Piecewise Defined Function

  • #1

Homework Statement


The equation [tex]\left| {x + a} \right| = \left| {x - b} \right|[/tex] has exactly one solution; at x=2. Find the value(s) of a and b

2. The attempt at a solution
Here is the way in which I have approached the situation:
[tex]
\begin{array}{l}
\left| {x + a} \right| = \left| {x - b} \right| \\
x + a = x + b \\
\end{array}
[/tex]
this solution is not possible as the x's on either side of the equation cancel each other out, and a is not equal to b, so;
[tex]
\begin{array}{l}
x + a = - x + b \\
2x + a = b \\
4 + a = b \\
a = k \\
a = k + 4 \\
\end{array}
[/tex]

However I'm not sure on my my final answer, as I have not evaluated the values for a or b, as i have left them in the term of k. However what I have interpreted from the question, I have not been given enough information. Thank you to those who reply, and correct me if anything I have stated is wrong

unique_pavadrin
 

Answers and Replies

  • #2
184
0

Homework Statement


The equation [tex]\left| {x + a} \right| = \left| {x - b} \right|[/tex] has exactly one solution; at x=2. Find the value(s) of a and b

2. The attempt at a solution
Here is the way in which I have approached the situation:
[tex]
\begin{array}{l}
\left| {x + a} \right| = \left| {x - b} \right| \\
x + a = x + b \\
\end{array}
[/tex]
this solution is not possible as the x's on either side of the equation cancel each other out, and a is not equal to b, so;
[tex]
\begin{array}{l}
x + a = - x + b \\
2x + a = b \\
4 + a = b \\
a = k \\
a = k + 4 \\
\end{array}
[/tex]

However I'm not sure on my my final answer, as I have not evaluated the values for a or b, as i have left them in the term of k. However what I have interpreted from the question, I have not been given enough information. Thank you to those who reply, and correct me if anything I have stated is wrong

unique_pavadrin
Not sure how you got this step:
[tex]
\begin{array}{l}
\left| {x + a} \right| = \left| {x - b} \right| \\
x + a = x + b \\
\end{array}
[/tex]
If you are taking both to be positive, it should be [tex]x + a = x - b [/tex]
Then you can kill the x's. so a=-b.
 
  • #3
thanks for the reply theperthvan

ive used those steps as a simply does not equal -b as the x can be anywhere on the number line. by stating that a=-b, x is a fixed point at zero, which is incorrect as z is given the value of 2

or is that completely wrong?
thanks
 

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