Find the "critical points" of dy/dx = [(x-2)(x+3)]/[x-1]?

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Find the "critical points" of dy/dx = [(x-2)(x+3)]/[x-1]?

I know that the critical points are x = 2 and -3.
BUT should x = 1 also be another critical point?

2006-09-26 04:38:46 · 3 answers · asked by babycatpriere 2 in Education & Reference ➔ Homework Help

3 answers

yes at x=1 you get an asymptote

2006-09-26 04:48:43 · answer #1 · answered by who be boo? 5 · 0⤊ 0⤋

You could consider x = 1 a critical point - this equation graphed would have a 'gap' at x = 1, so it could hypothetically be called a 'critical point'

2006-09-26 04:47:16 · answer #2 · answered by p_rutherford2003 5 · 0⤊ 0⤋

no, beacuse if you fill in x=1, then you'll have to divide by 0 and that's impossible, so critical points are x=2 and x=-3 AND x DOES NOT equal 1!!

2006-09-26 04:42:09 · answer #3 · answered by Anonymous · 0⤊ 0⤋