0
$\begingroup$

say if you have

$$ \begin{cases} \frac{Ax}{a} & 0\leq x\leq a \\ A\frac{b-x}{b-a} & a\leq x\leq b \\ 0 & \textrm{else} \end{cases} $$

how would I check if it is continuous or not? The graph would be a straight line upwards from 0 to a and then a straight line down from a to b. would that be considered continuous?

Improve this question
$\endgroup$
1
  • $\begingroup$ Please consider using Mathjax in the future. I edited your answer. $\endgroup$
    – user224659
    Commented Apr 6, 2020 at 9:16

1 Answer 1

Reset to default
2
$\begingroup$

You are dividing the $x$-coordinate into three intervals. Within each interval the function is linear, and so it is trivially continuous there. The only places where discontinuities may arise is at the boundaries where the function changes form. To finish checking continuity you need to individually look at these boundaries. For example, if the function is $f(x)$ to the left of a boundary at $x=a$, and is $g(x)$ to the right of the boundary, you need to verify that: $$ \lim_{x \rightarrow 0^+} f(a-x) = \lim_{x \rightarrow 0^+} g(a+x) \ .$$

Improve this answer
$\endgroup$
4
  • $\begingroup$ Indeed you would also need to check the continuity inside the three parts. But as those are rational functions the continuity is trivial. Still it is a nice exercise to show that with your defintion. $\endgroup$
    – user224659
    Commented Apr 6, 2020 at 9:11
  • 1
    $\begingroup$ Yes, that's true. I've edited my answer to include some discussion of this. $\endgroup$
    – Clara Díaz Sanchez
    Commented Apr 6, 2020 at 9:25
  • $\begingroup$ The way you write the limits, you should put $x\rightarrow 0^+$ in the limits. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Apr 6, 2020 at 13:20
  • $\begingroup$ Yes, that's fixed now. $\endgroup$
    – Clara Díaz Sanchez
    Commented Apr 6, 2020 at 15:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.