Prove that $f(x) = 4 - 15x$ is continuous

Question

Need help showing this.

Show that the function $f\colon \mathbb{R} \to \mathbb{R}:$ defined by $f(x) = 4 - 15x$ is continuous at every point $c\in\mathbb{R}$.

Thanks.

Answer 1

Below I try to describe one way of thinking about this. The actual proof is written in blue, the rest is just guidance. I did it with a lot of detail, to the point where it might actually confuse you, so beware.

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You are being asked to prove that $$\forall c\in \mathbb R\forall \varepsilon>0\,\exists \delta>0\,\forall x\in \mathbb R\left(|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon\right).$$

(This is just the definition, there is nothing to it).

(In words: given a real number $c$ and a positive real $\varepsilon$, there exists a positive real number $\delta$ with the property that for all real numbers $x$ such that $|x-c|<\delta$, it holds that $|(4-15x)-(4-15c)|<\varepsilon$).

(Since this is a universal statement, you begin by taking an arbitrary $c\in \mathbb R$.)

$\color{blue}{\text{Let $c\in \mathbb R$.}}$

(Now you wish to prove that $\forall \varepsilon>0\,\exists \delta>0\,\forall x\in \mathbb R\left(|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon\right)$. This again is a universal statement, so take an arbitrary $\varepsilon >0$).

$\color{blue}{\text{Let $\varepsilon >0$.}}$

(Now you wish to prove that $\exists \delta>0\,\forall x\in \mathbb R\left(|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon\right)$.

(In words: there exists a positive real number $\delta$ with the property that for all real numbers $x$ such that $|x-c|<\delta$, $|(4-15x)-(4-15c)|<\varepsilon$).

This is an existential statement, one way to do it is to find $\delta >0$ such that $\forall x\in \mathbb R\left(|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon\right)$. This is what I suggest you do, find a $\delta$ that works). To do this, I suggest the thought of process below which I do in a scratch work box).

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Scratch work: To prove that $\exists \delta>0\,\forall x\in \mathbb R\left(|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon\right)$, we can begin by assuming we already have such $\delta$ (just to see what happens).

So let $\delta>0$ be such that $\forall x\in \mathbb R(|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon)$ and take an arbitrary $x\in \mathbb R$.

We have the conditional statement $|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon$.

Now it's useful to look at the consequent to see what kind of restrictions we need to impose on $\delta$ for this conditional statement to hold. It is almost always necessary to relate the consequent with the inequality in the antecedent ($|x-c|<\delta$).

The following holds: $|(4-15x)-(4-15c)|=15|x-c|$. One notices that in the consequent occurs the number $|x-c|$ which also occurs in the antecedent, thus relating the consequent to the antecedent. Recalling that $|x-c|<\delta$ one gets $|(4-15x)-(4-15c)|=15|x-c|<15\delta$ . This suggest that letting $\delta$ be such that $15\delta =\varepsilon$ (or equivalently $\delta=\dfrac \varepsilon {15}$) should work.

Now the scratch work is abandoned and we get back to the proof.

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$\color{blue}{\text{Choose $\delta =\dfrac{\varepsilon}{15}$.}}$

(You wish to prove that $\forall x\in \mathbb R\left(|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon\right)$. It is a universal statement so you should take $x\in \mathbb R$ arbitrarily).

$\color{blue}{\text{Let $x\in \mathbb R$.}}$

(Now you want to prove the conditional statement $|x-c|<\delta \implies |(4-15x)-(4-15c)|<\varepsilon$. In words: if $|x-c|<\delta$, then $|(4-15x)-(4-15c)|<\varepsilon$). The most forward way to do this is to assume the antecedent and try to prove the consequent).

$\color{blue}{\text{Suppose $|x-c|<\delta$.}}$

(Now you want to prove that $|(4-15x)-(4-15c)|<\varepsilon$).

$\color{blue}{\text{One has}}$

$$\color{blue}{\text{$\begin{align}|(4-15x)-(4-15c)|&=|4-15x-4+15c|\\ &=|-15x+15c|\\ &=15|-x+c|\\ &=15|x-c|<\varepsilon .\end{align}$}}$$

$\color{blue}{\text{Since $|x-c|<\delta$ and $\delta =\dfrac {\varepsilon}{15}$ it follows that}}$ $$\color{blue}{|(4-15x)-(4-15c)|=15|x-c|<15\delta=15\dfrac{\varepsilon}{15}=\varepsilon.\,\,\square}$$

This concludes the proof.

Answer 2

Let $a\in \Bbb R$ be any point, to show that it is continuous at point $a$,

$$|f(x) - f(a)|< \epsilon \\ \mathrm{or, } |4 - 15x - a + 15a| < \epsilon \\ \mathrm{or, } 15 |x-a| < \epsilon$$

Choose your $\delta = \epsilon / 15$.

The $\epsilon-\delta $ definition of continuity says that you need to show that for every $\epsilon > 0$ there exists some $\delta > 0$ such that , $ |x-a|< \delta \implies |f(x) - f(a)| < \epsilon $. So by choosing $\delta = \frac \epsilon {15}$ we showed the existence of such delta.

Answer 3

Let $c \in \mathbb{R}$ be arbitrary. Let $\epsilon > 0 $ be given. To show continuity of $f(x)$ ar $c$, by definition, we have to find $\delta > 0$ such that if $|x-c|< \delta$, then $|f(x) - f(c) | < \epsilon $. So, our goal is to find $\delta $ that works. Let's see,

$$ |f(x) - f(c)| = | 4 - 15x - (c - 15c) | = |4 - 15x - 4 + 15c| = 15|x-c| $$

The last term if less than $\epsilon $ if and only if $ \delta = \frac{ \epsilon }{15} $

Answer 4

1. First of all $\varepsilon$-$\delta$ definition:
   $f$ is continuous at point $c$ if for every $\varepsilon > 0$ there exists a $\delta > 0$ such that for all $x$ $$| x - c | < \delta \Rightarrow | f(x) - f(c) | < \varepsilon. \,$$

2. We'll try to find $\delta$ which will be good enough: $$\varepsilon>|f(x)-f(c)|=|4-15x-4+15c|=15|x-c|.$$ We know that $|x-c|$ should be less than $\delta$ so we can use $\delta=\frac{\varepsilon}{15}$.

3. And now check the definition:
   let $c$ be arbitrary, for every $\varepsilon>0$ we will use $\delta=\frac{\varepsilon}{15}>0$.
   Now, for all $x$ $$| x - c | < \delta \Rightarrow |x-c|<\frac{\varepsilon}{15} \Rightarrow 15|x-c|<\varepsilon \Rightarrow |15c - 15x|<\varepsilon \Rightarrow |4-15x-4+15c|<\varepsilon \Rightarrow |f(x)-f(c)|<\varepsilon.$$ And that's what we wanted to prove.

It is pretty much like the other answers, so nothing new, but I hope it will help you.