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For what value of "a" is f(x)= x^2-1, x=<3 2ax , x >3 continuous over all Real numbers?
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replace x by 3 in both expressions set them equal and solve for a
\[f(3)=3^2-1=9-1=8=2a\times 3=6a\]
\[6a=8\] \[a=\frac{8}{6}=\frac{4}{3}\]
Brilliant, thank you very much!
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