Algebra
Calculus
Trigonometry
Matrix
Question
Solve the inequality
Solve the inequality by testing the values in the interval
Solve the inequality by separating into cases
x∈(−1,0)∪(23,+∞)
Evaluate
2x−x3×1>1
Find the domain
2x−x3×1>1,x=0
Calculate
2x−x3>1
Move the expression to the left side
2x−x3−1>0
Subtract the terms
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Evaluate
2x−x3−1
Reduce fractions to a common denominator
x2x×x−x3−xx
Write all numerators above the common denominator
x2x×x−3−x
Multiply the terms
x2x2−3−x
x2x2−3−x>0
Set the numerator and denominator of x2x2−3−x equal to 0 to find the values of x where sign changes may occur
2x2−3−x=0x=0
Calculate
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Evaluate
2x2−3−x=0
Factor the expression
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Evaluate
2x2−3−x
Reorder the terms
2x2−x−3
Rewrite the expression
2x2+(−3+2)x−3
Calculate
2x2−3x+2x−3
Rewrite the expression
x×2x−x×3+2x−3
Factor out x from the expression
x(2x−3)+2x−3
Factor out 2x−3 from the expression
(x+1)(2x−3)
(x+1)(2x−3)=0
When the product of factors equals 0,at least one factor is 0
x+1=02x−3=0
Solve the equation for x
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Evaluate
x+1=0
Move the constant to the right-hand side and change its sign
x=0−1
Removing 0 doesn't change the value,so remove it from the expression
x=−1
x=−12x−3=0
Solve the equation for x
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Evaluate
2x−3=0
Move the constant to the right-hand side and change its sign
2x=0+3
Removing 0 doesn't change the value,so remove it from the expression
2x=3
Divide both sides
22x=23
Divide the numbers
x=23
x=−1x=23
x=−1x=23x=0
Determine the test intervals using the critical values
x<−1−1<x<00<x<23x>23
Choose a value form each interval
x1=−2x2=−21x3=1x4=3
To determine if x<−1 is the solution to the inequality,test if the chosen value x=−2 satisfies the initial inequality
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Evaluate
2(−2)−−23>1
Simplify
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Evaluate
2(−2)−−23
Use b−a=−ba=−ba to rewrite the fraction
2(−2)−(−23)
Multiply the numbers
−4−(−23)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
−4+23
Reduce fractions to a common denominator
−24×2+23
Write all numerators above the common denominator
2−4×2+3
Multiply the numbers
2−8+3
Add the numbers
2−5
Use b−a=−ba=−ba to rewrite the fraction
−25
−25>1
Calculate
−2.5>1
Check the inequality
false
x<−1 is not a solutionx2=−21x3=1x4=3
To determine if −1<x<0 is the solution to the inequality,test if the chosen value x=−21 satisfies the initial inequality
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Evaluate
2(−21)−−213>1
Simplify
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Evaluate
2(−21)−−213
Divide the terms
2(−21)−(−6)
Multiply the numbers
−1−(−6)
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
−1+6
Add the numbers
5
5>1
Check the inequality
true
x<−1 is not a solution−1<x<0 is the solutionx3=1x4=3
To determine if 0<x<23 is the solution to the inequality,test if the chosen value x=1 satisfies the initial inequality
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Evaluate
2×1−13>1
Simplify
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Evaluate
2×1−13
Divide the terms
2×1−3
Any expression multiplied by 1 remains the same
2−3
Subtract the numbers
−1
−1>1
Check the inequality
false
x<−1 is not a solution−1<x<0 is the solution0<x<23 is not a solutionx4=3
To determine if x>23 is the solution to the inequality,test if the chosen value x=3 satisfies the initial inequality
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Evaluate
2×3−33>1
Simplify
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Evaluate
2×3−33
Divide the terms
2×3−1
Multiply the numbers
6−1
Subtract the numbers
5
5>1
Check the inequality
true
x<−1 is not a solution−1<x<0 is the solution0<x<23 is not a solutionx>23 is the solution
The original inequality is a strict inequality,so does not include the critical value ,the final solution is x∈(−1,0)∪(23,+∞)
x∈(−1,0)∪(23,+∞)
Check if the solution is in the defined range
x∈(−1,0)∪(23,+∞),x=0
Solution
x∈(−1,0)∪(23,+∞)
Show Solution
