Let $f \colon \mathbb{R} \to \mathbb{R}$ be differentiable on an interval $I$. Prove: $$f \prime \, \text{is bounded on $I$} \Rightarrow f \, \text{is uniformly continuous on I}$$
I'm supposed to solve this using Lagrange's theorem.
My attempt:
Recall the definition of uniform continuity: $$\forall \varepsilon>0 \exists \delta>0 \forall x,y \in I \colon(|x-y|< \delta \Rightarrow |f(x)-f(y)|<\varepsilon)$$ Now suppose $f \prime$ is bounded on $I$. Then: $$\exists x_1 \in I \colon \forall x_2 \in I \colon |f \prime(x_1)| \geq |f \prime (x_2)|$$ Let $x,y \in I$ be arbitrary. By Lagrange's theorem: $$\exists \xi \in (x,y):|f(x)-f(y)|=|f \prime(\xi)||x-y| \leq |f \prime(x_1)||x-y|< \varepsilon$$ So let's recap: $$|x-y|< \frac{\varepsilon}{|f \prime (x_1)|} \Rightarrow |f(x)-f(y)|< \varepsilon$$ So we have found $\delta$ and we use $\delta:= \frac{\varepsilon}{|f \prime (x_1)|}$
This concludes the proof. Is this valid?