Prove that $inf(S_1)=sup(S_2)$ 
$f:I \to \Bbb R$, $I=[0,1]$ continuous and differentiable on $(0,1)$ s.t $f(0)<0<f(1)$ and $f'(x)\neq 0$ $\forall x \in (0,1)$. Let $S_1=\{x \in I | f(x) >0\}$ and $S_2=\{x \in I | f(x) <0\}$. Prove that $inf(S_1)=sup(S_2)$.

My attempt:
If $\exists x\in I$ s.t $f(x)<f(0)$ then $\exists c\in I$ s.t $f(c)=f(0)$ contradiction by Rolle's thm.
So $f'(0)>0$
Claim: $f'(x)>0$, $\forall x \in (0,1)$
Suppose not then $\exists x_1\in I$ s.t $f'(x_1)<0$.
Consider, $d=inf\{x\in I|f'(x)<0\}$
If we can prove that $d \neq 0$ then $\exists a,b\in (d-\epsilon,d+\epsilon)\subseteq [0,1]$ s.t $f(a)=f(b)$ contradiction by Rolle's thm.
So,$f'(x)>0$, $\forall x \in (0,1)$ $\Rightarrow inf(S_1)=sup(S_2)$. Done.
My question is: How can we prove that $d \neq 0$?

Since $f'$ has IVP and $f'(x) \neq 0$ for all $x$ it follows that $f'(x) >0$ for all $x$ or $f'(x) <0$ for all $x$. In other words, $f$ is strictly increasing or strictly decreasing. But $f(0)<f(1)$ so the second possibility is ruled out and $f$ must be strictly increasing.
Now let $a$ be the infimum of $S_1$ and $b$ be the supremum of $S_2$. $b \leq a$ because $x \in S_2, y \in S_2$ implies $x <y$. Next note that $f(a)=0$. This is an easy consequence of continuity of $f$ and I will leave it to you to check this. Hence $f(t) <0$ (and $t \in S_2$) for all $t <a$. As a consequence $b \geq t$ for any $t <a$. Hence $b \geq a$. 
