function not having same value at every point Let $f : (−1,1) \to\Bbb R$ be a twice differentiable function such that $f''(0) > 0$. Show that there exists $n \in \Bbb N$ such that $f(\frac 1n)$ is not equals $1$.
I know that  $f$ has a minima at $0$ and $f'(0)$ is $0$ but how to proceed from here? 
 A: By definition of derivative you get $f'(0)=0$. Since $f''(0) >0$ there exits $\delta >0$ such that $f'(x)>f'(0)=0$ for $0<x<\delta$. This makes $f$ strictly increasing in $(0,\delta)$ making $f(\frac 1 {n+1}) <f(\frac 1 n)$ or $1<1$!
A: Assume that
$\forall n \in \Bbb N, \; f \left ( \dfrac{1}{n} \right ) = 1; \tag 1$
then by continuity, 
$f(0) = \displaystyle \lim_{n \to \infty} f \left ( \dfrac{1}{n} \right ) = 1; \tag 2$
also,
$f'(0) = \displaystyle \lim_{n \to \infty} \dfrac{f(1/n) - 1}{1/n} = 0; \tag 3$
since
$f''(0) > 0, \tag 4$
we find that $f(x)$ has a local minimum of value $f(0) = 1$ at $0$; now I claim that there is a fixed $\epsilon > 0$ such that $f'(\alpha) > 0$ for every $\alpha \in (0, \epsilon)$; otherwise, we could find sequences $\epsilon_n \to 0$ and $\alpha_n \in (0, \epsilon_n)$ with $f'(\alpha_n) \le 0$, and then
$f''(0) =  \displaystyle \lim_{n \to \infty} \dfrac{f'(\alpha_n) - f'(0)}{\alpha_n} = \lim_{\alpha_n \to \infty} \dfrac{f'(\alpha_n)}{\alpha_n} \le 0 \Rightarrow \Leftarrow f''(0) > 0; \tag 5$
now since $f'(x)$ is differentiable, it is also continuous and thus we may apply the mean value theorem to any $\beta \in (0, \epsilon)$ and obtain
$f(\beta) - 1 = f(\beta) - f(0) = f'(\gamma) (\beta - 0) = f'(\gamma) \beta, \; \gamma \in (0, \beta); \tag 6$
since $f'(\gamma) > 0$ this yields
$f(\beta) = 1 + f'(\gamma) \beta; \tag 7$
we see that every $\beta \in (0, \epsilon)$ satisfies
$f(\beta) > 1, \tag 8$
but this contradicts our assumption (1), and thus
$\exists n \in \Bbb N, \; f \left ( \dfrac{1}{n} \right ) \ne 1. \tag 9$
Note that we have in fact proved that $n$ in (9) may be taken arbitrarily large, and thus there are an infinite number of such $n$, which also follows from (7)-(8).  In fact, we have proved the stronger statement than (1),
$\exists M \in \Bbb N, \; n > M \Longrightarrow  f \left ( \dfrac{1}{n} \right ) \ne 1. \tag{10}$
