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Find k so that the following function is continuous on any interval: f(x)=kx if 0<=x<3 and f(x)=8x^2 if 3<=x K=?
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for x=3, the right piece of the function has value 72.. so to make this continuous, the left piece must have a value of 72 also... so, kx = 72 k(3) = 72 can you take it from here?
yup thanks
yw...:)
wait im lost plz continue ?
3k = 72 divide both sides by 3....
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where did u get 3k?
k(3) = 72 from the last post i made before i came back..
ok thanks now i got it
ok...
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