Solving absolute value inequalities in interval notation :
Here we are going to see how to solve absolute value inequalities in interval notation.
If the given inequalities are in the following form, we may represent the the function inside the absolute value sign between the range -r and r and solve for x.
Given function |
Solution |
(i) |x - a| < r (ii) |x - a| > r (iii) |x - a| ≤ r (iv) |x - a| ≥ r |
(a - r, a + r) (∞, a - r) U (a + r, ∞) (∞, a - r] U [a + r , ∞) (∞, a - r] U [a + r, ∞) |
Here we may have some different cases.
Given function |
Solution |
|x - a| > -r (or) |x - a| ≥ - r |
Hence the solution is all real numbers.Since we have modulus sign in the left side, always we get positive value as answer. |
|x - a| < -r (or) |x - a| ≤ - r |
There is no solution. By applying any positive and negative values for x, we get only positive answer. Because we have modulus sign in the left side. |
Let us look into some example problems to understand the concept.
Example 1 :
Solve 1/|2x−1| < 6 and express the solution using the interval notation.
Solution :
Now we have to split the given inequality into two branches.
1/(2x-1) > -6 Multiply by 2x - 1 throughout the equation 1 > -6(2x - 1) 1 < -12x + 6 Subtract 6 on both sides 1 - 6 < -12x + 6 - 6 -5 < -12x Divide by -12 on both sides 5/12 > x |
1/(2x-1) < 6 Multiply by 2x - 1 throughout the equation 1 > 6(2x - 1) 1 < 12x - 6 Add 6 on both sides 1 + 6 < 12x - 6 + 6 7 < 12x Divide by 12 on both sides 7/12 < x |
In the last step of first part, we divide -12 throughout the equation. So < sign becomes > sign
Hence the solution in interval notation is (-∞, 5/12) U (7/12, ∞).
Example 2 :
Solve −3|x| + 5 ≤ −2 and graph the solution set in a number line.
Solution :
Let us split the given inequality into two parts
−3(x) + 5 ≤ −2 -3x + 5 ≤ −2 Subtract 5 on both sides -3x + 5 - 5 ≤ −2 - 5 -3x ≤ −7 Divide by -3 on both sides x ≥ 7/3 |
−3(-x) + 5 ≤ −2 3x + 5 ≤ −2 Subtract 5 on both sides 3x + 5 - 5 ≤ −2 - 5 3x ≤ −7 Divide by 3 on both sides x ≤ −7/3 |
The interval notation of the above solution is (-∞, -7/3] U [7/3, ∞).
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