# Proving function in real analysis

A function $f : \mathbb R \to \mathbb R$ is continuous if for every $x \in \mathbb R$ and every $\varepsilon>0$ there exists a $\delta>0$ such that

$$|f(y)−f(x)|< \varepsilon \ \ \text{whenver} \ \ \ |y−x|< \delta$$

(a) Show that the function $f : \mathbb R \to \mathbb R$ given by

$$f(x) = 4x − 3$$

is continuous.

I am revising for my exam and I am unable to arrive at the answer. Does anyone know how I can tackle this.

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So you took an arbitrary $\epsilon>0$. Then you formed $|f(y)-f(x)|=4|y-x|$. What kept you from finding an appropriate $\delta$? – 1015 Apr 28 '13 at 20:44
also note, at least from how I've learned analysis, this is uniform continuity not general continuity at a point – DanZimm Apr 28 '13 at 21:00
@julien: Oops! Never mind. – Pete L. Clark Apr 28 '13 at 21:17
@DanZimm No. This is continuity at every $x$. Not uniform continuity. The $\delta$ depends on $x$ as it comes after. – 1015 Apr 28 '13 at 21:27

Define $f : \mathbb R \to \mathbb R$ by $f(x) = 4x - 3$.
Let $x_0 \in \mathbb R$ be arbitrary, we want to show that $f$ is continuous at $x_0$.
Let $\varepsilon > 0$ and choose $\displaystyle \delta = \frac{\varepsilon}{4}$, if $\vert x - x_0 \vert < \delta$, then $$\vert f(x) - f(x_0) \vert = \vert (4x - 3) - (4x_0 - 3) \vert = 4 \vert x - x_0 \vert < 4 \delta = 4 \frac{\varepsilon}{4} = \varepsilon.$$
Usually, the way I think through these problems is: I let $\varepsilon > 0$ and choose $\delta = \boxed{\text{to pick later}}$, then I go about my calculations, adding restraints on $\delta$ as I go along to make the proof work out. Then I go back and fill it in with the appropriate value.