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math456

  • 3 years ago

Need help! Explain why the function is discontinuous at the given number a. Sketch the graph of the function!

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  1. math456
    • 3 years ago
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  2. math456
    • 3 years ago
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    @cwrw238 @Hero

  3. math456
    • 3 years ago
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    @jim_thompson5910

  4. jim_thompson5910
    • 3 years ago
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    what is cos(0)

  5. math456
    • 3 years ago
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    its 1

  6. jim_thompson5910
    • 3 years ago
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    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?

  7. math456
    • 3 years ago
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    yea so that means \[\cos x \neq x=0\]

  8. jim_thompson5910
    • 3 years ago
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    do you mean cos(x) doesn't equal x when x = 0?

  9. math456
    • 3 years ago
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    yea because at x its 0 and at x<0 its 1

  10. jim_thompson5910
    • 3 years ago
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    then you are correct, and this shows us that there is a discontinuity

  11. math456
    • 3 years ago
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    alright kool thanks and does the graph looks like this?|dw:1348180882506:dw|

  12. jim_thompson5910
    • 3 years ago
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    you're on the right track

  13. jim_thompson5910
    • 3 years ago
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    however, you only graph cos(x) when x < 0

  14. jim_thompson5910
    • 3 years ago
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    and you only graph 1 - x^2 when x > 0

  15. jim_thompson5910
    • 3 years ago
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    make sure you draw a hole at (0,1) and a closed circle at (0,0)

  16. math456
    • 3 years ago
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    |dw:1348181069990:dw| like this?

  17. jim_thompson5910
    • 3 years ago
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    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

  18. math456
    • 3 years ago
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    |dw:1348181225073:dw|

  19. jim_thompson5910
    • 3 years ago
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    good, you got it

  20. math456
    • 3 years ago
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    alright thanks for ur help!

  21. jim_thompson5910
    • 3 years ago
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    np

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