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Eyad

  • 3 years ago

Need Help In Continuity of a function .

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  1. bahrom7893
    • 3 years ago
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    post the question

  2. Eyad
    • 3 years ago
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    F(x)= x+2m ,x<-3 3m(x^2)+k ,-3≤x<1 2(x^3)-m ,x>1 ___________________________ (this is da multi-rule) ^^ Where F(x) is continuous at x=-3 ,at x=1 Find m,k

  3. Eyad
    • 3 years ago
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    Can't u wait ?!

  4. bahrom7893
    • 3 years ago
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    NO! BAH DOES NOT WAIT!

  5. bahrom7893
    • 3 years ago
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    So basically plug in x=-3 into x+2m and into 3m(x^2)+k

  6. bahrom7893
    • 3 years ago
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    -3+2m 3m(9)+k

  7. Eyad
    • 3 years ago
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    Alright BAH (i wish if i was a mod >.>) Do BAH KNOW THE ANSWER ?

  8. bahrom7893
    • 3 years ago
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    Now equate them because you want them to meet at the same point for the function to be continuous -3+2m=27m+k 2m-27m=k+3 -25m=k+3 k = -(25m+3)

  9. Eyad
    • 3 years ago
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    i did that ,and ?

  10. bahrom7893
    • 3 years ago
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    Do the same thing for 3m(x^2)+k 2(x^3)-m, but at x=1

  11. bahrom7893
    • 3 years ago
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    3m+k=2-m

  12. bahrom7893
    • 3 years ago
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    Solve the two equations

  13. bahrom7893
    • 3 years ago
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    And everything will be revealed............................................

  14. Eyad
    • 3 years ago
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    well that was my answer but i got a diff. value :/

  15. bahrom7893
    • 3 years ago
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    :\

  16. Eyad
    • 3 years ago
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    ty Bahrom :D

  17. bahrom7893
    • 3 years ago
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    k = -(25m+3) 3m+k=2-m 3m-25m-3=2-m -22m+m=2+3 -21m=5 m=-5/21

  18. bahrom7893
    • 3 years ago
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    k = -(125/21 + 3) ... whatever that is

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