Where Do the Graphs of |3x-2| and 1/x Intersect and Diverge?

[AN630078]

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?

 

[haruspex]

For 1 I would start with |3x-2| is nonnegative so |3x-2|<1/x implies x>0.

I didn't follow your reasoning here:

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​

Is the third line supposed to be 3x^2-2x+1>0?
Write the LHS as 2x^2+(x^2-2x+1).

In this sequence

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​

Shouldn't the last line read "x>-1/3 and x<1"?

For 2, your graph is wrong.

 

[AN630078]

For 1 I would start with |3x-2| is nonnegative so |3x-2|<1/x implies x>0.

I didn't follow your reasoning here:

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​

Is the third line supposed to be 3x^2-2x+1>0?
Write the LHS as 2x^2+(x^2-2x+1).

In this sequence

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​

Shouldn't the last line read "x>-1/3 and x<1"?

For 2, your graph is wrong.

Thank you for your reply. For 1, yes sorry my workings should simplify -3x+2-1/x<0 by converting to a fractional form;
-3xx/x+2x/x-1/x<0
-3x^2+2x-1/x<0
Multiply both sides by -1 to reverse the inequality;
3x^2-2x+1/x>0

Combine the intervals x>0 and 0<x<2/3, the solution is 0<x<2/3.

Yes, I think you are correct I should have written x>-1/3 and x<1, but I was trying to identify the regions that satisfy the condition of < 0, being x<-1/3 or 0<x<1.

I have tried to correct my graph for part 2? Sorry, I think I accidentally graphed |x-1| instead of |x+1|?
Besides from this would my workings be correct? Is is there a preferable method to solve modulus inequalities like these instead?

 

[haruspex]

Yes, that's the right graph now.
Solving nonlinear inequalities does tend to be messy, often needing to be broken down into cases. Graphing is certainly helpful in guiding you.
There can be quicker ad hoc methods. In the second question you could have written it as (x+1)²>(4x-2)².
This quickly simplifies to (5x-1)(x-1)<0, whence 1/5<x<1.

 

[AN630078]

Yes, that's the right graph now.
Solving nonlinear inequalities does tend to be messy, often needing to be broken down into cases. Graphing is certainly helpful in guiding you.
There can be quicker ad hoc methods. In the second question you could have written it as (x+1)²>(4x-2)².
This quickly simplifies to (5x-1)(x-1)<0, whence 1/5<x<1.

Thank you for your reply. Yes, I found found that with nonlinear inequalities. Graphs certainly have been helpful to my understanding this topic better. Although, I still find it a little confusing.
Thank you for your suggestion for the alternative method of solving the second part to this problem, I found that very helpful and less time consuming!