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math456

  • 3 years ago

Alright last question need help! Q.51

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  1. math456
    • 3 years ago
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  2. math456
    • 3 years ago
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    @lgbasallote @jim_thompson5910

  3. nphuongsun93
    • 3 years ago
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    to me, the a and b are discontinuous at a, because they are undefined when x = a

  4. math456
    • 3 years ago
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    right so when I plug in 1 is get -1 and when i plug 2 i get 15.. so they aren't equal!

  5. nphuongsun93
    • 3 years ago
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    lol no a) if you plug in x = 1, the denominator will be 1-1 = 0 --> undefined b) if you plug in x = 2, the denominator will be 2-2 = 0 --> undefined now you have to find function g (try simplify the function f for a and b)

  6. nphuongsun93
    • 3 years ago
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    i will do a) for you \[f(x) = \frac{x^4-1}{x-1}\] using a^2 - b^2 = (a+b)(a-b) \[f(x) = \frac{(x^2+1)(x^2-1)}{x-1}\]and again\[f(x) = \frac{(x^2+1)(x+1)(x-1)}{x-1}\]simplify\[f(x) = (x^2+1)(x+1)\]^^^ your new function g

  7. math456
    • 3 years ago
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    but its hard. should i find it by the cal..

  8. math456
    • 3 years ago
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    ohhh i was asking for Question 51 not 47!

  9. math456
    • 3 years ago
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    the IVT theorem

  10. nphuongsun93
    • 3 years ago
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    Oh sheet my bad

  11. nphuongsun93
    • 3 years ago
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    x^4 + x - 3 = 0 1^4 + 1 - 3 = -2 2^4 + 2 - 3 = 15 to get from -2 to 15, you have to cross the x-axis somewhere, therefore there's a root between 1 and 2.

  12. math456
    • 3 years ago
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    so I was right b4 :P

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