Question

Continuous Functions 1

Answer 1

Check if the function is continuous:
\[f(x) =\] {
\[{x^2 - 1, x < 1}\]
\[4-x, x \ge 1\]
}

Answer 2

Okay - you'll see that the part of f that you use to evaluate the function will depend on whether x is less than 1, or greater than or equal to 1. To show that the function is continuous, you're being asked to show that the value you get for f(x) and you approach 1 from either side, is the same. I'll show you.

Answer 3

brb

Answer 4

ok, tks!

Answer 5

So this is about taking things called, "left" and "right" limits. When we approach a function from the left, it's a left-hand limit. When it's from the right, it's a right-hand limit.

When x<1, you to get to 1, you need to approach f(x) from the left. So we write\[\lim_{x^- \rightarrow 1}f(x)=\lim_{x^- \rightarrow 1}(x^2-1)=1-1=0\]

Answer 6

This is because f(x)=x^2-1 is only defined for x<1.

Answer 7

We couldn't use f(x) = 4-x since this is only true for x>=1.

Answer 8

*Now* we look at the right-hand limit.

Answer 9

I begin to understand.

Answer 10

\[\lim_{x^+ \rightarrow 1}f(x)=\lim_{x^+ \rightarrow 1}(4-x)=4-1=3\]

Answer 11

Now, the function can only be continuous if it has the same value when x approaches 1 from either side (otherwise, if you look at a plot, the graph will be cut in two).

Since your function does not reach the same value as we approach from the left and the right, it cannot be continuous. If you were to plot f(x), you could see it visually.

Answer 12

UNDERSTAND!
0 <> 3.
I have another problem of that kind, I'll try to do it. Thank you!

Answer 13

Excellent, I'm glad! Become a fan :p

Answer 14

I am already!
:)

Answer 15

:D

Answer 16

Ah... What kind is this discontinuity? Removable, infinite, ...?

Answer 17

First kind, irremovable discontinuity.

Answer 18

Otherwise known as a 'jump' discontinuity.

Answer 19

Thanks again!

Answer 20

np