- #1
AN630078
- 242
- 25
- Homework Statement
- Hello, I have been doing some further work on composite transformations, specifically solving equations and inequalities involving the modulus function. I have not actually been taught how to solve modulus inequalities but have found a method of doing so. In my textbook there are some exercises which I have tried to solve, and I was wondering whether anyone could evaluate my workings to see whether I have used an appropriate method. I am rather uncertain of my solutions so I would greatly appreciate any possible improvements I could implement.
By drawing a sketch of suitable graphs, or otherwise, find the sets of values for x which;
1. |3x-2|<1/x
2. |x+1|-|4x-2|>0
- Relevant Equations
- |3x-2|<1/x
|x+1|-|4x-2|>0
1. I think the question is asking where is the graph of |3x-2| below the graph of 1/x.
To sketch the graph of y= |3x-2| draw the line of y=3x-2 and reflect the section with negative y-values in the x-axis. Alternatively, I could set 3x-2 ≥0, meaning |3x-2|=3x-2 so draw the line of y=3x-2. Then find 3x-2≤0, meaning |3x-2|=-(3x-2)=2-3x so draw the line of y=2-3x.
Then plot the graph of y=1/x and find the point of intersection between the modulus graph and y=1/x.
This is equivalent to asking where does the modulus of 3x-2 meets the curve y=1/x;
|3x-2|=1/x
The critical values are 0 and 2/3
Consider three situations:
When x<0, rewrite |3x-2|<1/x as -(3x-2)=1/x
-(3x-2)x=1
-3x^2+2x-1=0
Evaluating the discriminant of the quadratic; b^2-4ac=-8
Since the discriminant is negative this indicates that the solutions are not real numbers, since the discriminant cannot be negative for x ∈ ℝ
So there is no solution for x<0.
When 0<x<2/3 rewrite |3x-2|<1/x as -(3x-2)<1/x
-3x+2-1/x<0
3x^2-2x+1/x>0
x>0
Combine the intervals x>0 and 0<x<2/3, the solution is 0<x<2/3
When x>2/3, rewrite |3x-2|<1/x as 3x-2<1/x
3x-2<1/x
3x^2-2x-1<0
Factor this to find the solution, (3x+1)(x-1)
Thus, x=-1/3 or x=1
x<-1/3 or 0<x<1
Combine the intervals x<-1/3, 0<x<1 and x>2/3, the solution is 2/3<x<1
Combining all three regions we see that the inequality is satisfied
when: 0<x<2/3 or 2/3<x<1
So the solution is 0<x<1
One is also able to evaluate this by sketching both of the graphs. The point of intersection is found to be (1,1). Since the graph of y=1/x has a horizontal asymptote of y=0, meaning when x is negative the graph of y=1/x is below the graph of y= |3x-2|
Therefore, combining the solutions to find the set of values for which |3x-2|<1/x this would be when 0<x<1.
2. |x+1|-|4x-2|>0
Which is equivalent to;|x+1|>|4x-2|
The critical values are -1 and 1/2.
Consider three situations; x<-1
When x<-1, rewrite |x+1|>|4x-2| as -(x+1)>-(4x-2)
-x-1>-4x+2
-x>-4x+3
3x>3
Combining the intervals x>1 and x<0 demonstrates there is no solution.
Consider the second situation; -1<x<1/2
When -1<x<1/2, rewrite |x+1|>|4x-2| as (x+1)>-(4x-2)
x+1>-4x+2
x>-4x+1
5x>1
x>1/5
So the inequality is true for all x>1/5
Merge the overlapping intervals x>1/5 and -1<x<1/2, thus 1/5<x<1/2
Consider the third situation; x>1/2
When x>1/2, rewrite |x+1|>|4x-2| as (x+1)>(4x-2)
x+1>4x-2
x>4x-3
3x<3
x<1
So the inequality is true for all x<1
Merge the overlapping intervals x>1/2 and x<1, thus 1/2<x<1
Combining all three regions we see that the inequality is satisfied
when: 1/5<x<1/2 or 1/2<x<1
So the solution is 1/5<x<1
Would my method of solving the inequalities be correct, or is there a preferable option instead?
To sketch the graph of y= |3x-2| draw the line of y=3x-2 and reflect the section with negative y-values in the x-axis. Alternatively, I could set 3x-2 ≥0, meaning |3x-2|=3x-2 so draw the line of y=3x-2. Then find 3x-2≤0, meaning |3x-2|=-(3x-2)=2-3x so draw the line of y=2-3x.
Then plot the graph of y=1/x and find the point of intersection between the modulus graph and y=1/x.
This is equivalent to asking where does the modulus of 3x-2 meets the curve y=1/x;
|3x-2|=1/x
The critical values are 0 and 2/3
Consider three situations:
When x<0, rewrite |3x-2|<1/x as -(3x-2)=1/x
-(3x-2)x=1
-3x^2+2x-1=0
Evaluating the discriminant of the quadratic; b^2-4ac=-8
Since the discriminant is negative this indicates that the solutions are not real numbers, since the discriminant cannot be negative for x ∈ ℝ
So there is no solution for x<0.
When 0<x<2/3 rewrite |3x-2|<1/x as -(3x-2)<1/x
-3x+2-1/x<0
3x^2-2x+1/x>0
x>0
Combine the intervals x>0 and 0<x<2/3, the solution is 0<x<2/3
When x>2/3, rewrite |3x-2|<1/x as 3x-2<1/x
3x-2<1/x
3x^2-2x-1<0
Factor this to find the solution, (3x+1)(x-1)
Thus, x=-1/3 or x=1
x<-1/3 or 0<x<1
Combine the intervals x<-1/3, 0<x<1 and x>2/3, the solution is 2/3<x<1
Combining all three regions we see that the inequality is satisfied
when: 0<x<2/3 or 2/3<x<1
So the solution is 0<x<1
One is also able to evaluate this by sketching both of the graphs. The point of intersection is found to be (1,1). Since the graph of y=1/x has a horizontal asymptote of y=0, meaning when x is negative the graph of y=1/x is below the graph of y= |3x-2|
Therefore, combining the solutions to find the set of values for which |3x-2|<1/x this would be when 0<x<1.
2. |x+1|-|4x-2|>0
Which is equivalent to;|x+1|>|4x-2|
The critical values are -1 and 1/2.
Consider three situations; x<-1
When x<-1, rewrite |x+1|>|4x-2| as -(x+1)>-(4x-2)
-x-1>-4x+2
-x>-4x+3
3x>3
Combining the intervals x>1 and x<0 demonstrates there is no solution.
Consider the second situation; -1<x<1/2
When -1<x<1/2, rewrite |x+1|>|4x-2| as (x+1)>-(4x-2)
x+1>-4x+2
x>-4x+1
5x>1
x>1/5
So the inequality is true for all x>1/5
Merge the overlapping intervals x>1/5 and -1<x<1/2, thus 1/5<x<1/2
Consider the third situation; x>1/2
When x>1/2, rewrite |x+1|>|4x-2| as (x+1)>(4x-2)
x+1>4x-2
x>4x-3
3x<3
x<1
So the inequality is true for all x<1
Merge the overlapping intervals x>1/2 and x<1, thus 1/2<x<1
Combining all three regions we see that the inequality is satisfied
when: 1/5<x<1/2 or 1/2<x<1
So the solution is 1/5<x<1
Would my method of solving the inequalities be correct, or is there a preferable option instead?
