Derivative is Zero Implies Constant Function

See Mean Value Theorem (MVT), which gives the following results:

Corollary

If $g : I \to R$ is differentiable on an interval $I$ and satisfies $g^{'} (x) = 0$ for all $x \in I$ then $g (x) = k$ for some constant $k \in R$ .

Proof

Take $x, y \in I$ and suppose $x < y$ . Apply the Mean Value Theorem (MVT) to $g$ on the interval $[x, y]$ to get that:

g^{'} (c) = \frac{g (y) - g (x)}{y - x} = 0

Thus $g (y) = g (x)$ , but since $x, y$ were arbitrary then set $k = g (x) = g (y) = \dots$ , then it follows that $g (x) = k$ for all $x \in I$ .

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Corollary

If $f, g$ are differentiable on an interval $I$ and satisfy $f^{'} (x) = g^{'} (x)$ for all $x \in I$ then $f (x) = g (x) + k$ for some constant $k \in R$ .

Proof

Let $h (x) = f (x) - g (x)$ and apply the previous corollary to the differentiable function $h$ .

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