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EyadBest ResponseYou've already chosen the best response.1
F(x)= x+2m ,x<3 3m(x^2)+k ,3≤x<1 2(x^3)m ,x>1 ___________________________ (this is da multirule) ^^ Where F(x) is continuous at x=3 ,at x=1 Find m,k
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.1
NO! BAH DOES NOT WAIT!
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.1
So basically plug in x=3 into x+2m and into 3m(x^2)+k
 one year ago

EyadBest ResponseYou've already chosen the best response.1
Alright BAH (i wish if i was a mod >.>) Do BAH KNOW THE ANSWER ?
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.1
Now equate them because you want them to meet at the same point for the function to be continuous 3+2m=27m+k 2m27m=k+3 25m=k+3 k = (25m+3)
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.1
Do the same thing for 3m(x^2)+k 2(x^3)m, but at x=1
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.1
Solve the two equations
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.1
And everything will be revealed............................................
 one year ago

EyadBest ResponseYou've already chosen the best response.1
well that was my answer but i got a diff. value :/
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.1
k = (25m+3) 3m+k=2m 3m25m3=2m 22m+m=2+3 21m=5 m=5/21
 one year ago

bahrom7893Best ResponseYou've already chosen the best response.1
k = (125/21 + 3) ... whatever that is
 one year ago
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