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- Dec 2nd 2012, 11:52 AM #1

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- Jul 2011
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- 80

## MVT

Suppose f:[a,b]->R satisfies the following properties:

1. f is continuous on [a,b] and continuously differentiable on (a,b)

2. f(a) and f(b) are positive numbers

3. f(x)=0 for some x in (a,b)

Prove that there exists some c in (a,b) such that f'(c)=0.

I understand what the problem is asking and can draw a descriptive picture, but I'm having trouble putting it in general terms. Since f(a) and f(b) are both positive numbers, and there is some x such that f(x)=0, then the slope from f(a) to f(x) must be negative, and the slope from f(x) to f(b) must be positive. Since the function is continuous and continuously differentiable, then the derivative function must also be continuous.

Can you show me how to prove that f'(c)=0?

- Dec 2nd 2012, 12:11 PM #2

- Dec 2nd 2012, 12:18 PM #3

- Joined
- Jul 2011
- Posts
- 80

## Re: MVT

Why isn't f'(p)=(f(d)-f(a))/(d-a)? Isn't that the definition of derivative? It would have to be less than zero since a<d, which means it has a negative slope. The same would be true for f'(q) except that it would be positive because d<b. So, f'(q)=((f(b)-f(d))/(b-d).

- Dec 2nd 2012, 12:25 PM #4