anonymous
  • anonymous
Continuous Functions 1
Mathematics
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anonymous
  • anonymous
Continuous Functions 1
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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anonymous
  • anonymous
Check if the function is continuous: \[f(x) =\] { \[{x^2 - 1, x < 1}\] \[4-x, x \ge 1\] }
anonymous
  • anonymous
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.
anonymous
  • anonymous
brb

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anonymous
  • anonymous
ok, tks!
anonymous
  • anonymous
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\[\lim_{x^- \rightarrow 1}f(x)=\lim_{x^- \rightarrow 1}(x^2-1)=1-1=0\]
anonymous
  • anonymous
This is because f(x)=x^2-1 is only defined for x<1.
anonymous
  • anonymous
We couldn't use f(x) = 4-x since this is only true for x>=1.
anonymous
  • anonymous
*Now* we look at the right-hand limit.
anonymous
  • anonymous
I begin to understand.
anonymous
  • anonymous
\[\lim_{x^+ \rightarrow 1}f(x)=\lim_{x^+ \rightarrow 1}(4-x)=4-1=3\]
anonymous
  • anonymous
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.
anonymous
  • anonymous
UNDERSTAND! 0 <> 3. I have another problem of that kind, I'll try to do it. Thank you!
anonymous
  • anonymous
Excellent, I'm glad! Become a fan :p
anonymous
  • anonymous
I am already! :)
anonymous
  • anonymous
:D
anonymous
  • anonymous
Ah... What kind is this discontinuity? Removable, infinite, ...?
anonymous
  • anonymous
First kind, irremovable discontinuity.
anonymous
  • anonymous
Otherwise known as a 'jump' discontinuity.
anonymous
  • anonymous
Thanks again!
anonymous
  • anonymous
np

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