# How do you use the (analysis) definition of continuity to prove the following function f(x) = 3x +5 is continuous for all x in R?

Mar 5, 2016

We will use the Epsilon-Delta definition of continuity for $f \left(x\right) = 3 x + 5.$ See answer below.

#### Explanation:

Epsilon-Delta definition of continuity on all ${x}_{0} \in \mathbb{R}$ :

$\forall {x}_{0} \in \mathbb{R} , \forall \epsilon > 0 \exists {\delta}_{\epsilon , {x}_{0}} \text{ such that} \forall x \in \mathbb{R} :$

$| x - {x}_{0} | < {\delta}_{\epsilon , {x}_{0}} \Rightarrow | f \left(x\right) - f \left({x}_{0}\right) | < \epsilon .$

So we must find ${\delta}_{\epsilon , {x}_{0}}$ with respect to $\epsilon$ and ${x}_{0}$ so that the implication will be true.

$| f \left(x\right) - f \left({x}_{0}\right) | = | \left(3 x + 5\right) - \left(3 {x}_{0} + 5\right) | = | 3 x - 3 {x}_{0} | = | 3 \left(x - {x}_{0}\right) | = 3 | x - {x}_{0} | < 3 \cdot {\delta}_{\epsilon , {x}_{0}} = \epsilon$ if we set down ${\delta}_{\epsilon , {x}_{0}} = \frac{\epsilon}{3.}$

Therefore, $f \left(x\right)$ is continuous $\forall {x}_{0} \in \mathbb{R} .$

Notice that our $\delta$ only depends on $\epsilon$ and not on ${x}_{0}$, we call that uniform continuity.