Calculus TA

Problem Let $f$ be a continuous real function on $\mathbb{R}^1$., of which it is known that $f'(x)$ exists for all $x \neq 0$ and that $f'(x) \rightarrow 3$ as $x \rightarrow 0$. Does it follow that $f'(0)$ exists? Answer $f'(0) = \lim_{h \rightarrow 0}\frac{f(0+h)-f(0)}{h} = \lim_{h \rightarrow 0}\frac{\frac{d}{dh}(f(0+h)-f(0))}{\frac{d}{dh}h}=\lim_{h \rightarrow 0} f'(h) = 3$. 

Problem If $$ C_0 + \frac{C_1}{2} + \cdots + \frac{C_{n-1}}{n} + \frac{C_n}{n+1} =0, $$ where $C_0,\cdots,C_n$ are real constants, prove that the equation $$ C_0 + C_1x + \cdots + C_{n-1}x^{n-1} + C_nx^n = 0 $$ has at least one real root between 0 and 1. Answer Define a function $p:[0,1] \rightarrow \mathbb{R}$ as $$ p(x) = C_0x + \frac{C_1}{2}x^2 + \cdots + \frac{C_n}{n+1}x^{n+1}. $$ Then, $p$ is differentiable on $[0,1]$ and $p(0) = p(1) = 0$. By the mean value theorem, there exists $x \in (0,1)$ such that $$ C_0 + C_1x + \cdots + C_{n-1}x^{n-1} + C_nx^n = p'(x) = \frac{p(1) - p(0)}{1-0} = 0.$$

Problem If $f:\mathbb{R} \rightarrow \mathbb{R}$ be differentiable. Assume there is no $x \in \mathbb{R}$ such that $f(x)=0=f'(x)$. Show that $S=\{x\;|\;0\leq x \leq 1,f(x)=0\}$ is finite. Answer Note that $S=f^{-1}(0)$ is closed since $f$ is continuous and $\{0\}$ is closed. Also, $S$ is bounded since $S \subset [0,1]$. These imply $S$ is compact. If $S$ is infinite, there is a sequence $\{x_n\} \subset S$ and its subsequence $x_{n_k} \rightarrow x \in S$. Then $$ f'(x) = \lim_{k \rightarrow \infty} \frac{f(x) - f(x_{n_k})}{x-x_{n_k}} =0. $$ Hence, $S$ has to be finite.