A non constant function satisfying the condition of Rolle's theorem cannot be monotonic.

Question

I'm reading a book of elementary calculus there a found the statement "A non constant function satisfying the condition of Rolle's theorem cannot be monotonic."

But i'm finding myself unable to prove it.I need help in proving this

Answer 1

Without loss of generality, let's use a monotonically increasing function. It means that for any $y>x$ you have $f(y)\ge f(x)$. Now choose the ends of the interval, so you get $f(b)\ge f(a)$. Since we want Rolles' theorem conditions, $f(b)=f(a)$. Obviously this is true for the constant function (you can always call a constant function monotonically increasing). Now let's assume that the function is not constant, so that means that there exists at least one $c$ in $(a,b)$ where $f(c)\ne f(a)$. Since we use monotonically increasing function, it means that $f(c)\gt f(a)$ (strict inequality). Now apply monotonic definition between $b$ and $c$ and you get $f(b)\ge f(c) \gt f(a)$, or $f(b)>f(a)$. This does not obey the Rolles' theorem definition.

Answer 2

We have $f$ continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)$.

We assuming that $f$ is not constant hence there exists $x_0\in(a,b)$ such that $f(x_0)\ne f(a)$(I will assume that $f(x_0)> f(a)$, but it is all the same if it is $<$).

So we have $a<x_0<b$ but $f(a)<f(x_0)>f(b)$, i.e. $f(a)<f(x_0)$ and $f(x_0)>f(b)$.