Let $f:[0,2]\to\mathbb{R}$ be a continuous function and $f$ is differentiable on $(0,2)$, and let $f(0)=f(2)=0$. Now, suppose that there is a point $c\in(0,2)$ such that $f(c)=1$. Then, there is a point $k\in(0,2)$ such that $\vert f'(k)\vert>1$.
Intuitively, it is pretty trivial. But, I can't find any way to prove it.
How to prove it using the M.V.T. or another popular theorem?
Give some idea or advice. Thank you!
On
On the interval $[0,c]$ by MVT we have $\exists\, s \in (0,c)$ such that $$f'(s)=\frac{f(c)-f(0)}{c-0}=\frac{1}{c}.$$
Likewise on the interval $[c,2]$ by MVT we have $\exists\, t \in (c,2)$ such that $$f'(t)=\frac{f(2)-f(c)}{2-c}=\frac{1}{c-2}.$$
If $c <1$, then $f'(s)>1$ and if $1 < c<2$, then $|f'(t)|>1$. The only case left is when $c=1$. Can you complete it now.
On
$c=1$ case: If $|f(k)|>1$ for some $k$, then we are done by the mean value theorem.
Therefore assume $|f| \leq 1$. Then $c = 1$ is a point of maximum $\implies f'(1)=0$. Assume towards as contradiction that $f'$ is never greater than 1. Then we have $f(x) \leq x$ $\forall x$.
By the mean value theorem, we know that $f'(k) = 1$ for some $k \in [0,1]$. By Darboux's theorem (the derivative has the intermediate value property), $0 < f'(a) = \frac{1}{2} < 1$ for some $a\in (k,1) \implies \exists x \in(0,1) s.t. f(x) < x$, and we get a contradiction by the mean value thorem as $\frac{1-f(x)}{1-x} > \frac{1-x}{1-x} = 1$.
The $c<1$ and $c>1$ cases are easy.
On
Simplest Answer.
Case 1
Assume $0< c \leq 1$ then by MVT $ \exists$ $k\in (0,c)$ satisfying
$ f'(k)=\frac{f(0)-f(c)}{0-c}$
$ \implies f'(k)=\frac{0-1}{-c}=\frac{1}{c} \geq 1$
Case 2
Assume $1<c<2$ then by MVT $ \exists$ $k\in (c,2)$ satisfying
$ f'(k)=\frac{f(c)-f(2)}{c-2}$
$ \implies f'(k)=\frac{1-0}{c-2}=\frac{1}{c-2}<-1$
Can you see how this solution clicked?
Given two distinct points $p,q$ in the plane, let $S(p,q)$ be the slope of the line through $p$ and $q.$
If $f(x)>x$ for some $x\in (0,1),$ then $S((0,f(0)),(x,f(x))>1.$ Apply the MVT to see that $f'>1$ somewhere in $(0,x).$ If $f(x)<x$ for some $x\in (0,1),$ then $S((x,f(x)),(1,f(1))>1.$ This time we find $f'>1$ somewhere in $(x,1).$
Similary, if $f(x)\ne 2-x$ for some $x$ in the interval $(1,2),$ $f'<-1$ somewhere in $(1,2).$
We are done unless $f(x)=x$ in $[0,1]$ and $f(x)=2-x$ in $[1,2].$ But this case doesn't occur; if it did, $f'(1)$ would not exist.