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twirlere

  • 3 years ago

f(x)=(x^2-3x-4)/(x-2) Find points of increase and decrease and all relative extrema.

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  1. anonymous
    • 3 years ago
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    you don't really need any calculus for this one

  2. Xavier
    • 3 years ago
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    Try factoring the numerator

  3. anonymous
    • 3 years ago
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    it is a rational function with a slant asymptote at \(x-1\)

  4. anonymous
    • 3 years ago
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    it is always increasing

  5. anonymous
    • 3 years ago
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    you could take the derivative and get \[\frac{x^2-4x+10}{(x-2)^2}\] but the denominator is never negative, and neither is the numerator ( you can check that it has no zeros)

  6. anonymous
    • 3 years ago
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    if you remember plotting rational functions in pre calc you may remember what something like this looks like

  7. twirlere
    • 3 years ago
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    That's the derivative that I got also. I was trying to find the critical points from this...

  8. twirlere
    • 3 years ago
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    So what would the critical points be?

  9. saloniiigupta95
    • 3 years ago
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    Factorise the Numerator... Can you??? Then the critical points for an expression like this... \[(x-a)(x-b)/ (x-c) \] are a,b and c...

  10. twirlere
    • 3 years ago
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    I can't factor \[x^2-4x-10\]

  11. saloniiigupta95
    • 3 years ago
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    You are supposed to factor x^2 - 3x-4 ... I think your question says so...

  12. twirlere
    • 3 years ago
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    Into... (x-2)(x-2)...?

  13. saloniiigupta95
    • 3 years ago
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    x^2 - 3x - 4 =x^2 - 4x +x -4 = x(x-4) +1(x-4) = (x-4) (x+1)

  14. twirlere
    • 3 years ago
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    And critical points are then x=4 and x=-1

  15. twirlere
    • 3 years ago
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    I have to find the first derivative then factor

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