anonymous
  • anonymous
Let f be the function defined by: f(x)= {sin x, x<0 {x^2, 02 For what values of x is f not continuous? A. 0 only B. 1 only C. 2 only (teacher said this was it) D. 0 and 2 only E. 0, 1, and 2
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
It is C. To check continuity for these piecewise functions, the upper limit for the function must equal the lower limit for where the next functions starts. In English...sin (0) must equal 0 ^2, 1^2 must equal 2 - 1, and so on. Pretty much, where each part ends, the next function must pick up at the same point. The only one that doesn't is at x = 2. Because 2 - 2 = 0, but 2 - 3 = -1. Hope you understood my messy explanation.
anonymous
  • anonymous
How would you go about doing the problem? I get overwhelmed with all of it - especially the sin part.
anonymous
  • anonymous
@dtcrane

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anonymous
  • anonymous
Yes, it's hard to explain over the internet. Look at it this way. F is not continuous if you have to lift your pencil off the graph. For instance, y = x. You can draw this line forever, and ever, and ever, and...and never have to lift your pencil off the page. This function, you have to at x = 2. f(x)= {sin x, x<0 {x^2, 02 Imagine drawing sin x where x < 0. You're drawing, and drawing, and as you come closer to x = 0, sin (x) = 0. Then once you get close to 0, you have to start drawing x^2. Do you have to lift the pencil off the paper? No! Because sin(x) ended near 0, and x^2 picked up right where sin(x) left off. Again, now you have to draw x^2 from 0 to 1. So you're drawing, and drawing...as you get very very close to x = 1, x^2 = 1, right? Then, you have to start drawing 2-x. Do you have to take your pencil off the graph? No again! Because x^2 at 1 = 1, and so does 2 - x! Now try it for 2 - x up to 2, and see what happens when you have to start drawing x-3. Let me know if you understand this better please!
anonymous
  • anonymous
I don't understand this at all :3
anonymous
  • anonymous
I understand the part about not lifting your pencil - my teacher explained that concept. But when you said "This function, you have to at x = 2" I get confused. How did you know to do that?
anonymous
  • anonymous
I need to be able to do this algebraically - without drawing or using a calculator
anonymous
  • anonymous
Because 2 - x is only for 1< x < 2, right? That means when x = 2, plugging in, 2 - x ends at 0. Because 2 - x = 2 - 2 = 0 at x = 2. And x - 3 is for 2 < x < 3, right? So at 2, x - 3 = 2 - 3 = -1. So the endpoints of each tiny function have to be equal for it to be continuous. Since these two aren't equal at x = 2, it is not continuous.
anonymous
  • anonymous
How did you plug it in? - the 2-x I mean
anonymous
  • anonymous
We know the function is 2 - x, right? So when x = 2, you can plug it in for x. so 2 - x AT x = 2, is 2 - 2 = 0.
anonymous
  • anonymous
why did you choose to work with 2-x? what about the other three pieces?
anonymous
  • anonymous
You have to do all of them. You have to plug in all the endpoints to make sure they are equal. we did it for x = 0 ( sin(x) and x ^2), x = 1 (x^2 and 2 - x). Those were both equal at 0 or 1, so they were continuous. I'm having a hard time trying to explain this better, so if you don't understand after this, I am very sorry!

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