $f(x) = f(y)$, then $x^5 = y^5$ and so $x=y$ and we have that this is a 1-1 function. However at $x=0$ on $(-3,3), f'(x) = 0$.