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Show a full solution to the inequality [(x+2)/(x-4)] \leqslant 3 Can someone please tell me what I did wrong?
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Oct 8, 2019
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Show a full solution to the inequality [(x+2)/(x-4)]
3
Can someone please tell me what I did wrong?
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Your sign chart is a bit off. The first two rows are correct. However keep in mind that you factored out a negative (when you factored -2x+14) so you need to negate the signs of the f(x) row.
Another thing to keep in mind: x = 4 is not allowed as its the vertical asymptote. However, x = 7 is valid. So you'll have an open circle at x = 4 and a close circle at x = 7
The answer in interval notation is
(-infinity, 4) U [7, infinity)
which translates to this disjoint set of inequalities:
x < 4 or x >= 7
Mike Marks
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Oct 13, 2019
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