# HW 7: Challenge 3

**Problem:** Let be a differentiable function on an interval . Show that if on , then is one-to-one. Provide an example to show that the converse of the statement need not be true.

**Proof: ** Suppose is a differentiable function on some interval . Let for some arbitrary . By Theorem 5.2.3, is also continuous on and so by the Mean Value Theorem, such that . By some algebra, . Since , . So either or . However, by assumption, so , so and thus is one-to-one.

**Corrected Proof: (Proof by Contradicition) **Suppose is a differentiable function on some interval with the condition that . Let for some arbitrary . By Theorem 5.2.3, is also continuous on and so by the Mean Value Theorem, such that (WLOG) for , . By some algebra, . Since , . Since by contruction of the mean value thereorem, However, this is a contradiction to the assumption that , thus and thus is one-to-one.

**Example: **Let on the interval . We see that . If , then and so and we have that this is a 1-1 function. However at on .

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