Let g(x) = [c(x^2 - 9)/(x-3)] + 4 if x does not equal 3 and g(x) = 6 when x = 3... for what value(s) of c, if any, is g continuous at 3?
2007-12-11 07:26:24 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
limit c(x^2-9)/(x-3) +4 =limtc (x-3)(x+3)/(x-3) +4 =limit c( x+3) +4=c(3+3)+4=6c +4
x-->3
so 6c+4=6 so 6c=2 c= 1/3
so g(3)+ =g(3) =g(3)-
so it is continuous when c=1/3
2007-12-11 07:35:33 · answer #1 · answered by mbdwy 5 · 0⤊ 0⤋
for x not 3 g(x) = c(x+3)+4 which has lim 6c+4 if x==>3
6c+4=6 so c= 1/3
2007-12-11 07:34:49 · answer #2 · answered by santmann2002 7 · 0⤊ 0⤋
idk but by looking at it wouldn that problem be undefined since you can not divide by zero
This sucks I have to take technical calculus next semster :(
2007-12-11 07:36:21 · answer #3 · answered by the man the myth the answerer 5 · 0⤊ 1⤋