How do you solve: #(x^2-9)/(x^2-1) < 0#?

Answer 1

#x in (-3, -1) uu (1, 3)#

Your inequality looks like this

#(x^2 - 9)/(x^2 - 1) < 0 #
Right from the start, you know that any solution set that you might come up with cannot include the values of #x# that will make the denominator equal to zero.

More specifically, you need to have

#x^2 - 1 != 0 implies x != +- 1#

Now, in order for this inequality to be true, you need to have

#x^2 - 9 < 0" "# and #" "x^2 - 1 > 0#

or

#x^2 - 9 >0" "# and #" "x^2 - 1 < 0#

For the fist set of conditions to be true, you need to have

#{(x^2 - 9 < 0 implies x < +- 3 implies x in (-3, 3)), (x^2 - 1 > 0 implies x > +- 1 implies x in (-oo, -1) uu (1, + oo)) :}#
This means that you need #x in (-3, -1) uu (1, 3)#.

For the second set of conditions, you need to have

#{(x^2 - 9 > 0 implies x > +- 3 implies x in (-oo, -3) uu (3, + oo)), (x^2 - 1 < 0 implies x < +- 1 implies x in (-1, 1)) :}#
This time, those two intervals will not produce a valid solution set, or #x in O/#.
The only option left to you is #x in (-3, -1) uu (1,3)#. The values of #x# that belong to this interval will make the numerator negative and the denominator positive, which in turn will make the fraction negative.

graph{(x^2 - 9)/(x^2 - 1) [-18.02, 18.01, -9.01, 9.01]}

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Answer 2

To solve the inequality (\frac{x^2 - 9}{x^2 - 1} < 0), we first need to find the critical points where the expression equals zero or is undefined. The critical points occur where the numerator or denominator equals zero.

The numerator (x^2 - 9) equals zero when (x = \pm 3), and the denominator (x^2 - 1) equals zero when (x = \pm 1).

So, the critical points are (x = -3), (x = -1), (x = 1), and (x = 3).

Next, we need to test each interval created by these critical points to determine where the expression (\frac{x^2 - 9}{x^2 - 1}) is negative. We can do this by choosing test points in each interval and evaluating the expression.

  • Test a point in the interval ((- \infty, -3)), for example, (x = -4): (\frac{(-4)^2 - 9}{(-4)^2 - 1} > 0)
  • Test a point in the interval ((-3, -1)), for example, (x = -2): (\frac{(-2)^2 - 9}{(-2)^2 - 1} < 0)
  • Test a point in the interval ((-1, 1)), for example, (x = 0): (\frac{(0)^2 - 9}{(0)^2 - 1} > 0)
  • Test a point in the interval ((1, 3)), for example, (x = 2): (\frac{(2)^2 - 9}{(2)^2 - 1} < 0)
  • Test a point in the interval ((3, \infty)), for example, (x = 4): (\frac{(4)^2 - 9}{(4)^2 - 1} > 0)

So, the solution to the inequality (\frac{x^2 - 9}{x^2 - 1} < 0) is (x \in (-3, -1) \cup (1, 3)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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