# Second mean value theorem proof (differentiation)

Since $F(x)$ is continuous in the closed interval [a,a+h] and differentiable in the open interval (a,a+h). Also f(a) = f(b) so by Rolle's theorem we get

$$F'(a+\theta h)=0,a<a+\theta h<a+h$$

As you seen in the quoted description (I took it from my book) $f(a)=f(b)$ But, I saw in the same book in my language they wrote that $F(a)=F(a+h)$ instead of $f(a)=f(b)$. I think it's a mistake. But, I can't understand which one is correct. Even, I can't see any derivation of the equation $F'(a+\theta h)=0,a<a+\theta h<a+h$

## book

Let $$F(x)=f(x)+(a+h-x)f'(x)+A(a+h-x)^2$$-----1

$$=>F(a)=f(a)+(h)f'(a)+A(h)^2$$--------2

Differentiate (1) respect to $x$ :

$$F'(x)=f'(x)-f'(x)+(a+h-x)f''(x)-2A(a+h-x)$$ $$=>F'(x)=(a+h-x)f''(x)-2A(a+h-x)$$ $$=>F'(a+\theta h)=(h-\theta h)f''(h+\theta h)-2A(h-\theta h)$$

As we seen, from that book $F'(a+\theta h)=0$.

So,

$$=>2A(h-\theta h)=(h-\theta h)f''(h+\theta h)$$

$$=>A=\frac{f''(h+\theta h)}{2}$$

Put the value in (1). Then, you will get second mean value theorem. Sorry! I can't edit it now for shortness of time.

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