Algebra
Calculus
Trigonometry
Matrix
Question
Solve the inequality
−1.52138≤x≤1
Alternative Form
x∈[−1.52138,1]
Evaluate
∣x−2∣≥x3
Rearrange the terms
∣x−2∣−x3≥0
Separate the inequality into 4 possible cases
x−2−x3≥0,x−2≥0,x3≥0x−2−(−x3)≥0,x−2≥0,x3<0−(x−2)−x3≥0,x−2<0,x3≥0−(x−2)−(−x3)≥0,x−2<0,x3<0
Evaluate
More Steps
Evaluate
x−2−x3≥0
Rewrite the expression
x−2−x3=0
Find the critical values by solving the corresponding equation
x≈−1.52138
Determine the test intervals using the critical values
x<−1.52138x>−1.52138
Choose a value form each interval
x1=−3x2=−1
To determine if x<−1.52138 is the solution to the inequality,test if the chosen value x=−3 satisfies the initial inequality
More Steps
Evaluate
−3−2−(−3)3≥0
Subtract the numbers
22≥0
Check the inequality
true
x<−1.52138 is the solutionx2=−1
To determine if x>−1.52138 is the solution to the inequality,test if the chosen value x=−1 satisfies the initial inequality
More Steps
Evaluate
−1−2−(−1)3≥0
Apply the inverse property of addition
−2≥0
Check the inequality
false
x<−1.52138 is the solutionx>−1.52138 is not a solution
The original inequality is a nonstrict inequality,so include the critical value in the solution
x≤−1.52138 is the solution
The final solution of the original inequality is x≤−1.52138
x≤−1.52138
x≤−1.52138,x−2≥0,x3≥0x−2−(−x3)≥0,x−2≥0,x3<0−(x−2)−x3≥0,x−2<0,x3≥0−(x−2)−(−x3)≥0,x−2<0,x3<0
Evaluate
More Steps
Evaluate
x−2≥0
Move the constant to the right side
x≥0+2
Removing 0 doesn't change the value,so remove it from the expression
x≥2
x≤−1.52138,x≥2,x3≥0x−2−(−x3)≥0,x−2≥0,x3<0−(x−2)−x3≥0,x−2<0,x3≥0−(x−2)−(−x3)≥0,x−2<0,x3<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≤−1.52138,x≥2,x≥0x−2−(−x3)≥0,x−2≥0,x3<0−(x−2)−x3≥0,x−2<0,x3≥0−(x−2)−(−x3)≥0,x−2<0,x3<0
Evaluate
More Steps
Evaluate
x−2−(−x3)≥0
Remove the parentheses
x−2+x3≥0
Factor the expression
(−1+x)(x2+x+2)≥0
Separate the inequality into 2 possible cases
{−1+x≥0x2+x+2≥0{−1+x≤0x2+x+2≤0
Solve the inequality
More Steps
Evaluate
−1+x≥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+2≥0{−1+x≤0x2+x+2≤0
Solve the inequality
More Steps
Evaluate
x2+x+2≥0
Move the constant to the right side
x2+x≥0−2
Add the terms
x2+x≥−2
Add the same value to both sides
x2+x+41≥−2+41
Evaluate
x2+x+41≥−47
Evaluate
(x+21)2≥−47
Calculate
x∈R
{x≥1x∈R{−1+x≤0x2+x+2≤0
Solve the inequality
More Steps
Evaluate
−1+x≤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∈R{x≤1x2+x+2≤0
Solve the inequality
More Steps
Evaluate
x2+x+2≤0
Move the constant to the right side
x2+x≤0−2
Add the terms
x2+x≤−2
Add the same value to both sides
x2+x+41≤−2+41
Evaluate
x2+x+41≤−47
Evaluate
(x+21)2≤−47
Calculate
x∈/R
{x≥1x∈R{x≤1x∈/R
Find the intersection
x≥1{x≤1x∈/R
Find the intersection
x≥1x∈/R
Find the union
x≥1
x≤−1.52138,x≥2,x≥0x≥1,x−2≥0,x3<0−(x−2)−x3≥0,x−2<0,x3≥0−(x−2)−(−x3)≥0,x−2<0,x3<0
Evaluate
More Steps
Evaluate
x−2≥0
Move the constant to the right side
x≥0+2
Removing 0 doesn't change the value,so remove it from the expression
x≥2
x≤−1.52138,x≥2,x≥0x≥1,x≥2,x3<0−(x−2)−x3≥0,x−2<0,x3≥0−(x−2)−(−x3)≥0,x−2<0,x3<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≤−1.52138,x≥2,x≥0x≥1,x≥2,x<0−(x−2)−x3≥0,x−2<0,x3≥0−(x−2)−(−x3)≥0,x−2<0,x3<0
Evaluate
More Steps
Evaluate
−(x−2)−x3≥0
Remove the parentheses
−x+2−x3≥0
Factor the expression
(1−x)(x2+x+2)≥0
Separate the inequality into 2 possible cases
{1−x≥0x2+x+2≥0{1−x≤0x2+x+2≤0
Solve the inequality
More Steps
Evaluate
1−x≥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+2≥0{1−x≤0x2+x+2≤0
Solve the inequality
More Steps
Evaluate
x2+x+2≥0
Move the constant to the right side
x2+x≥0−2
Add the terms
x2+x≥−2
Add the same value to both sides
x2+x+41≥−2+41
Evaluate
x2+x+41≥−47
Evaluate
(x+21)2≥−47
Calculate
x∈R
{x≤1x∈R{1−x≤0x2+x+2≤0
Solve the inequality
More Steps
Evaluate
1−x≤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∈R{x≥1x2+x+2≤0
Solve the inequality
More Steps
Evaluate
x2+x+2≤0
Move the constant to the right side
x2+x≤0−2
Add the terms
x2+x≤−2
Add the same value to both sides
x2+x+41≤−2+41
Evaluate
x2+x+41≤−47
Evaluate
(x+21)2≤−47
Calculate
x∈/R
{x≤1x∈R{x≥1x∈/R
Find the intersection
x≤1{x≥1x∈/R
Find the intersection
x≤1x∈/R
Find the union
x≤1
x≤−1.52138,x≥2,x≥0x≥1,x≥2,x<0x≤1,x−2<0,x3≥0−(x−2)−(−x3)≥0,x−2<0,x3<0
Evaluate
More Steps
Evaluate
x−2<0
Move the constant to the right side
x<0+2
Removing 0 doesn't change the value,so remove it from the expression
x<2
x≤−1.52138,x≥2,x≥0x≥1,x≥2,x<0x≤1,x<2,x3≥0−(x−2)−(−x3)≥0,x−2<0,x3<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≤−1.52138,x≥2,x≥0x≥1,x≥2,x<0x≤1,x<2,x≥0−(x−2)−(−x3)≥0,x−2<0,x3<0
Evaluate
More Steps
Evaluate
−(x−2)−(−x3)≥0
Remove the parentheses
−x+2+x3≥0
Rewrite the expression
−x+2+x3=0
Find the critical values by solving the corresponding equation
x≈−1.52138
Determine the test intervals using the critical values
x<−1.52138x>−1.52138
Choose a value form each interval
x1=−3x2=−1
To determine if x<−1.52138 is the solution to the inequality,test if the chosen value x=−3 satisfies the initial inequality
More Steps
Evaluate
−(−3)+2+(−3)3≥0
Add the numbers
−22≥0
Check the inequality
false
x<−1.52138 is not a solutionx2=−1
To determine if x>−1.52138 is the solution to the inequality,test if the chosen value x=−1 satisfies the initial inequality
More Steps
Evaluate
−(−1)+2+(−1)3≥0
Apply the inverse property of addition
2≥0
Check the inequality
true
x<−1.52138 is not a solutionx>−1.52138 is the solution
The original inequality is a nonstrict inequality,so include the critical value in the solution
x≥−1.52138 is the solution
The final solution of the original inequality is x≥−1.52138
x≥−1.52138
x≤−1.52138,x≥2,x≥0x≥1,x≥2,x<0x≤1,x<2,x≥0x≥−1.52138,x−2<0,x3<0
Evaluate
More Steps
Evaluate
x−2<0
Move the constant to the right side
x<0+2
Removing 0 doesn't change the value,so remove it from the expression
x<2
x≤−1.52138,x≥2,x≥0x≥1,x≥2,x<0x≤1,x<2,x≥0x≥−1.52138,x<2,x3<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≤−1.52138,x≥2,x≥0x≥1,x≥2,x<0x≤1,x<2,x≥0x≥−1.52138,x<2,x<0
Find the intersection
x∈∅x≥1,x≥2,x<0x≤1,x<2,x≥0x≥−1.52138,x<2,x<0
Find the intersection
x∈∅x∈∅x≤1,x<2,x≥0x≥−1.52138,x<2,x<0
Find the intersection
x∈∅x∈∅0≤x≤1x≥−1.52138,x<2,x<0
Find the intersection
x∈∅x∈∅0≤x≤1−1.52138≤x<0
Solution
−1.52138≤x≤1
Alternative Form
x∈[−1.52138,1]
Show Solution
