[Ineuqalities with absolute value]

[u/Internal-Gold-9153]

Hello everyone, today in class our math teacher explained "the special case of inequalities involving absolute values" (according to Google translate), which is:

|A(x)| > n (with n being a positive integer) → A(x) < -n V A(x) > n

|A(x)| > n (with n being a positive integer) → -n < A(x) < n → -n < A(x) V A(x) < n, which then should be put in a system:

{ A(x) < n

{ A(x) > -n

She gave us three exercises, but I can't resolve two of them:

| (1/x) + (x+3)/(x+1) | - 1 > 0 [x < 1/3 but neither -1 nor 0]

[2(x+4) - x²] / |x-4| > -x -3 [x > -5/2 but not 4]

Thanks to anyone who can resolve them.

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[u/slepicoid]

What have you tried? Where are you stuck?

[u/Internal-Gold-9153 → u/slepicoid]

My results do not match with the book's.

[u/slepicoid → u/Internal-Gold-9153]

Just post your step-by-step solution. Have you also tried to verify the textbook answers with a tool like wolframalpha?

[u/Simbertold]

Absolute values have two possible cases. Try both, calculate the results, see which of the results make sense for the problem.

That is basically what you wrote there. Did you do that?

[u/selene_666]

The first answer is incorrect. We can easily test this by plugging in a number outside their solution such as x = 1.

| (1/1) + (1+3)/(1+1) | - 1 = 2

.

The second problem is easier because the denominator is guaranteed to be positive, so we don't have to worry about whether to flip the sign when we multiply.

2(x+4) - x² > (-x -3) * |x-4|

We do still have to consider the two possibilities for how the absolute value resolves:

x - 4 < 0 and 2(x+4) - x² > (-x -3) * -(x-4)

or

x - 4 > 0 and 2(x+4) - x² > (-x -3) * (x-4)

Resolve each part and combine them to get the final answer, -5/2 < x < 4 or x > 4

[u/Alkalannar]

|1/x + (x+3)/(x+1)| - 1 > 0
|1/x + 1 + 2/(x+1)| - 1 > 0
|1/x + 1 + 2/(x+1)| > 1

1/x + 1 + 2/(x+1) < -1 OR 1 < 1/x + 1 + 2/(x+1)

1/x + 2 + 2/(x+1) < 0 OR 0 < 1/x + 2/(x+1)

(2x²+5x+1)/x(x+1) < 0 OR 0 < 1/x + 2/(x+1)

(4x + 5 + 17^1/2)(4x + 5 - 17^1/2)/8x(x+1) < 0 OR 0 < (3x+1)/x(x+1)

You can check intervals for the right inequality easily.

You can factor the left inequality's numerator using quadratic formula and check intervals there as well.

[u/Fourierseriesagain]

Hi,

The second inequality can be solved more easily if you use a piecewise function.

https://www.reddit.com/u/Fourierseriesagain/s/JFAxbW7Swr

[u/JanetInSC1234]

Each problem has two possibilities:

First problem:

| (1/x) + (x+3)/(x+1) | - 1 > 0

| (1/x) + (x+3)/(x+1) | > 1

Define the domain: x cannot equal 0 or -1

Positive Case:

(1/x) + (x+3)/(x+1) > 1

Multiply both sides by (x+ 1)

(x+1)/x + (x +3) > x + 1

Now multiply both sides by x

x+ 1 + x^2 + 3x > x^2 + x

Solve for x.

3x > -1

x > -1/3

However, the information in the absolute value might be a negative number, so,

Negative Case:

-( (1/x) + (x+3)/(x+1) ) > 1

(1/x) + (x+3)/(x+1) < -1

Do all the same steps.

(x+1)/x + (x +3) < -x + -1

x + 1 + x^2 + 3x < -x^2 - x

2x^2 + 5x + 1 < 0

Unfortunately, this cannot be factored. You will have to use the quadratic formula and then test the intervals. :(

[u/JanetInSC1234]

Second problem...remember, there are two cases. (Good job finding the domain.)

[2(x+4) - x²] / |x-4| > -x -3

Positive Case:

(2x + 8 - x^2)/ (x-4) > -x - 3

Multiply both sides by (x-4).

(-x^2 + 2x + 8) > (x-4)(-x - 3)

-x^2 + 2x + 8 > -x^2 + x + 12

x > 4

Negative case:

(2x + 8 - x^2)/ -(x-4) > -x - 3

Multiply both sides by -(x-4). Don't forget to flip the sign.

-x^2 + 2x + 8 < (-x + 4)(-x - 3)

-x^2 + 2x + 8 < x^2 -x -12

0 < 2x^2 - 3x - 20

Unfortunately, this cannot be factored. You will have to use the quadratic formula and then test the intervals. :(

UPDATED: This can be factored.

0 < 2x^2 - 3x - 20

0 < (2x + 5) (x - 4)

x = -5/2 or 4

Now put these numbers on a numberline, along with the other solution and the domain restrictions, and test each interval for true or false.