Algebra
Calculus
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Matrix
Question
x3>∣2x−1∣
Solve the inequality
x∈(−∞,−25+1)∪(0.453398,25−1)∪(1,+∞)
Evaluate
x3>∣2x−1∣
Rearrange the terms
x3−∣2x−1∣>0
Separate the inequality into 4 possible cases
x3−(2x−1)>0,x3≥0,2x−1≥0x3−(−(2x−1))>0,x3≥0,2x−1<0−x3−(2x−1)>0,x3<0,2x−1≥0−x3−(−(2x−1))>0,x3<0,2x−1<0
Evaluate
More Steps
Evaluate
x3−(2x−1)>0
Remove the parentheses
x3−2x+1>0
Factor the expression
(x−1)(x2+x−1)>0
Separate the inequality into 2 possible cases
{x−1>0x2+x−1>0{x−1<0x2+x−1<0
Solve the inequality
More Steps
Evaluate
x−1>0
Move the constant to the right side
x>0+1
Removing 0 doesn't change the value,so remove it from the expression
x>1
{x>1x2+x−1>0{x−1<0x2+x−1<0
Solve the inequality
More Steps
Evaluate
x2+x−1>0
Move the constant to the right side
x2+x>0−(−1)
Add the terms
x2+x>1
Add the same value to both sides
x2+x+41>1+41
Evaluate
x2+x+41>45
Evaluate
(x+21)2>45
Take the 2-th root on both sides of the inequality
(x+21)2>45
Calculate
x+21>25
Separate the inequality into 2 possible cases
x+21>25x+21<−25
Calculate
x>25−1x+21<−25
Calculate
x>25−1x<−25+1
Find the union
x∈(−∞,−25+1)∪(25−1,+∞)
{x>1x∈(−∞,−25+1)∪(25−1,+∞){x−1<0x2+x−1<0
Solve the inequality
More Steps
Evaluate
x−1<0
Move the constant to the right side
x<0+1
Removing 0 doesn't change the value,so remove it from the expression
x<1
{x>1x∈(−∞,−25+1)∪(25−1,+∞){x<1x2+x−1<0
Solve the inequality
More Steps
Evaluate
x2+x−1<0
Move the constant to the right side
x2+x<0−(−1)
Add the terms
x2+x<1
Add the same value to both sides
x2+x+41<1+41
Evaluate
x2+x+41<45
Evaluate
(x+21)2<45
Take the 2-th root on both sides of the inequality
(x+21)2<45
Calculate
x+21<25
Separate the inequality into 2 possible cases
{x+21<25x+21>−25
Calculate
{x<25−1x+21>−25
Calculate
{x<25−1x>−25+1
Find the intersection
−25+1<x<25−1
{x>1x∈(−∞,−25+1)∪(25−1,+∞){x<1−25+1<x<25−1
Find the intersection
x>1{x<1−25+1<x<25−1
Find the intersection
x>1−25+1<x<25−1
Find the union
x∈(−25+1,25−1)∪(1,+∞)
x∈(−25+1,25−1)∪(1,+∞),x3≥0,2x−1≥0x3−(−(2x−1))>0,x3≥0,2x−1<0−x3−(2x−1)>0,x3<0,2x−1≥0−x3−(−(2x−1))>0,x3<0,2x−1<0
The only way a base raised to an odd power can be greater than or equal to 0 is if the base is greater than or equal to 0
x∈(−25+1,25−1)∪(1,+∞),x≥0,2x−1≥0x3−(−(2x−1))>0,x3≥0,2x−1<0−x3−(2x−1)>0,x3<0,2x−1≥0−x3−(−(2x−1))>0,x3<0,2x−1<0
Evaluate
More Steps
Evaluate
2x−1≥0
Move the constant to the right side
2x≥0+1
Removing 0 doesn't change the value,so remove it from the expression
2x≥1
Divide both sides
22x≥21
Divide the numbers
x≥21
x∈(−25+1,25−1)∪(1,+∞),x≥0,x≥21x3−(−(2x−1))>0,x3≥0,2x−1<0−x3−(2x−1)>0,x3<0,2x−1≥0−x3−(−(2x−1))>0,x3<0,2x−1<0
Evaluate
More Steps
Evaluate
x3−(−(2x−1))>0
Remove the parentheses
x3+2x−1>0
Rewrite the expression
x3+2x−1=0
Find the critical values by solving the corresponding equation
x≈0.453398
Determine the test intervals using the critical values
x<0.453398x>0.453398
Choose a value form each interval
x1=−1x2=1
To determine if x<0.453398 is the solution to the inequality,test if the chosen value x=−1 satisfies the initial inequality
More Steps
Evaluate
(−1)3+2(−1)−1>0
Simplify
−4>0
Check the inequality
false
x<0.453398 is not a solutionx2=1
To determine if x>0.453398 is the solution to the inequality,test if the chosen value x=1 satisfies the initial inequality
More Steps
Evaluate
13+2×1−1>0
Simplify
2>0
Check the inequality
true
x<0.453398 is not a solutionx>0.453398 is the solution
The original inequality is a strict inequality,so does not include the critical value ,the final solution is x>0.453398
x>0.453398
x∈(−25+1,25−1)∪(1,+∞),x≥0,x≥21x>0.453398,x3≥0,2x−1<0−x3−(2x−1)>0,x3<0,2x−1≥0−x3−(−(2x−1))>0,x3<0,2x−1<0
The only way a base raised to an odd power can be greater than or equal to 0 is if the base is greater than or equal to 0
x∈(−25+1,25−1)∪(1,+∞),x≥0,x≥21x>0.453398,x≥0,2x−1<0−x3−(2x−1)>0,x3<0,2x−1≥0−x3−(−(2x−1))>0,x3<0,2x−1<0
Evaluate
More Steps
Evaluate
2x−1<0
Move the constant to the right side
2x<0+1
Removing 0 doesn't change the value,so remove it from the expression
2x<1
Divide both sides
22x<21
Divide the numbers
x<21
x∈(−25+1,25−1)∪(1,+∞),x≥0,x≥21x>0.453398,x≥0,x<21−x3−(2x−1)>0,x3<0,2x−1≥0−x3−(−(2x−1))>0,x3<0,2x−1<0
Evaluate
More Steps
Evaluate
−x3−(2x−1)>0
Remove the parentheses
−x3−2x+1>0
Rewrite the expression
−x3−2x+1=0
Find the critical values by solving the corresponding equation
x≈0.453398
Determine the test intervals using the critical values
x<0.453398x>0.453398
Choose a value form each interval
x1=−1x2=1
To determine if x<0.453398 is the solution to the inequality,test if the chosen value x=−1 satisfies the initial inequality
More Steps
Evaluate
−(−1)3−2(−1)+1>0
Simplify
4>0
Check the inequality
true
x<0.453398 is the solutionx2=1
To determine if x>0.453398 is the solution to the inequality,test if the chosen value x=1 satisfies the initial inequality
More Steps
Evaluate
−13−2×1+1>0
Simplify
−2>0
Check the inequality
false
x<0.453398 is the solutionx>0.453398 is not a solution
The original inequality is a strict inequality,so does not include the critical value ,the final solution is x<0.453398
x<0.453398
x∈(−25+1,25−1)∪(1,+∞),x≥0,x≥21x>0.453398,x≥0,x<21x<0.453398,x3<0,2x−1≥0−x3−(−(2x−1))>0,x3<0,2x−1<0
The only way a base raised to an odd power can be less than 0 is if the base is less than 0
x∈(−25+1,25−1)∪(1,+∞),x≥0,x≥21x>0.453398,x≥0,x<21x<0.453398,x<0,2x−1≥0−x3−(−(2x−1))>0,x3<0,2x−1<0
Evaluate
More Steps
Evaluate
2x−1≥0
Move the constant to the right side
2x≥0+1
Removing 0 doesn't change the value,so remove it from the expression
2x≥1
Divide both sides
22x≥21
Divide the numbers
x≥21
x∈(−25+1,25−1)∪(1,+∞),x≥0,x≥21x>0.453398,x≥0,x<21x<0.453398,x<0,x≥21−x3−(−(2x−1))>0,x3<0,2x−1<0
Evaluate
More Steps
Evaluate
−x3−(−(2x−1))>0
Remove the parentheses
−x3+2x−1>0
Factor the expression
(−x+1)(x2+x−1)>0
Separate the inequality into 2 possible cases
{−x+1>0x2+x−1>0{−x+1<0x2+x−1<0
Solve the inequality
More Steps
Evaluate
−x+1>0
Move the constant to the right side
−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 inequality and flip the inequality sign
x<1
{x<1x2+x−1>0{−x+1<0x2+x−1<0
Solve the inequality
More Steps
Evaluate
x2+x−1>0
Move the constant to the right side
x2+x>0−(−1)
Add the terms
x2+x>1
Add the same value to both sides
x2+x+41>1+41
Evaluate
x2+x+41>45
Evaluate
(x+21)2>45
Take the 2-th root on both sides of the inequality
(x+21)2>45
Calculate
x+21>25
Separate the inequality into 2 possible cases
x+21>25x+21<−25
Calculate
x>25−1x+21<−25
Calculate
x>25−1x<−25+1
Find the union
x∈(−∞,−25+1)∪(25−1,+∞)
{x<1x∈(−∞,−25+1)∪(25−1,+∞){−x+1<0x2+x−1<0
Solve the inequality
More Steps
Evaluate
−x+1<0
Move the constant to the right side
−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 inequality and flip the inequality sign
x>1
{x<1x∈(−∞,−25+1)∪(25−1,+∞){x>1x2+x−1<0
Solve the inequality
More Steps
Evaluate
x2+x−1<0
Move the constant to the right side
x2+x<0−(−1)
Add the terms
x2+x<1
Add the same value to both sides
x2+x+41<1+41
Evaluate
x2+x+41<45
Evaluate
(x+21)2<45
Take the 2-th root on both sides of the inequality
(x+21)2<45
Calculate
x+21<25
Separate the inequality into 2 possible cases
{x+21<25x+21>−25
Calculate
{x<25−1x+21>−25
Calculate
{x<25−1x>−25+1
Find the intersection
−25+1<x<25−1
{x<1x∈(−∞,−25+1)∪(25−1,+∞){x>1−25+1<x<25−1
Find the intersection
x∈(−∞,−25+1)∪(25−1,1){x>1−25+1<x<25−1
Find the intersection
x∈(−∞,−25+1)∪(25−1,1)x∈∅
Find the union
x∈(−∞,−25+1)∪(25−1,1)
x∈(−25+1,25−1)∪(1,+∞),x≥0,x≥21x>0.453398,x≥0,x<21x<0.453398,x<0,x≥21x∈(−∞,−25+1)∪(25−1,1),x3<0,2x−1<0
The only way a base raised to an odd power can be less than 0 is if the base is less than 0
x∈(−25+1,25−1)∪(1,+∞),x≥0,x≥21x>0.453398,x≥0,x<21x<0.453398,x<0,x≥21x∈(−∞,−25+1)∪(25−1,1),x<0,2x−1<0
Evaluate
More Steps
Evaluate
2x−1<0
Move the constant to the right side
2x<0+1
Removing 0 doesn't change the value,so remove it from the expression
2x<1
Divide both sides
22x<21
Divide the numbers
x<21
x∈(−25+1,25−1)∪(1,+∞),x≥0,x≥21x>0.453398,x≥0,x<21x<0.453398,x<0,x≥21x∈(−∞,−25+1)∪(25−1,1),x<0,x<21
Find the intersection
x∈[21,25−1)∪(1,+∞)x>0.453398,x≥0,x<21x<0.453398,x<0,x≥21x∈(−∞,−25+1)∪(25−1,1),x<0,x<21
Find the intersection
x∈[21,25−1)∪(1,+∞)0.453398<x<21x<0.453398,x<0,x≥21x∈(−∞,−25+1)∪(25−1,1),x<0,x<21
Find the intersection
x∈[21,25−1)∪(1,+∞)0.453398<x<21x∈∅x∈(−∞,−25+1)∪(25−1,1),x<0,x<21
Find the intersection
x∈[21,25−1)∪(1,+∞)0.453398<x<21x∈∅x<−25+1
Solution
x∈(−∞,−25+1)∪(0.453398,25−1)∪(1,+∞)
Show Solution
