1.Let f be a real-valued differentiable function defined on [0, 1]. If $f(0)=0$ and $f(1)=1$, prove that there exists two numbers $x_1,x_2 \in [0, 1]$ such that \frac{1}{f'(x_1)}+\frac{1}{f'(x_2)}=2.
2.Let f be a real-valued differentiable function defined on (0,$\infty$). If $\lim_{x\rightarrow0^+}f(x)=\lim_{x\rightarrow\infty}f(x)=1$, prove that there exists $c\in(1,\infty)$ such that f'(c)=0.
1.From the question's statement, I wasn't sure if $x_1$ could equal $x_2$. If it can, is the following proof valid? Clearly, it suffices to show f'(x)=1 for some $x \in [0,1]$.
We proceed by contradiction.
Suppose f'(x)\neq1 for all $x \in [0,1]$.
Then either f'(x)>1 or f'(x)<1 for all relevant x (otherwise, $\exists x_1,x_2$ s.t. f'(x_1)>1, f'(x_2)<1 which contradicts the IVT for derivatives).
Assume, wlog, f'(x)>1 for all x.
Then 1=f(1)=\int_{0}^{1}f'(x)dx>\int_{0}^{1}dx=1, contradiction.
2. Again we proceed by contradiction; suppose f'(x)\neq0 for all x. By an identical argument to the one used in 1, either f'(x)>0 or f'(x)<0 for all x. Wlog, assume f'(x)>0. Then f(x) is strictly increasing $\forall x$.
Then, if x>1, $1=\lim_{t\rightarrow0^+}f(t)
First, if my proofs are valid, are there simpler proofs? Second, can someone provide me with a proof of the case where $x_1\neq x_2$ in 1?