Question
Solve the inequality
Solve the inequality by testing the values in the interval
Solve the inequality by separating into cases
x∈(−∞,0)∪(0,1)∪[2,3]
Evaluate
((x−3)(x−2))÷(x3×x2)(−(x−1))÷x≥0
Find the domain
More Steps

Evaluate
{x3×x2(−(x−1))=0x=0
Calculate
More Steps

Evaluate
x3×x2(−(x−1))=0
Simplify
x5(−x+1)=0
Apply the zero product property
{x5=0−x+1=0
The only way a power can not be 0 is when the base not equals 0
{x=0−x+1=0
Solve the inequality
{x=0x=1
Find the intersection
x∈(−∞,0)∪(0,1)∪(1,+∞)
{x∈(−∞,0)∪(0,1)∪(1,+∞)x=0
Find the intersection
x∈(−∞,0)∪(0,1)∪(1,+∞)
((x−3)(x−2))÷(x3×x2)(−(x−1))÷x≥0,x∈(−∞,0)∪(0,1)∪(1,+∞)
Calculate
((x−3)(x−2))÷(x3×x2(−(x−1)))÷x≥0
Simplify
More Steps

Evaluate
((x−3)(x−2))÷(x3×x2(−(x−1)))÷x
Multiply the terms
(x−3)(x−2)÷(x3×x2(−(x−1)))÷x
Calculate
(x−3)(x−2)÷(x3×x2(−x+1))÷x
Multiply
More Steps

Multiply the terms
x3×x2(−x+1)
Multiply the terms with the same base by adding their exponents
x3+2(−x+1)
Add the numbers
x5(−x+1)
(x−3)(x−2)÷x5(−x+1)÷x
Rewrite the expression
x5(−x+1)(x−3)(x−2)÷x
Multiply by the reciprocal
x5(−x+1)(x−3)(x−2)×x1
Multiply the terms
x5(−x+1)x(x−3)(x−2)
Multiply the terms
More Steps

Evaluate
x5×x
Use the product rule an×am=an+m to simplify the expression
x5+1
Add the numbers
x6
x6(−x+1)(x−3)(x−2)
x6(−x+1)(x−3)(x−2)≥0
Set the numerator and denominator of x6(−x+1)(x−3)(x−2) equal to 0 to find the values of x where sign changes may occur
(x−3)(x−2)=0x6(−x+1)=0
Calculate
More Steps

Evaluate
(x−3)(x−2)=0
Separate the equation into 2 possible cases
x−3=0x−2=0
Solve the equation
More Steps

Evaluate
x−3=0
Move the constant to the right-hand side and change its sign
x=0+3
Removing 0 doesn't change the value,so remove it from the expression
x=3
x=3x−2=0
Solve the equation
More Steps

Evaluate
x−2=0
Move the constant to the right-hand side and change its sign
x=0+2
Removing 0 doesn't change the value,so remove it from the expression
x=2
x=3x=2
x=3x=2x6(−x+1)=0
Calculate
More Steps

Evaluate
x6(−x+1)=0
Separate the equation into 2 possible cases
x6=0−x+1=0
The only way a power can be 0 is when the base equals 0
x=0−x+1=0
Solve the equation
More Steps

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
Change the signs on both sides of the equation
x=1
x=0x=1
x=3x=2x=0x=1
Determine the test intervals using the critical values
x<00<x<11<x<22<x<3x>3
Choose a value form each interval
x1=−1x2=21x3=23x4=25x5=4
To determine if x<0 is the solution to the inequality,test if the chosen value x=−1 satisfies the initial inequality
More Steps

Evaluate
(−1)6(−(−1)+1)(−1−3)(−1−2)≥0
Simplify
More Steps

Evaluate
(−1)6(−(−1)+1)(−1−3)(−1−2)
Subtract the numbers
(−1)6(−(−1)+1)(−4)(−1−2)
Remove the parentheses
(−1)6(−(−1)+1)−4(−1−2)
Subtract the numbers
(−1)6(−(−1)+1)−4(−3)
Add the numbers
(−1)6×2−4(−3)
Evaluate the power
1×2−4(−3)
Multiply the terms
1×212
Rewrite the expression
212
Reduce the fraction
6
6≥0
Check the inequality
true
x<0 is the solutionx2=21x3=23x4=25x5=4
To determine if 0<x<1 is the solution to the inequality,test if the chosen value x=21 satisfies the initial inequality
More Steps

Evaluate
(21)6(−21+1)(21−3)(21−2)≥0
Simplify
More Steps

Evaluate
(21)6(−21+1)(21−3)(21−2)
Subtract the numbers
(21)6(−21+1)(−25)(21−2)
Remove the parentheses
(21)6(−21+1)−25(21−2)
Subtract the numbers
(21)6(−21+1)−25(−23)
Add the numbers
(21)6×21−25(−23)
Multiply the numbers
(21)6×21415
Multiply the numbers
271415
Evaluate the power
1281415
Multiply by the reciprocal
415×128
Reduce the numbers
15×32
Multiply the numbers
480
480≥0
Check the inequality
true
x<0 is the solution0<x<1 is the solutionx3=23x4=25x5=4
To determine if 1<x<2 is the solution to the inequality,test if the chosen value x=23 satisfies the initial inequality
More Steps

Evaluate
(23)6(−23+1)(23−3)(23−2)≥0
Simplify
More Steps

Evaluate
(23)6(−23+1)(23−3)(23−2)
Subtract the numbers
(23)6(−23+1)(−23)(23−2)
Remove the parentheses
(23)6(−23+1)−23(23−2)
Subtract the numbers
(23)6(−23+1)−23(−21)
Add the numbers
(23)6(−21)−23(−21)
Reduce the fraction
(23)6−23
Evaluate the power
64729−23
Multiply by the reciprocal
−23×72964
Reduce the numbers
−21×24364
Reduce the numbers
−1×24332
Multiply the numbers
−24332
−24332≥0
Calculate
−0.131687≥0
Check the inequality
false
x<0 is the solution0<x<1 is the solution1<x<2 is not a solutionx4=25x5=4
To determine if 2<x<3 is the solution to the inequality,test if the chosen value x=25 satisfies the initial inequality
More Steps

Evaluate
(25)6(−25+1)(25−3)(25−2)≥0
Simplify
More Steps

Evaluate
(25)6(−25+1)(25−3)(25−2)
Subtract the numbers
(25)6(−25+1)(−21)(25−2)
Remove the parentheses
(25)6(−25+1)−21(25−2)
Subtract the numbers
(25)6(−25+1)−21×21
Add the numbers
(25)6(−23)−21×21
Multiply the numbers
(25)6(−23)−41
Multiply the numbers
−2746875−41
Evaluate the power
−12846875−41
Multiply by the reciprocal
−41(−46875128)
Multiplying or dividing an even number of negative terms equals a positive
41×46875128
Reduce the numbers
1×4687532
Multiply the numbers
4687532
4687532≥0
Calculate
0.000683≥0
Check the inequality
true
x<0 is the solution0<x<1 is the solution1<x<2 is not a solution2<x<3 is the solutionx5=4
To determine if x>3 is the solution to the inequality,test if the chosen value x=4 satisfies the initial inequality
More Steps

Evaluate
46(−4+1)(4−3)(4−2)≥0
Simplify
More Steps

Evaluate
46(−4+1)(4−3)(4−2)
Subtract the numbers
46(−4+1)1×(4−2)
Subtract the numbers
46(−4+1)1×2
Add the numbers
46(−3)1×2
Rewrite the expression
46(−3)2
Reduce the fraction
46×3−2
Factor the expression
2×2048×3−2
Reduce the fraction
2048×3−1
Calculate
−61441
−61441≥0
Calculate
−0.000163≥0
Check the inequality
false
x<0 is the solution0<x<1 is the solution1<x<2 is not a solution2<x<3 is the solutionx>3 is not a solution
The original inequality is a nonstrict inequality,so include the critical value in the solution
x<0 is the solution0<x<1 is the solution2≤x≤3 is the solution
The final solution of the original inequality is x∈(−∞,0)∪(0,1)∪[2,3]
x∈(−∞,0)∪(0,1)∪[2,3]
Check if the solution is in the defined range
x∈(−∞,0)∪(0,1)∪[2,3],x∈(−∞,0)∪(0,1)∪(1,+∞)
Solution
x∈(−∞,0)∪(0,1)∪[2,3]
Show Solution
