2006-11-15 15:41:13 · 9 answers · asked by saima y 1 in Science & Mathematics ➔ Mathematics
x = 3. Try entering it on your calculator.
2006-11-15 15:42:57 · answer #1 · answered by Anonymous · 0⤊ 0⤋
When the function is undefined, that is the value for which it is discontinuous. At x=3, the function would be a division by 0, which is undefined, and hence discontinuous.
2006-11-15 15:43:50 · answer #2 · answered by Brian B 4 · 0⤊ 0⤋
y=1/x is discontinuous at x=0, and -3 has no effect
y=1/(x-3) is discontinuous for x=3
2006-11-15 15:44:42 · answer #3 · answered by questor_2001 3 · 1⤊ 0⤋
the function is discontinuous when x equals three since three would make the denominator equal zero and produce a vertical asymptote. So the answer is:
X=3
2006-11-15 15:43:57 · answer #4 · answered by AnswersGuru 3 · 0⤊ 0⤋
the function has a spinoff (= 2) on the period (0, infinity) so the function is continuous on that period. * theorem: a differentiable function is a continuous function. likewise, the function has a spinoff (= -2x) on the period (-infinity, 0) so the function is continuous on that period. at x=0, the left-sided reduce (= 3) equals the right-sided reduce (= 3), and also equals to f(0)=3. subsequently through definition of continuity, f is continuous at x=0. f is continuous everywhere.
2016-11-24 21:56:45 · answer #5 · answered by fahner 4 · 0⤊ 0⤋
There is a vertical asymptote at the point at which the function is undefined. It is undefined when the denominator = 0, which would be caused by x=3.
2006-11-15 16:11:29 · answer #6 · answered by Anonymous · 0⤊ 0⤋
3
nothing can be divided by zero.
Hence when x = 3 then the equation would be y=1/0 which is impossible
2006-11-15 15:42:48 · answer #7 · answered by Metaspy 3 · 0⤊ 0⤋
0 if y=(1/x) -3
3 if y=1/(x-3)
2006-11-15 15:46:26 · answer #8 · answered by Ahkay 2 · 0⤊ 0⤋
No its continuous
2006-11-15 15:43:01 · answer #9 · answered by Anonymous · 0⤊ 0⤋