SOLUTION: Did I do this table of signs problem correctly? http://i.imgur.com/o5Mjd7o.png Thanks

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Question 904505: Did I do this table of signs problem correctly?
http://i.imgur.com/o5Mjd7o.png
Thanks

Answer by jim_thompson5910(35256)   (Show Source): You can put this solution on YOUR website!
I've marked all of the incorrect answers in red boxes. I've also labeled the red boxes (A through G) to reference them below


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Explanation for Box A (column 1):
Pick a number that is in the interval (-infinty, -4). I'll pick x = -5
x+4 = -5+4 = -1 which is negative, therefore x+4 < 0 on the interval (-infinty, -4)
Further proof: x+4 < 0 ------> x < -4
So you should have a negative sign in box A.
----------------------------------------------------------------------
Explanation for Box B (column 1):
The signs above this last box in column 1 should be: +, -, - from top to bottom in that order
The two negatives multiply and/or divide to get a positive, so we're left with positive * positive = positive
You should have a "+" in box B
----------------------------------------------------------------------
Explanation for Box C (column 4):
plug x = -1 into x+1 to get x+1 = -1+1 = 0
The number 0 should be in box C
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Explanation for Box D (column 4):
since x+1 is zero when x = -1, and x+1 is in the numerator, the whole expression is 0
you should have a 0 in box D
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Explanation for Box E (column 5):
x = 0 is in the interval (-1,7)
7 - x = 7 - 0 = 7
So you have have a plus sign in box E
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Explanation for Box F (column 5):
x = 0 is in the interval (-1,7)
x+1 = 0+1 = 1
You should have a plus sign in box F
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Explanation for Box G (column 5):
The expression/function is defined on the interval (-1,7). It is only undefined when x = -4 since this causes a division by zero error.
In the boxes above box G, all of them are filled with a + sign
So the overall function is positive on the interval (-1,7)
You should have a + in box G
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Here is what the chart should look like. My corrections are in blue



And here is the graph of (the dashed blue line is the vertical asymptote)



This confirms the sign chart. Using either the graph or the sign chart, we see that the answer to in interval notation is

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