# $M(x)$ and $L(x)$ interpolate $f(x)$ on $n+1$ points. Show, that $f(x)$ lies between $L(x)$ and $M(x)$

We have $n+2$ points $x_0 \lt x_1 \lt x_2 ... \lt x_{n+1}$.

We have two polynomials - $L$ and $M$. $L(x)$ interpolates $f(x)$ on points $x_0,...,x_n$ and $M(x)$ does so on $x_1,...,x_{n+1}$.

The $n+1$ derivative of $f$ is continuous and does change it's sign in interval $[x_0,x_{n+1}]$.

Show, that for all $x \in [x_0,x_{n+1}]$ $f(x)$ lies between $M(x)$ and $L(x)$.

My guess since we know what we know about $f$ then by theorem about remainder of interpolation we have :

$f(x)-M(x)=\dfrac{1}{(n+1)!}f^{(n+1)}(\theta_M)\prod\limits_{k=0}^{n}(x-x_k)$

and

$f(x)-L(x)=\dfrac{1}{(n+1)!}f^{(n+1)}(\theta_L)\prod\limits_{k=0}^{n}(x-x_k)$

for some $\theta_L$ and $\theta_M$ in $[x_0,x_{n+1}]$.

For nodes we obviously have what we want, because then $M(x_k)=L(x_k)=f(x_k)$.

Now for some other points I guess I would need to somehow prove that the first equality is negative, and the one below is positive.

But how to go about that?

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