Zero Derivative implies Constant Function
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Theorem
Let $f$ be a real function which is continuous on the closed interval $\closedint a b$ and differentiable on the open interval $\openint a b$.
Suppose that:
- $\forall x \in \openint a b: \map {f'} x = 0$
Then $f$ is constant on $\closedint a b$.
Proof
When $x = a$ then $\map f x = \map f a$ by definition of mapping.
Otherwise, let $x \in \hointl a b$.
We have that:
- $f$ is continuous on the closed interval $\closedint a b$
- $f$ is differentiable on the open interval $\openint a b$
Hence it satisfies the conditions of the Mean Value Theorem on $\closedint a b$.
Hence:
- $\exists \xi \in \openint a x: \map {f'} \xi = \dfrac {\map f x - \map f a} {x - a}$
But by our supposition:
- $\forall x \in \openint a b: \map {f'} x = 0$
which means:
- $\forall x \in \openint a b: \map f x - \map f a = 0$
and hence:
- $\forall x \in \openint a b: \map f x = \map f a$
$\blacksquare$
Also see
- Derivative of Constant: This is the converse. Thus we see that $f$ is a constant function if and only if $\forall x: \map {f'} x = 0$.
- Zero Derivative implies Constant Complex Function
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 11.7$