anonymous
  • anonymous
Alright last question need help! Q.51
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
@lgbasallote @jim_thompson5910
anonymous
  • anonymous
to me, the a and b are discontinuous at a, because they are undefined when x = a

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anonymous
  • anonymous
right so when I plug in 1 is get -1 and when i plug 2 i get 15.. so they aren't equal!
anonymous
  • anonymous
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)
anonymous
  • anonymous
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
anonymous
  • anonymous
but its hard. should i find it by the cal..
anonymous
  • anonymous
ohhh i was asking for Question 51 not 47!
anonymous
  • anonymous
the IVT theorem
anonymous
  • anonymous
Oh sheet my bad
anonymous
  • anonymous
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.
anonymous
  • anonymous
so I was right b4 :P

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