1. ## Mean Value Theorem

Let f:R--->R twice differentiable on R. Suppose that f(a)=f(b)=f(c) for real numbers a<b<c. show that there exists cE(a,c) such that $f''(d)=0$

do I apply Mean Value Theorem to the interval (a,c)? why do i need the function to be twice differentiable to answer the question?

thanks

2. Originally Posted by charikaar
Let f:R--->R twice differentiable on R. Suppose that f(a)=f(b)=f(c) for real numbers a<b<c. show that there exists cE(a,c) such that $f''(d)=0$

do I apply Mean Value Theorem to the interval (a,c)? why do i need the function to be twice differentiable to answer the question?

thanks
since $f$ is differentiable, and $f(a) = f(b)$ , then by the MVT, there exists an $x_1 \in (a,b)$ such that $f'(x_1) = 0$ ... same argument for an $x_2 \in (b,c)$ such that $f'(x_2) = 0$

since f is twice differentiable, then the MVT also applies to the function $f'(x)$ , and since $f'(x_1) = f'(x_2)$ , then there exists a value $d \in (x_1,x_2)$ such that $f''(d) = 0$

3. Originally Posted by charikaar
Let f:R--->R twice differentiable on R. Suppose that f(a)=f(b)=f(c) for real numbers a<b<c. show that there exists cE(a,c) such that $f''(d)=0$

do I apply Mean Value Theorem to the interval (a,c)? why do i need the function to be twice differentiable to answer the question?
Apply Rolle's theoremto show that the first derivative f' has a zero in each of the intervals (a,b) and (b,c). Then apply Rolle's theorem again, this time to the function f', to show that f" has a zero in the interval (a,c).