1. ## 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?

2. ## Re: MVT

Originally Posted by lovesmath
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.

The fact that $f$ is continuously differentiable means that $f'$ is continuous.

Let $f(d)=0$ where $a

$(\exists p\in (a,d))\left[f'(p)=\frac{f(d)-f(a)}{d-a}<0\right]$ Why?

Can you finish? Find $f'(q)>0$.

3. ## 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).

4. ## Re: MVT

Originally Posted by lovesmath
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).
$f'(p)$ comes from the mean value theorem as does $f'(q)~.$

Use the intermediate value theorem on $f'(p)<0