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mathslover

find the minimum value of \( 4-5x-4x^2\).

  • one year ago
  • one year ago

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  1. FoolForMath
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    For any parabola of the form \(ax^2+bx+c=0 \) the vertex will be at \( \large \left(-\frac b {2a} , -\frac {b^2-4ac}{4a}\right) \).Equation of line of symmetry will be at \(x=-\frac b {2a} \) Maximum and minimum values will be at \( \large \left(-\frac b {2a} , -\frac {b^2-4ac}{4a}\right) \) and \( \large \left(-\frac b {2a} , -\frac {b^2-4ac}{4a}\right) \) for \(a<0\) and \(a>0\) respectively.

    • one year ago
  2. mathslover
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    @FoolForMath but that does not lead to the correct answer

    • one year ago
  3. RaphaelFilgueiras
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    i think there is no minimum value

    • one year ago
  4. RaphaelFilgueiras
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    it's unbounded

    • one year ago
  5. FoolForMath
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    For this problem there will be no minimum values.

    • one year ago
  6. mathslover
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    can we do this like : 4-5x-4x^2 4-4(x^2+5x/4) 4-4(x^2+5x4 +25/64) + 25/16 ?

    • one year ago
  7. mathslover
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    is this the correct method ?

    • one year ago
  8. RaphaelFilgueiras
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    |dw:1340428823754:dw|

    • one year ago
  9. RaphaelFilgueiras
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    just maximum value

    • one year ago
  10. mathslover
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    4-4(x^2+5x4 +25/64) + 25/16 89/16-4((x+5/8)^2) hence it as min. value of 89/16 at x =-5/8 is it right ?

    • one year ago
  11. RaphaelFilgueiras
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    nope

    • one year ago
  12. mathslover
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    why ?

    • one year ago
  13. RaphaelFilgueiras
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    THIS FUNCTION HAS NO MINIMUM VALUE

    • one year ago
  14. mathslover
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    http://www.wolframalpha.com/input/?i=4-5x-4x%5E2&t=crmtb01 i am getting right according to this but it may be wrong please have a look

    • one year ago
  15. FoolForMath
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    Lets try to plot this function graphically,

    • one year ago
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  16. mathslover
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    k

    • one year ago
  17. RaphaelFilgueiras
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    ok try x = 10000

    • one year ago
  18. FoolForMath
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    when do you think the minimum value will be achieved?

    • one year ago
  19. FoolForMath
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    why?

    • one year ago
  20. mathslover
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    can u tell m not soo good

    • one year ago
  21. RaphaelFilgueiras
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    every x you get i will pick one,that that my function will be less than yours

    • one year ago
  22. FoolForMath
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    Yes, for every minimum you find I can find another x which gives a lesser value ...

    • one year ago
  23. mathslover
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    thanks a lot every one

    • one year ago
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