Verify the Rolle's theorem when $f(x)=(x+1)^m(x-1)^n,\;\;-1\leq x\leq 1$

Question

Verify the Rolle's theorem when $f(x)=(x+1)^m(x-1)^n,\;\;-1\leq x\leq 1$ and show this result is not true for $f(x)=2x^{-2},\;\;\text{on}\;[-1,1]$ and $g(x)=|x|,\;\;\text{on}\;[-1,1].$

I know that the Rolle's Thoerem states that: Suppose

1. $f$ is continuous on $[\alpha,\beta];$
2. $f(\alpha)=f(\beta);$
3. $f'$ exists on $(\alpha,\beta).$

Then, $f$ has a local maximum or minimum at some $c\in (\alpha,\beta),$ thus $f'(c)=0.$

Please, can anyone help me out?

Answer 1

$f'(x)=(x+1)^{m-1} (x-)^{n-1} (m(x-1)+n(x+1))$. Hence $f'(x)=0$ precisely when $x=\frac {m-n} {m+n}$. Note that this number is indeed in $[-1,1]$. If $n$ is even then $f$ is non-negative and 0 at the end points, so $f$ has maximum at this point. If $n$ is odd then $f \leq 0$ and $f$ has a minimum at this point. For the the functions $2x^{-2}$ and $|x|$ take $\alpha =-1, \beta =1$