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For what value of "a" is f(x)= x²-1 if x≤3 2ax, if x>3 continuous over Real numbers?
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-2 <= x <= 2 for the 1st one
\[\lim_{x \rightarrow 3-}x^2-1=8=\lim_{x \rightarrow 3+}2ax=\lim_{x \rightarrow 3+}6a=8, a=4/3\]
It's given that f(x) is continuous at 3 so limit at (3-h) is equal to (3+h) where \(h\to0\) limit at x=3-h=> (3-h)^2-1=> 9-1=8 at x=3+h f(3+h)=2a(3+h)=6a+2ah=6a f(3+h)=f(3-h) 6a=8 a=4/3
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