Question

suppose f(x) has second derivative in a, b, and f(a) = f(b) = 0. prove that for every x ∈ (a, b), there exists a ξ ∈ (a, b), such that f(x) = \frac{f(ξ)}{2}(x - a)(x - b).

Explanation:

Step1: Define Auxiliary Function

Let \( g(t) = f(t) - \frac{f''(\xi)}{2}(t - a)(t - b) \). But wait, actually, we should construct a function to apply Taylor's theorem or Mean Value Theorems. Alternatively, consider using the Taylor expansion or constructing a function \( F(t) = f(t) - \frac{f''(\xi)}{2}(t - a)(t - b) \), but maybe a better approach is to use the Mean Value Theorem for derivatives (Rolle's theorem or Taylor's formula). Wait, let's start over. Let's define \( h(t) = f(t) - \frac{f''(\xi)}{2}(t - a)(t - b) \), but actually, we need to prove for a given \( x \in (a,b) \), there exists \( \xi \in (a,b) \) such that \( f(x) = \frac{f''(\xi)}{2}(x - a)(x - b) \). Let's consider the function \( F(t) = f(t) - \frac{f''(\xi)}{2}(t - a)(t - b) \), but maybe use Taylor's formula around \( a \) and \( b \), but since \( f(a) = f(b) = 0 \), let's use the second - order Taylor expansion. Wait, another approach: Let's define \( \varphi(t)=f(t)-\frac{f(x)}{(x - a)(x - b)}(t - a)(t - b) \). Then \( \varphi(a)=f(a)-0 = 0 \), \( \varphi(b)=f(b)-0 = 0 \), and \( \varphi(x)=f(x)-f(x)=0 \). So \( \varphi(t) \) has three roots: \( t = a \), \( t = x \), \( t = b \) in the interval \([a,b]\) (assuming \( a<x < b \)). By Rolle's theorem, \( \varphi'(t) \) has two roots in \((a,x)\) and \((x,b)\), say \( t_1\in(a,x) \) and \( t_2\in(x,b) \). Then \( \varphi''(t) \) has a root \( \xi\in(t_1,t_2)\subset(a,b) \) by Rolle's theorem again. Now, compute \( \varphi''(t) \): \( \varphi'(t)=f'(t)-\frac{f(x)}{(x - a)(x - b)}[2t-(a + b)] \), and \( \varphi''(t)=f''(t)-\frac{2f(x)}{(x - a)(x - b)} \). Since \( \varphi''(\xi)=0 \) (because \( \xi \) is a root of \( \varphi''(t) \)), we have \( f''(\xi)-\frac{2f(x)}{(x - a)(x - b)} = 0 \), which implies \( f(x)=\frac{f''(\xi)}{2}(x - a)(x - b) \).

Step2: Verify Rolle's Theorem Applications

- First, define \( \varphi(t)=f(t)-\frac{f(x)}{(x - a)(x - b)}(t - a)(t - b) \). We check the values of \( \varphi(t) \) at \( t = a \), \( t = x \), and \( t = b \):
- When \( t = a \): \( \varphi(a)=f(a)-\frac{f(x)}{(x - a)(x - b)}(a - a)(a - b)=f(a)=0 \) (since \( f(a) = 0 \)).
- When \( t = x \): \( \varphi(x)=f(x)-\frac{f(x)}{(x - a)(x - b)}(x - a)(x - b)=f(x)-f(x)=0 \).
- When \( t = b \): \( \varphi(b)=f(b)-\frac{f(x)}{(x - a)(x - b)}(b - a)(b - b)=f(b)=0 \) (since \( f(b) = 0 \)).
- By Rolle's theorem, since \( \varphi(t) \) is continuous on \([a,b]\) (because \( f(t) \) is twice - differentiable, so continuous) and differentiable on \((a,b)\), and \( \varphi(a)=\varphi(x) \), there exists \( c_1\in(a,x) \) such that \( \varphi'(c_1)=0 \). Also, since \( \varphi(x)=\varphi(b) \), there exists \( c_2\in(x,b) \) such that \( \varphi'(c_2)=0 \).
- Now, consider the function \( \varphi'(t) \). It is differentiable on \((a,b)\) (because \( f(t) \) is twice - differentiable) and \( \varphi'(c_1)=\varphi'(c_2)=0 \). By Rolle's theorem again, there exists \( \xi\in(c_1,c_2)\subset(a,b) \) such that \( \varphi''(\xi)=0 \).
- Compute \( \varphi''(t) \):
- We know that \( \varphi(t)=f(t)-\frac{f(x)}{(x - a)(x - b)}(t - a)(t - b) \).
- First derivative: \( \varphi'(t)=f'(t)-\frac{f(x)}{(x - a)(x - b)}[2t-(a + b)] \) (using the product rule: derivative of \((t - a)(t - b)=t^2-(a + b)t+ab\), so derivative is \( 2t-(a + b) \)).
- Second derivative: \( \varphi''(t)=f''(t)-\frac{2f(x)}{(x - a)(x - b)} \).
- Since \( \varphi''(\xi)=0 \), we have \( f''(\xi)-\frac{2f(x)}{(x - a)(x - b)} = 0 \).
- Rearranging the equation \( f''(\xi)-\frac{2f(x)}{(x - a)(x - b)} = 0 \), we get \( f(x)=\frac{f''(\xi)}{2}(x - a)(x - b) \).

Answer:

For a given \( x\in(a,b) \), by constructing the function \( \varphi(t)=f(t)-\frac{f(x)}{(x - a)(x - b)}(t - a)(t - b) \), applying Rolle's theorem twice (first to \( \varphi(t) \) to get two roots of \( \varphi'(t) \), then to \( \varphi'(t) \) to get a root \( \xi \) of \( \varphi''(t) \)), and then computing the second - derivative of \( \varphi(t) \), we can show that there exists \( \xi\in(a,b) \) such that \( f(x)=\frac{f''(\xi)}{2}(x - a)(x - b) \).