Exercise 5.3.2

Question

Let $f$ be differentiable on an interval $A$. If $f^{\prime}(x) 
eq 0$ on $A$, show that $f$ is one-to-one on $A$. Provide an example to show that the converse statement need not be true.

Ulisse Mini · Accepted Answer

Let $x,y$ be in $A$ with $x < y$, to show $f(x) 
e f(y)$ apply the Mean Value Theorem on $[x,y]$ to get $c \in (x,y)$ with
  $$
  f'(c) = \frac{f(x) - f(y)}{x-y}
  $$
  Now since $f'(c) 
e 0$ we must have $f(x) - f(y) 
e 0$, and thus $f(x) 
e f(y)$.

To see the converse is false consider how $f(x) = x^3$ is 1-1 but has $f'(0) = 0$.

Exercise 5.3.2

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Exercise 5.3.2

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