Rolle's theorem for postive integer

Question

Verify Rolle's theorem for the function $f(x)=(x-a)^m (x-b)^n$ on $[a,b]$, where $m$ and $n$ are positive integers.

Answer 1

If $m$ and $n$ are continuous, then this is an ordinary polynomial, that is, it is differentiable and continuous at every point for $x \in \mathbb R$.

We have that Rolle's Theorem holds if $f(x)$ is differentiable and continuous on $(a,b)$, and that $f(a)=f(b)$.

In THIS case, we can easily see that $f(a)=f(b)=0$.

Therefore, because $f(x)$ is differentiable and continuous on the interval, and $f(a)=f(b)$, Rolle's Theorem guarantees a $c$ such that $a<c<b$ and $f'(c)=0$.

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We have that $f'(x) = m(x-a)^{m-1}(x-b)^n + n(x-a)^m(x-b)^{n-1}$.

This is equal to $f'(x)=(x-a)^{m-1}(x-b)^{n-1}(m(x-b)+n(x-a))$

At point $c$, we have $f'(c)=0$. Because neither $(x-a)$ nor $(x-b)$ is $0$ at $x=c$, we divide to get $m(x-b)+n(x-a)=0$.

We then have $m(x-b)=-n(x-a)$, so $mx-mb=-nx+na$. Therefore, $(m+n)x=mb+na$.

Thus, $\displaystyle x=\boxed{c=\frac{mb+na}{m+n}}$.

We know $c$ exists because $m \neq n$.

Answer 2

Hint:

$$f'(x)=m(x-a)^{m-1}×(x-b)^n+n(x-b)^{n-1}×(x-a)^m$$ Then show that there exists one $c\in(a,b)$ such that $f'(c)=0$.