anonymous
  • anonymous
Need help! Explain why the function is discontinuous at the given number a. Sketch the graph of the function!
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
@cwrw238 @Hero
anonymous
  • anonymous
@jim_thompson5910

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jim_thompson5910
  • jim_thompson5910
what is cos(0)
anonymous
  • anonymous
its 1
jim_thompson5910
  • jim_thompson5910
so notice how in the first piece, the y value at x = 0 is 1 but the y value in the second piece at x = 0 is 0 Can you visualize what I'm describing?
anonymous
  • anonymous
yea so that means \[\cos x \neq x=0\]
jim_thompson5910
  • jim_thompson5910
do you mean cos(x) doesn't equal x when x = 0?
anonymous
  • anonymous
yea because at x its 0 and at x<0 its 1
jim_thompson5910
  • jim_thompson5910
then you are correct, and this shows us that there is a discontinuity
anonymous
  • anonymous
alright kool thanks and does the graph looks like this?|dw:1348180882506:dw|
jim_thompson5910
  • jim_thompson5910
you're on the right track
jim_thompson5910
  • jim_thompson5910
however, you only graph cos(x) when x < 0
jim_thompson5910
  • jim_thompson5910
and you only graph 1 - x^2 when x > 0
jim_thompson5910
  • jim_thompson5910
make sure you draw a hole at (0,1) and a closed circle at (0,0)
anonymous
  • anonymous
|dw:1348181069990:dw| like this?
jim_thompson5910
  • jim_thompson5910
closer, but now erase the piece of cos(x) that's to the right of x = 0 and erase the piece of 1 - x^2 that's to the left of x = 0
anonymous
  • anonymous
|dw:1348181225073:dw|
jim_thompson5910
  • jim_thompson5910
good, you got it
anonymous
  • anonymous
alright thanks for ur help!
jim_thompson5910
  • jim_thompson5910
np

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