At what point is the function continuous
y=(1/(x-2)-3x)
2006-10-01 23:10:48 · 10 answers · asked by Bernard 1 in Science & Mathematics ➔ Mathematics
Realize, when X-2 in the denominator equals ZERO, the function is invalid. (devide by zero). That means x-2=0, x=2.
The answer is all real number except x=2
2006-10-01 23:22:44 · answer #1 · answered by tkquestion 7 · 1⤊ 0⤋
the function has a singularity at x=2 because 1/0 is undefined.
the function is continuous everywhere else.
2006-10-02 10:26:18 · answer #2 · answered by michaell 6 · 0⤊ 0⤋
y=(1/(x-2)-3x)
x=2 the function is discontinous
2006-10-02 09:13:36 · answer #3 · answered by openpsychy 6 · 0⤊ 0⤋
(x-2)
2006-10-02 06:55:57 · answer #4 · answered by Steph_kinse 2 · 0⤊ 0⤋
(x-2)
2006-10-02 06:21:08 · answer #5 · answered by sani 1 · 0⤊ 0⤋
(x-2)
2006-10-02 06:17:18 · answer #6 · answered by Asher 3 · 0⤊ 0⤋
the function is continious at all points except when not defined that is not defined at x-2
so it is continuous at all points except 2
2006-10-02 06:23:01 · answer #7 · answered by Mein Hoon Na 7 · 1⤊ 0⤋
That function is continuous everywhere it is defined.
2006-10-02 06:44:08 · answer #8 · answered by tgypoi 5 · 0⤊ 0⤋
(minus infinity, 2) or (2,infinity)
2006-10-02 06:38:56 · answer #9 · answered by iyiogrenci 6 · 0⤊ 0⤋
You should try studying more.
2006-10-02 06:19:10 · answer #10 · answered by Linzy Rae 4 · 0⤊ 0⤋