If $h \gt 0$ and $f'(x)$ exist for $x \in (a-h,a+h)$ and $f$ is continuous in $[a-h,a+h]$. Prove that $$\frac{f(a+h)-2f(a)+f(a-h)}{h}=f'(a+ \lambda h)-f'(a-\lambda h)$$ with $0 \lt \lambda \lt 1$.
Until now I get $$\frac{f(a+h)-2f(a)+f(a-h)}{h}=f'(a+ \lambda_1 h)-f'(a-\lambda_2 h)$$ with $0\lt \lambda_1, \lambda_2 \lt 1$ taking $c_1 \in (a-h,a)$ and $c_2 \in (a,a+h)$ by MVT.
But I must get a unique $\lambda$ and I don't know how to get it.
I would really appreciate any help to take it from here.
Set $\forall x \in [a, a+h] $ $$ g(x) = f(x) + f(2a-x) $$
thus $g$ is continuous in its domain and differentiable in $]a, a+h[$. Notice that $g'(x) = f'(x) - f'(2a-x)$ by the chain rule. By MVT, we get that there exists a real number $b \in ]a, a+h[$ such that
$$ g'(b) = \frac{g(a+h) - g(a)}{h}.$$
Set $\lambda = (b-a)/h$. Then $ 0<\lambda < 1$ and $b = a + \lambda h$. Thus
$$\frac{f(a+h) + f(a-h) -2f(a)}{h} = \frac{g(a+h) - g(a)}{h} = g'(a + \lambda h) = f'(a+\lambda h) - f'(a - \lambda h).$$
The technique for solving many problem sets involving MVT is declaring an auxilary function whose derivative will crack the problem.