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anonymous
 5 years ago
Continuous Functions 1
anonymous
 5 years ago
Continuous Functions 1

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Check if the function is continuous: \[f(x) =\] { \[{x^2  1, x < 1}\] \[4x, x \ge 1\] }

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay  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
 5 years ago
Best ResponseYou've already chosen the best response.0So this is about taking things called, "left" and "right" limits. When we approach a function from the left, it's a lefthand limit. When it's from the right, it's a righthand limit. When x<1, you to get to 1, you need to approach f(x) from the left. So we write\[\lim_{x^ \rightarrow 1}f(x)=\lim_{x^ \rightarrow 1}(x^21)=11=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0This is because f(x)=x^21 is only defined for x<1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0We couldn't use f(x) = 4x since this is only true for x>=1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0*Now* we look at the righthand limit.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I begin to understand.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{x^+ \rightarrow 1}f(x)=\lim_{x^+ \rightarrow 1}(4x)=41=3\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, 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
 5 years ago
Best ResponseYou've already chosen the best response.0UNDERSTAND! 0 <> 3. I have another problem of that kind, I'll try to do it. Thank you!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Excellent, I'm glad! Become a fan :p

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah... What kind is this discontinuity? Removable, infinite, ...?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0First kind, irremovable discontinuity.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Otherwise known as a 'jump' discontinuity.
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