# Calculus

For what value of a is
f(x)=(x^2)-a x<3 f(x)=2ax x(> or =)3
continuous at every x?

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1. correction its f(x)=(x^2)-1 not f(x)=(x^2)-a

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posted by Mary
2. To be continuous, the two graph must be "linked" at x = 3
that is,
x^2 - 1 = 2ax for x = 3
9-1 = 6a
a = 8/6 = 4/3

so the two functions are
f(x) = x^2 - 1 for x < 3 and f(x) = (8/3)x for x≥ 3

(note that (3,8) would be on both graphs, ignoring the restrictions)

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posted by Reiny

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