# Find all possible values for the inequality.

• Sep 29th 2013, 03:21 PM
turbozz
Find all possible values for the inequality.
Question: What are the possible values of |2x3| when 0<|x1|<2?

My solution:

We know
x1
becomes x-1 if x-1≥0 and -(x-1) if x-1<0.
Now consider two cases.
Case 1:
0<x-1<2
1<x<3
-1<2x-3<3.

Case 2:
0<-x+1<2
-1<x<1
-2<2x<2
-5<2x-3<-1

Then the possible value include |2x-3|<3. Am I solving this problem correctly ??
Note, -5<|2x-3|<-1 would not work since the |2x-3| is bounded between two negative values.

Am I solving this problem correctly ?? The solution seems so incomplete?
• Sep 29th 2013, 04:59 PM
HallsofIvy
Re: Find all possible values for the inequality.
If 0< |x- 2|< 2 then -2< x- 2< 2 (with $x- 2\ne 0$). Adding 2 to each part, 0< x< 4 (with $x\ne 2$). Multiplying by 2, 0< 2x< 8 (with $2x\ne 4$). Subtracting 3, -3< 2x- 3< 5 (with 2x- 3\ne 1). That is, if 0< |x- 2|< 2, 2x- 3 can be any number strictly between -3 and 5, except 1.
• Sep 29th 2013, 05:11 PM
chiro
Re: Find all possible values for the inequality.
Hey turbozz.

Take a look at this thread:

http://mathhelpforum.com/calculus/22...conundrum.html
• Sep 29th 2013, 05:13 PM
Soroban
Re: Find all possible values for the inequality.
Hello, turbozz!

Quote:

$\text{What are the possible values of }|2x-3|\,\text{ when }\,0 < |x-1| < 2\,?$

$\begin{array}{ccccccc}\text{Given: } & 0 & < & |x-1| & < & 2 \\ \\ \text{Then:} & \text{-}2 & < & x-1 & < & 2 \\ \\ \text{Add 1:} & \text{-}1 & < & x & < & 3 \\ \\ \text{Times 2} & \text{-}2 & < & 2x & < & 6 \\ \\ \text{Minus 3:} & \text{-}5 &<& 2x-3 &<& 3 \end{array}$
• Sep 29th 2013, 05:19 PM
turbozz
Re: Find all possible values for the inequality.
Quote:

Originally Posted by HallsofIvy
If 0< |x- 2|< 2 then -2< x- 2< 2 (with $x- 2\ne 0$). Adding 2 to each part, 0< x< 4 (with $x\ne 2$). Multiplying by 2, 0< 2x< 8 (with $2x\ne 4$). Subtracting 3, -3< 2x- 3< 5 (with 2x- 3\ne 1). That is, if 0< |x- 2|< 2, 2x- 3 can be any number strictly between -3 and 5, except 1.

I don't understand why your using 0<|x-2|<2 ?? How does this relate to the inequality at hand?
• Sep 29th 2013, 05:23 PM
turbozz
Re: Find all possible values for the inequality.
Quote:

Originally Posted by Soroban
Hello, turbozz!

$\begin{array}{ccccccc}\text{Given: } & 0 & < & |x-1| & < & 2 \\ \\ \text{Then:} & \text{-}2 & < & x-1 & < & 2 \\ \\ \text{Add 1:} & \text{-}1 & < & x & < & 3 \\ \\ \text{Times 2} & \text{-}2 & < & 2x & < & 6 \\ \\ \text{Minus 3:} & \text{-}5 &<& 2x-3 &<& 3 \end{array}$

how to do go from 0 <|x-1|<2 to -2<x-1<2 ? And x=1 wouldn't work since it wouldn't satisfy 0<|x-1|<2.
• Sep 29th 2013, 11:50 PM
Prove It
Re: Find all possible values for the inequality.
Quote:

Originally Posted by turbozz
Question: What are the possible values of |2x3| when 0<|x1|<2?

My solution:

We know
x1
becomes x-1 if x-1≥0 and -(x-1) if x-1<0.
Now consider two cases.
Case 1:
0<x-1<2
1<x<3
-1<2x-3<3.

Case 2:
0<-x+1<2
-1<x<1
-2<2x<2
-5<2x-3<-1

Then the possible value include |2x-3|<3. Am I solving this problem correctly ??
Note, -5<|2x-3|<-1 would not work since the |2x-3| is bounded between two negative values.

Am I solving this problem correctly ?? The solution seems so incomplete?

The triangle inequality works nicely here to get an upper bound: \displaystyle \begin{align*} |a + b| \leq |a| + |b| \end{align*}

So \displaystyle \begin{align*} |2x - 3| &= |2x - 2 - 1| \\ &\leq |2x - 2| + |-1| \\ &= 2|x - 1| + 1 \\ &< 2(2) + 1 \textrm{ since } |x - 1| < 2 \\ &= 5 \end{align*}

So we can say \displaystyle \begin{align*} |2x - 3| < 5 \end{align*}.

Also, the reverse triangle inequality is useful here to get a lower bound. \displaystyle \begin{align*} \left| |a| - |b| \right| \leq |a - b| \end{align*}. So that means

\displaystyle \begin{align*} |2x - 3| &= |2x - 2 - 1| \\ &\geq \left| |2x - 2| - |1| \right| \textrm{ by the Reverse Triangle Inequality} \\ &= \left| 2|x - 1| - 1 \right| \\ &> \left| 2(0) - 1 \right| \textrm{ since } |x - 1| > 0 \\ &= |-1| \\ &= 1 \end{align*}

So that means we have \displaystyle \begin{align*} 1 < |2x - 3| < 5 \end{align*}.
• Sep 30th 2013, 03:48 AM
Plato
Re: Find all possible values for the inequality.
Quote:

Originally Posted by turbozz
Question: What are the possible values of |2x−3| when 0<|x−1|<2?

This question and the resulting discussion need a comment on the logic of the question.
In my view reply #4 is correct.

Here is the reason. This is a simple implication, an if-then question.
If $0<|x-1|<2$ then what are the possible values of $|2 x-3|~?$
If $P$ is true then $Q$ must be true.

Note that means that if $0<|x-1|<2$ then $|2x-3|<5$.

If $a=1.5$ then $0<|a-1|<2$ as well as $|2a-3|=0<5$

Note that if $0<|x-1|<2$ it is possible that $|2x-3|<1$, example $x=1.8$.

It is also worth noting that if $x=1$ then $0<|x-1|<2$ is a false statement.
BUT that means that if $0<|x-1|<2$ then $|2x-3|<5$ is still a true statement.
• Sep 30th 2013, 06:24 AM
turbozz
Re: Find all possible values for the inequality.
Quote:

Originally Posted by Prove It
The triangle inequality works nicely here to get an upper bound: \displaystyle \begin{align*} |a + b| \leq |a| + |b| \end{align*}

So \displaystyle \begin{align*} |2x - 3| &= |2x - 2 - 1| \\ &\leq |2x - 2| + |-1| \\ &= 2|x - 1| + 1 \\ &< 2(2) + 1 \textrm{ since } |x - 1| < 2 \\ &= 5 \end{align*}

So we can say \displaystyle \begin{align*} |2x - 3| < 5 \end{align*}.

Also, the reverse triangle inequality is useful here to get a lower bound. \displaystyle \begin{align*} \left| |a| - |b| \right| \leq |a - b| \end{align*}. So that means

\displaystyle \begin{align*} |2x - 3| &= |2x - 2 - 1| \\ &\geq \left| |2x - 2| - |1| \right| \textrm{ by the Reverse Triangle Inequality} \\ &= \left| 2|x - 1| - 1 \right| \\ &> \left| 2(0) - 1 \right| \textrm{ since } |x - 1| > 0 \\ &= |-1| \\ &= 1 \end{align*}

So that means we have \displaystyle \begin{align*} 1 < |2x - 3| < 5 \end{align*}.

I like the way you showed the solution a lot. Only one issue though shouldn't 0<|2x-3|<5 ?
• Sep 30th 2013, 12:43 PM
Plato
Re: Find all possible values for the inequality.
Quote:

Originally Posted by turbozz
Only one issue though shouldn't 0<|2x-3|<5 ?

The answer is $|2x-3|<5$.
Look at this: $0<|1.5-1|<2$ and $0=|2(1.5)-3|<5$.