For what value of a is
f(x)=(x^2)-a x<3 f(x)=2ax x(> or =)3
continuous at every x?
2 answers
correction its f(x)=(x^2)-1 not f(x)=(x^2)-a
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)
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)