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.

Posted on by Brittney D'Oleo
Categories: Challenges, Peer Problems

One Response

  1. Jeremy LeCrone says:
    November 4, 2017 at 2:12 pm

    https://richmond.qualtrics.com/jfe/form/SV_2uiXuKjommZWBpP

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