How do you solve #|x+1|≤8#?

Answer 1

#-9<=x<=7#

Considering

#|f(x)|\=kharr -k\=f(x)\=k#,

In this instance, you have:

8 harr -8\=x+1\=8# #|x+1|
#-8-1<=x<=8-1#
#-9<=x<=7#
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Answer 2

To solve the inequality (|x + 1| \leq 8), follow these steps:

  1. Split the inequality into two cases: one for when (x + 1) is positive or zero, and one for when (x + 1) is negative.

  2. For the case where (x + 1 \geq 0), solve (x + 1 \leq 8).

  3. For the case where (x + 1 < 0), solve (-(x + 1) \leq 8).

  4. Solve each inequality separately to find the values of (x).

  5. After solving each case, check if the solutions are valid by plugging them back into the original inequality.

  6. Once you have the valid solutions, write them as an ordered pair or in interval notation, depending on the context.

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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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