Graph |(2x-1)/x|<2

Math
|2x-1x|<2
Write |2x-1x|<2 as a piecewise.
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To find the interval for the first piece, find where the inside of the absolute value is non-negative.
2x-1x≥0
Solve the inequality.
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Find all the values where the expression switches from negative to positive by setting each factor equal to 0 and solving.
x=0
2x-1=0
Add 1 to both sides of the equation.
2x=1
Divide each term by 2 and simplify.
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Divide each term in 2x=1 by 2.
2×2=12
Cancel the common factor of 2.
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Cancel the common factor.
2×2=12
Divide x by 1.
x=12
x=12
x=12
Solve for each factor to find the values where the absolute value expression goes from negative to positive.
x=0
x=12
Consolidate the solutions.
x=0,12
Find the domain of |2x-1x|-2.
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Set the denominator in 2x-1x equal to 0 to find where the expression is undefined.
x=0
The domain is all values of x that make the expression defined.
(-∞,0)∪(0,∞)
(-∞,0)∪(0,∞)
Use each root to create test intervals.
x<0
0<x<12
x>12
Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.
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Test a value on the interval x<0 to see if it makes the inequality true.
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Choose a value on the interval x<0 and see if this value makes the original inequality true.
x=-2
Replace x with -2 in the original inequality.
2(-2)-1-2≥0
The left side 2.5 is greater than the right side 0, which means that the given statement is always true.
True
True
Test a value on the interval 0<x<12 to see if it makes the inequality true.
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Choose a value on the interval 0<x<12 and see if this value makes the original inequality true.
x=0.25
Replace x with 0.25 in the original inequality.
2(0.25)-10.25≥0
The left side -2 is less than the right side 0, which means that the given statement is false.
False
False
Test a value on the interval x>12 to see if it makes the inequality true.
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Choose a value on the interval x>12 and see if this value makes the original inequality true.
x=3
Replace x with 3 in the original inequality.
2(3)-13≥0
The left side 1.6‾ is greater than the right side 0, which means that the given statement is always true.
True
True
Compare the intervals to determine which ones satisfy the original inequality.
x<0 True
0<x<12 False
x>12 True
x<0 True
0<x<12 False
x>12 True
The solution consists of all of the true intervals.
x<0 or x≥12
x<0 or x≥12
In the piece where 2x-1x is non-negative, remove the absolute value.
2x-1x<2
Find the domain of 2x-1x<2 and find the intersection with x<0 or x≥12.
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Find the domain of 2x-1x<2.
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Set the denominator in 2x-1x equal to 0 to find where the expression is undefined.
x=0
The domain is all values of x that make the expression defined.
(-∞,0)∪(0,∞)
(-∞,0)∪(0,∞)
Find the intersection of x<0 or x≥12 and (-∞,0)∪(0,∞).
x<0 or x≥12
x<0 or x≥12
To find the interval for the second piece, find where the inside of the absolute value is negative.
2x-1x<0
Solve the inequality.
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Find all the values where the expression switches from negative to positive by setting each factor equal to 0 and solving.
x=0
2x-1=0
Add 1 to both sides of the equation.
2x=1
Divide each term by 2 and simplify.
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Divide each term in 2x=1 by 2.
2×2=12
Cancel the common factor of 2.
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Cancel the common factor.
2×2=12
Divide x by 1.
x=12
x=12
x=12
Solve for each factor to find the values where the absolute value expression goes from negative to positive.
x=0
x=12
Consolidate the solutions.
x=0,12
Find the domain of |2x-1x|-2.
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Set the denominator in 2x-1x equal to 0 to find where the expression is undefined.
x=0
The domain is all values of x that make the expression defined.
(-∞,0)∪(0,∞)
(-∞,0)∪(0,∞)
Use each root to create test intervals.
x<0
0<x<12
x>12
Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.
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Test a value on the interval x<0 to see if it makes the inequality true.
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Choose a value on the interval x<0 and see if this value makes the original inequality true.
x=-2
Replace x with -2 in the original inequality.
2(-2)-1-2<0
The left side 2.5 is not less than the right side 0, which means that the given statement is false.
False
False
Test a value on the interval 0<x<12 to see if it makes the inequality true.
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Choose a value on the interval 0<x<12 and see if this value makes the original inequality true.
x=0.25
Replace x with 0.25 in the original inequality.
2(0.25)-10.25<0
The left side -2 is less than the right side 0, which means that the given statement is always true.
True
True
Test a value on the interval x>12 to see if it makes the inequality true.
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Choose a value on the interval x>12 and see if this value makes the original inequality true.
x=3
Replace x with 3 in the original inequality.
2(3)-13<0
The left side 1.6‾ is not less than the right side 0, which means that the given statement is false.
False
False
Compare the intervals to determine which ones satisfy the original inequality.
x<0 False
0<x<12 True
x>12 False
x<0 False
0<x<12 True
x>12 False
The solution consists of all of the true intervals.
0<x<12
0<x<12
In the piece where 2x-1x is negative, remove the absolute value and multiply by -1.
-2x-1x<2
Find the domain of -2x-1x<2 and find the intersection with 0<x<12.
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Find the domain of -2x-1x<2.
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Set the denominator in 2x-1x equal to 0 to find where the expression is undefined.
x=0
The domain is all values of x that make the expression defined.
(-∞,0)∪(0,∞)
(-∞,0)∪(0,∞)
Find the intersection of 0<x<12 and (-∞,0)∪(0,∞).
0<x<12
0<x<12
Write as a piecewise.
{2x-1x<2x<0 or x≥12-2x-1x<20<x<12
{2x-1x<2x<0 or x≥12-2x-1x<20<x<12
Solve 2x-1x<2 when x<0 or x≥12.
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Solve 2x-1x<2 for x.
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Move 2 to the left side of the equation by subtracting it from both sides.
2x-1x-2<0
Simplify 2x-1x-2.
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To write -2 as a fraction with a common denominator, multiply by xx.
2x-1x-2⋅xx<0
Simplify terms.
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Combine -2 and xx.
2x-1x+-2xx<0
Combine the numerators over the common denominator.
2x-1-2xx<0
2x-1-2xx<0
Simplify the numerator.
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Subtract 2x from 2x.
0-1x<0
Subtract 1 from 0.
-1x<0
-1x<0
Move the negative in front of the fraction.
-1x<0
-1x<0
Find all the values where the expression switches from negative to positive by setting each factor equal to 0 and solving.
x=0
Find the domain of -1x.
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Set the denominator in 1x equal to 0 to find where the expression is undefined.
x=0
The domain is all values of x that make the expression defined.
(-∞,0)∪(0,∞)
(-∞,0)∪(0,∞)
The solution consists of all of the true intervals.
x>0
x>0
Find the intersection of x>0 and x<0 or x≥12.
x≥12
x≥12
Solve -2x-1x<2 when 0<x<12.
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Solve -2x-1x<2 for x.
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Move 2 to the left side of the equation by subtracting it from both sides.
-2x-1x-2<0
Simplify -2x-1x-2.
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To write -2 as a fraction with a common denominator, multiply by xx.
-2x-1x-2⋅xx<0
Simplify terms.
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Combine -2 and xx.
-2x-1x+-2xx<0
Combine the numerators over the common denominator.
-(2x-1)-2xx<0
-(2x-1)-2xx<0
Simplify the numerator.
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Apply the distributive property.
-(2x)–1-2xx<0
Multiply 2 by -1.
-2x–1-2xx<0
Multiply -1 by -1.
-2x+1-2xx<0
Subtract 2x from -2x.
-4x+1x<0
-4x+1x<0
Simplify with factoring out.
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Factor -1 out of -4x.
-(4x)+1x<0
Rewrite 1 as -1(-1).
-(4x)-1(-1)x<0
Factor -1 out of -(4x)-1(-1).
-(4x-1)x<0
Simplify the expression.
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Rewrite -(4x-1) as -1(4x-1).
-1(4x-1)x<0
Move the negative in front of the fraction.
-4x-1x<0
-4x-1x<0
-4x-1x<0
-4x-1x<0
Find all the values where the expression switches from negative to positive by setting each factor equal to 0 and solving.
x=0
4x-1=0
Add 1 to both sides of the equation.
4x=1
Divide each term by 4 and simplify.
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Divide each term in 4x=1 by 4.
4×4=14
Cancel the common factor of 4.
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Cancel the common factor.
4×4=14
Divide x by 1.
x=14
x=14
x=14
Solve for each factor to find the values where the absolute value expression goes from negative to positive.
x=0
x=14
Consolidate the solutions.
x=0,14
Find the domain of -4x-1x.
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Set the denominator in 4x-1x equal to 0 to find where the expression is undefined.
x=0
The domain is all values of x that make the expression defined.
(-∞,0)∪(0,∞)
(-∞,0)∪(0,∞)
Use each root to create test intervals.
x<0
0<x<14
x>14
Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.
Tap for more steps…
Test a value on the interval x<0 to see if it makes the inequality true.
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Choose a value on the interval x<0 and see if this value makes the original inequality true.
x=-2
Replace x with -2 in the original inequality.
-2(-2)-1-2<2
The left side -2.5 is less than the right side 2, which means that the given statement is always true.
True
True
Test a value on the interval 0<x<14 to see if it makes the inequality true.
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Choose a value on the interval 0<x<14 and see if this value makes the original inequality true.
x=0.12
Replace x with 0.12 in the original inequality.
-2(0.12)-10.12<2
The left side 6.3‾ is not less than the right side 2, which means that the given statement is false.
False
False
Test a value on the interval x>14 to see if it makes the inequality true.
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Choose a value on the interval x>14 and see if this value makes the original inequality true.
x=3
Replace x with 3 in the original inequality.
-2(3)-13<2
The left side -1.6‾ is less than the right side 2, which means that the given statement is always true.
True
True
Compare the intervals to determine which ones satisfy the original inequality.
x<0 True
0<x<14 False
x>14 True
x<0 True
0<x<14 False
x>14 True
The solution consists of all of the true intervals.
x<0 or x>14
x<0 or x>14
Find the intersection of x<0 or x>14 and 0<x<12.
14<x<12
14<x<12
Find the union of the solutions.
x>14
Find the domain of |2x-1x|-2.
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Set the denominator in 2x-1x equal to 0 to find where the expression is undefined.
x=0
The domain is all values of x that make the expression defined.
(-∞,0)∪(0,∞)
(-∞,0)∪(0,∞)
The solution consists of all of the true intervals.
x>14
(1)/(4)","color":0,"isGrey":false,"dashed":false,"holes":[]}],"asymptotes":[],"segments":[],"areaBetweenCurves":[],"points":[],"HasGraphInput":true}” class=”GraphWrapper”>
Graph |(2x-1)/x|<2

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