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mathslover

  • 3 years ago

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

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  1. FoolForMath
    • 3 years ago
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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.

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

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

  4. RaphaelFilgueiras
    • 3 years ago
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    it's unbounded

  5. FoolForMath
    • 3 years ago
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    For this problem there will be no minimum values.

  6. mathslover
    • 3 years ago
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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 ?

  7. mathslover
    • 3 years ago
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    is this the correct method ?

  8. RaphaelFilgueiras
    • 3 years ago
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    |dw:1340428823754:dw|

  9. RaphaelFilgueiras
    • 3 years ago
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    just maximum value

  10. mathslover
    • 3 years ago
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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 ?

  11. RaphaelFilgueiras
    • 3 years ago
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    nope

  12. mathslover
    • 3 years ago
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    why ?

  13. RaphaelFilgueiras
    • 3 years ago
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    THIS FUNCTION HAS NO MINIMUM VALUE

  14. mathslover
    • 3 years ago
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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

  15. FoolForMath
    • 3 years ago
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    Lets try to plot this function graphically,

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  16. mathslover
    • 3 years ago
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    k

  17. RaphaelFilgueiras
    • 3 years ago
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    ok try x = 10000

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

  19. FoolForMath
    • 3 years ago
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    why?

  20. mathslover
    • 3 years ago
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    can u tell m not soo good

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

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

  23. mathslover
    • 3 years ago
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    thanks a lot every one

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