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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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Your question is phrased as an isolated problem, without any further information or context. This does not match MSE quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide more context, and include your work and thoughts on the problem. Making these improvements will attract more appropriate answers and make the question more valuable for future MSE visitors. – Eric Naslund Apr 28 '13 at 20:44
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.

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