- #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?