How do you solve and write the following in interval notation:#((x+8)(x-5) )/ (x-1) ≥ 0#?

Answer 1

The solution is #x in [-8,1 [uu [5, +oo [#

Let #f(x)=((x+8)(x-5))/(x-1)#

Let's build the sign chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaa)##-8##color(white)(aaaaaaa)##1##color(white)(aaaaaaaa)##5##color(white)(aaaaaa)##+oo#
#color(white)(aaaa)##x+8##color(white)(aaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaaa)##+#
#color(white)(aaaa)##x-1##color(white)(aaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaaa)##+#
#color(white)(aaaa)##x-5##color(white)(aaaaaaa)##-##color(white)(aaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##-##color(white)(aaaaa)##+#
#color(white)(aaaa)##f(x)##color(white)(aaaaaaaa)##-##color(white)(aaaa)##+##color(white)(aaaa)##||##color(white)(aaaa)##-##color(white)(aaaaa)##+#

Therefore,

#f(x)>=0# when #x in [-8,1 [uu [5, +oo [#
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Answer 2

To solve and write the inequality (\frac{{(x+8)(x-5)}}{{x-1}} \geq 0) in interval notation, follow these steps:

  1. Find the critical points by setting the numerator and denominator equal to zero and solving for (x). ((x+8)(x-5) = 0) gives (x = -8) and (x = 5). (x - 1 = 0) gives (x = 1).

  2. Plot these critical points on a number line.

  3. Test intervals between the critical points by choosing test points in each interval and determining if the expression is positive or negative.

  4. Based on the signs of the expression in each interval, determine the intervals where the expression is greater than or equal to zero.

  5. Write the solution in interval notation.

The critical points divide the number line into four intervals: ((-∞, -8)), ((-8, 1)), ((1, 5)), and ((5, +∞)).

Testing:

  • Choose (x = -9) in ((-∞, -8)): (\frac{{(-9+8)(-9-5)}}{{-9-1}} = \frac{{(-1)(-14)}}{{-10}} = \frac{{14}}{{10}} > 0).
  • Choose (x = 0) in ((-8, 1)): (\frac{{(0+8)(0-5)}}{{0-1}} = \frac{{(8)(-5)}}{{-1}} = \frac{{-40}}{{-1}} < 0).
  • Choose (x = 2) in ((1, 5)): (\frac{{(2+8)(2-5)}}{{2-1}} = \frac{{(10)(-3)}}{{1}} = \frac{{-30}}{{1}} < 0).
  • Choose (x = 6) in ((5, +∞)): (\frac{{(6+8)(6-5)}}{{6-1}} = \frac{{(14)(1)}}{{5}} > 0).

The expression is greater than or equal to zero in the intervals ((-∞, -8)) and ((5, +∞)).

Therefore, the solution in interval notation is: ((-∞, -8) \cup (5, +∞)).

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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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