For what values of c is the function continuous
f(x)=(cx^2-1) , x<2
and
f(x)=(c^2x^2-1) x≧2
謝啦謝啦!!!!
Function continuous?!
2006-12-08 6:51 pm
回答 (2)
2006-12-08 7:49 pm
✔ 最佳答案
At x=2f(2)=(c^2x^2-1)=4c^2-1
when x tends to 2 for x<2
lim(x ->2) f(x) = 4c-1
if the function continuous, lim (x ->2) f(x) = f(2)
4c^2-1=4c-1
c(c-1)=0
c = 0 or 1
** Note f(x) is a constant function which f(x)= -1 if c=0
2006-12-09 2:25 am
to make f continuous,
left limx->2 with f(x<2) = f(x=2)
(c(2)^2 - 1) = (c^2*(2)^2 - 1)
4c-1 = 4c^2 - 1
c = c^2
0 = c^2 - c = c * (c-1)
c= 0 or c =1
if c = 0
f(x) = -1 for all x.
if c= 1,
f(x) = x^2 - 1 for all x.
left limx->2 with f(x<2) = f(x=2)
(c(2)^2 - 1) = (c^2*(2)^2 - 1)
4c-1 = 4c^2 - 1
c = c^2
0 = c^2 - c = c * (c-1)
c= 0 or c =1
if c = 0
f(x) = -1 for all x.
if c= 1,
f(x) = x^2 - 1 for all x.
收錄日期: 2021-04-12 22:57:05
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061208000051KK00857