Proving a continuous function is a constant Let $f$ be a function that is continuous on an interval $[a,b]$ and differentiable  on $(a,b)$. Assume $f'(x)=0$ for all $x\in (a,b)$. Show f is a constant.
I feel like the obvious set up is by contradiction. If I assume $f$ is not a constant, then it must have a variable, but I'm unsure how to construct an $f$ (without loss of generality) which I can then differentiate to contradict $f'(x)=0$.
any suggestions to start this proof?
 A: Let, $x,y\in [a,b]$ then by Mean value theorem we have, 
$$f(x)-f(y)=(x-y)f'(\xi)$$ for some $x<\xi<y$ and as given $f'(t)=0,\forall t\in(a,b)$ so, $f'(\xi)=0\implies f(x)=f(y),\forall x,y\in[a,b]$
Hence, $f$- is constant.
A: A proof by contradiction:
We have to show: $$(f'(x)=0 \implies f(x)=C) \ni \text{C is a constant}$$
Which is nothing but, $$P\implies Q$$. 
Suppose: $$\lnot (P\implies Q)$$
Which is nothing but, $$P \land \lnot Q$$
Therefore, suppose: 
$$f'(x)=0 \land f(x)=h(x) \ni \text{h(x) is an arbitrary non-constant function}$$ 
[$(h'(x)\neq 0)(\forall x\neq C^p_n)$; where $C^p$ is the critical point, and $n$ is an arbitrary index.] This bit is not really relevant since, we are comparing $f(x)$ and $h(x)$ on $A\subseteq D$ where the functions are identical: f(x)=h(x).
Notice, however, $$\frac{d}{dx}f(x)=\frac{d}{dx}h(x)$$
$$\frac{df}{dx}=\frac{dh}{dx}$$
Recall that we assumed: $h(x) \text{ is an arbitrary non-constant function}$
Therefore, 
$$\frac{dh}{dx}=[k(x) \lor L]$$
Where $k(x)$ is an arbitrary function not equal to zero and $L$ is an arbitrary non-zero constant. 
However, in either case:
$$\frac{dh}{dx}\neq0=\frac{df}{dx}=f'(x)$$ 
Therefore, since $\lnot (P\implies Q) \implies \bot$, then:
$$(P\implies Q)\implies \top$$
A: 
Let f be a function that is continuous on an interval [a,b] and differentiable on (a,b)$

1) that means for all $a \le a' < b' \le b$ then $f$ is continuous on the interval $[a', b']\subset [a,b]$ and differentiable on $(a',b') \subset (a,b)$.
and 
2) The mean value thereom applies.  
So for any $a \le a' < b' \le b$ we have that there is a $c; a' < c < b'$ where $f'(c) = \frac {f(b') - f(a')}2$. 
But $f'(c) = 0$ so $f(b') = f(a')$ for all $a \le a' < b' \le b$.
So $f$ is constant in the interval $[a,b]$ (can't say anythin about $f$ outside that interval.)
A: Here is an alternate proof which bypasses the Mean Value Theorem, and which I hope might be more conducive to visualization.  First, I'll start off with a small topological lemma to abstract out that part of the argument:
Lemma: Suppose $A \subseteq [0, 1]$ satisfies $0 \in A$, $A$ is closed under the left half-interval topology, and $A$ is open under the right half-interval topology.  Then $A = [0, 1]$.
Proof (outline): Let $B := \{ x \in [0, 1] \mid [0, x] \subseteq A \}$ and $c := \sup B$ (using the fact that $0 \in B$ so $B$ is nonempty) and suppose that $c < 1$.  Then from the condition that $A$ is closed under the left half-interval topology, we can conclude that $c \in B$.  On the other hand, the fact that $A$ is open under the right half-interval topology will now give a contradiction.  Therefore, $c = 1$; and again, from the condition that $A$ is closed under the left half-interval topology, this will imply that $1 \in B$, so $A = [0, 1]$ as desired. $\square$
Now, to use this: to illustrate the essential points of the argument, I will restrict to the case where the domain of $f$ is $[0, 1]$ and $f(0) = 0$.  The more general case should be straightforward to prove in a similar manner.  Now, fix any $\epsilon > 0$, and let $A_\epsilon := \{ x \in [0, 1] \mid |f(x)| \le \epsilon x \}$.  Then $0 \in A_\epsilon$; and it should be straightforward to see that $A_\epsilon$ is closed in the usual topology, and therefore also closed in the left half-interval topology.  On the other hand, suppose $x \in A_\epsilon$; then from the hypothesis $f'(x) = 0$, we see that there exists $\delta > 0$ such that whenever $0 < |y-x| < \delta$, then $\left| \frac{f(y) - f(x)}{y - x} \right| < \epsilon$.  Then, if in addition $y > x$, it follows that $|f(y)| \le |f(x)| + |f(y) - f(x)| < \epsilon x + \epsilon |y-x| = \epsilon y$; this shows that $A_\epsilon$ is open in the right half-interval topology.  Now from the lemma, it follows that $A_\epsilon = [0, 1]$.
In summary, we have shown that $\forall \epsilon > 0, \forall x \in [0, 1], |f(x)| \le \epsilon x$.  We may now interchange the two quantifiers to conclude $\forall x \in [0, 1], \forall \epsilon > 0, |f(x)| \le \epsilon x$.  However, this easily implies $\forall x \in [0, 1], f(x) = 0$.
A: Well, $f'(x)$ gives the rate of change of $f$ at a point $x$. If $f'(x)=0 \forall x \in (a,b)$, then $f(x)$ isn't changing, i.e. it is constant, on $(a,b)$.
That's the idea, you just need to show it mathematically.
