Eyad
Need Help In Continuity of a function .
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bahrom7893
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post the question
Eyad
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F(x)= x+2m ,x<-3
3m(x^2)+k ,-3≤x<1
2(x^3)-m ,x>1
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(this is da multi-rule) ^^
Where F(x) is continuous at x=-3 ,at x=1
Find m,k
Eyad
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Can't u wait ?!
bahrom7893
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NO! BAH DOES NOT WAIT!
bahrom7893
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So basically plug in x=-3 into x+2m and into 3m(x^2)+k
bahrom7893
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-3+2m
3m(9)+k
Eyad
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Alright BAH (i wish if i was a mod >.>)
Do BAH KNOW THE ANSWER ?
bahrom7893
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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)
Eyad
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i did that ,and ?
bahrom7893
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Do the same thing for
3m(x^2)+k
2(x^3)-m, but at x=1
bahrom7893
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3m+k=2-m
bahrom7893
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Solve the two equations
bahrom7893
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And everything will be revealed............................................
Eyad
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well that was my answer but i got a diff. value :/
bahrom7893
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:\
Eyad
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ty Bahrom :D
bahrom7893
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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
bahrom7893
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k = -(125/21 + 3) ... whatever that is