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Why am I getting a wrong answer on solving $|x-1|+|x-2|=1$
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I'm solving the equation,
$$|x-1| + |x-2| = 1$$
I'm making cases,
$C-1, , x in [2, infty) $
So, $ x-1 + x-2 = 1 Rightarrow x= 2$
$C-2, , x in [1, 2) $
$x-1 - x + 2 = 1 Rightarrow 1 =1 Rightarrow xin [1,2) $
$C-3, , x in (- infty, 1)$
$ - x + 1 - x+2 = 1 Rightarrow x= 1 notin (-infty, 1) Rightarrow x = phi$ (null set)
Taking common of all three solution set, I get $x= phi$ because of the last case. But the answer is supposed to be $x in [1,2]$
But when I write this equation in graphing calculator, it shows $2$ lines $x=1$ and $ x= 2$ rather than a region between $[1,2]$
Someone explain this too?
absolute-value
asked Jul 24 at 14:07
William
753214
 |Â
show 1 more comment
up vote
2
down vote
favorite
I'm solving the equation,
$$|x-1| + |x-2| = 1$$
I'm making cases,
$C-1, , x in [2, infty) $
So, $ x-1 + x-2 = 1 Rightarrow x= 2$
$C-2, , x in [1, 2) $
$x-1 - x + 2 = 1 Rightarrow 1 =1 Rightarrow xin [1,2) $
$C-3, , x in (- infty, 1)$
$ - x + 1 - x+2 = 1 Rightarrow x= 1 notin (-infty, 1) Rightarrow x = phi$ (null set)
Taking common of all three solution set, I get $x= phi$ because of the last case. But the answer is supposed to be $x in [1,2]$
But when I write this equation in graphing calculator, it shows $2$ lines $x=1$ and $ x= 2$ rather than a region between $[1,2]$
Someone explain this too?
absolute-value
asked Jul 24 at 14:07
William
753214
2
You are supposed to take the union (not the intersection) of the solution sets in the three cases: $2cup[1,2)cupemptyset=[1,2]$. And it is a bad notation to write $x=emptyset$. It would be weird, but look better if you write $xinemptyset$ (but it's best to just say that there is no solution in the third case).
– Batominovski
Jul 24 at 14:10
You should take the union of all three cases.
– xarles
Jul 24 at 14:11
@Batominovski why union though? Shouldn't the x values be those which satisfy all the three cases?
– William
Jul 24 at 14:12
1
@William No, because you are examining three different cases with three different restrictions on $x$. For example, the equation $x^2+3x+2$ has solutions $xin -1cup -2$, not $xin -1cap -2=varnothing$.
– Crosby
Jul 24 at 14:13
@Crosby Owwww!!!! Right ,I get it, sort of, also explain the graph too please! Why is it showing only 2 values of x and not the entire region between 1 and 2 inclusive?
– William
Jul 24 at 14:16
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm solving the equation,
$$|x-1| + |x-2| = 1$$
I'm making cases,
$C-1, , x in [2, infty) $
So, $ x-1 + x-2 = 1 Rightarrow x= 2$
$C-2, , x in [1, 2) $
$x-1 - x + 2 = 1 Rightarrow 1 =1 Rightarrow xin [1,2) $
$C-3, , x in (- infty, 1)$
$ - x + 1 - x+2 = 1 Rightarrow x= 1 notin (-infty, 1) Rightarrow x = phi$ (null set)
Taking common of all three solution set, I get $x= phi$ because of the last case. But the answer is supposed to be $x in [1,2]$
But when I write this equation in graphing calculator, it shows $2$ lines $x=1$ and $ x= 2$ rather than a region between $[1,2]$
Someone explain this too?
absolute-value
asked Jul 24 at 14:07
William
753214
I'm solving the equation,
$$|x-1| + |x-2| = 1$$
I'm making cases,
$C-1, , x in [2, infty) $
So, $ x-1 + x-2 = 1 Rightarrow x= 2$
$C-2, , x in [1, 2) $
$x-1 - x + 2 = 1 Rightarrow 1 =1 Rightarrow xin [1,2) $
$C-3, , x in (- infty, 1)$
$ - x + 1 - x+2 = 1 Rightarrow x= 1 notin (-infty, 1) Rightarrow x = phi$ (null set)
Taking common of all three solution set, I get $x= phi$ because of the last case. But the answer is supposed to be $x in [1,2]$
But when I write this equation in graphing calculator, it shows $2$ lines $x=1$ and $ x= 2$ rather than a region between $[1,2]$
Someone explain this too?
absolute-value
asked Jul 24 at 14:07
William
753214
edited Jul 24 at 14:20
asked Jul 24 at 14:07
William
753214
asked Jul 24 at 14:07
William
753214
asked Jul 24 at 14:07
William
753214
753214
2
You are supposed to take the union (not the intersection) of the solution sets in the three cases: $2cup[1,2)cupemptyset=[1,2]$. And it is a bad notation to write $x=emptyset$. It would be weird, but look better if you write $xinemptyset$ (but it's best to just say that there is no solution in the third case).
– Batominovski
Jul 24 at 14:10
You should take the union of all three cases.
– xarles
Jul 24 at 14:11
@Batominovski why union though? Shouldn't the x values be those which satisfy all the three cases?
– William
Jul 24 at 14:12
1
@William No, because you are examining three different cases with three different restrictions on $x$. For example, the equation $x^2+3x+2$ has solutions $xin -1cup -2$, not $xin -1cap -2=varnothing$.
– Crosby
Jul 24 at 14:13
@Crosby Owwww!!!! Right ,I get it, sort of, also explain the graph too please! Why is it showing only 2 values of x and not the entire region between 1 and 2 inclusive?
– William
Jul 24 at 14:16
 |Â
show 1 more comment
2
You are supposed to take the union (not the intersection) of the solution sets in the three cases: $2cup[1,2)cupemptyset=[1,2]$. And it is a bad notation to write $x=emptyset$. It would be weird, but look better if you write $xinemptyset$ (but it's best to just say that there is no solution in the third case).
– Batominovski
Jul 24 at 14:10
You should take the union of all three cases.
– xarles
Jul 24 at 14:11
@Batominovski why union though? Shouldn't the x values be those which satisfy all the three cases?
– William
Jul 24 at 14:12
1
@William No, because you are examining three different cases with three different restrictions on $x$. For example, the equation $x^2+3x+2$ has solutions $xin -1cup -2$, not $xin -1cap -2=varnothing$.
– Crosby
Jul 24 at 14:13
@Crosby Owwww!!!! Right ,I get it, sort of, also explain the graph too please! Why is it showing only 2 values of x and not the entire region between 1 and 2 inclusive?
– William
Jul 24 at 14:16
2
2
You are supposed to take the union (not the intersection) of the solution sets in the three cases: $2cup[1,2)cupemptyset=[1,2]$. And it is a bad notation to write $x=emptyset$. It would be weird, but look better if you write $xinemptyset$ (but it's best to just say that there is no solution in the third case).
– Batominovski
Jul 24 at 14:10
You are supposed to take the union (not the intersection) of the solution sets in the three cases: $2cup[1,2)cupemptyset=[1,2]$. And it is a bad notation to write $x=emptyset$. It would be weird, but look better if you write $xinemptyset$ (but it's best to just say that there is no solution in the third case).
– Batominovski
Jul 24 at 14:10
You should take the union of all three cases.
– xarles
Jul 24 at 14:11
You should take the union of all three cases.
– xarles
Jul 24 at 14:11
@Batominovski why union though? Shouldn't the x values be those which satisfy all the three cases?
– William
Jul 24 at 14:12
@Batominovski why union though? Shouldn't the x values be those which satisfy all the three cases?
– William
Jul 24 at 14:12
1
1
@William No, because you are examining three different cases with three different restrictions on $x$. For example, the equation $x^2+3x+2$ has solutions $xin -1cup -2$, not $xin -1cap -2=varnothing$.
– Crosby
Jul 24 at 14:13
@William No, because you are examining three different cases with three different restrictions on $x$. For example, the equation $x^2+3x+2$ has solutions $xin -1cup -2$, not $xin -1cap -2=varnothing$.
– Crosby
Jul 24 at 14:13
@Crosby Owwww!!!! Right ,I get it, sort of, also explain the graph too please! Why is it showing only 2 values of x and not the entire region between 1 and 2 inclusive?
– William
Jul 24 at 14:16
@Crosby Owwww!!!! Right ,I get it, sort of, also explain the graph too please! Why is it showing only 2 values of x and not the entire region between 1 and 2 inclusive?
– William
Jul 24 at 14:16
 |Â
show 1 more comment
4 Answers
4
active
oldest
votes
up vote
3
down vote
accepted
Case 1 tells you, that the only $x$ in $[2,infty)$ satisfying the equation is $x=2$.
Case 2 tells you, that all $x$ in $[1,2)$ satisfy the equation.
Case 3 tells you that no $x$ in $(-infty,1)$ satisfy the equation.
So the union of those, $[1,2]$ is the set of all $x$ satisfying the equation.
In other words: Let $x in mathbbR$ satisfy the equation. Then either C1) $xin [2,infty)$ or C2) $xin [1,2)$ or C3) $xin (-infty,1)$. You tackle all three cases, but $x$ only has to fulfill one of the three cases.
Apparently, desmos graphing calculator shows a wrong plot. See two other examples. The green curve is plottet the right way, the red one is obviously false.
answered Jul 24 at 14:14
Babelfish
418112
Thanks that makes sense! But why do you suggest the graphing calculator shows only two values of x instead of showing the entire region between 1 and 2 inclusive?
– William
Jul 24 at 14:18
I don't really know, wolframalpha shows a meaningful plot. What graphing calculator is this?
– Babelfish
Jul 24 at 14:20
desmos graphing calculator
– William
Jul 24 at 14:21
That's strange indeed. Even funnier: If you put in "$left|x-1right|+left|x-2right|=y$", the result is correct. $yleft(left|x-1right|+left|x-2right|-1right)=0$ also obviously has the wrong plot.
– Babelfish
Jul 24 at 14:28
@William If you are happy with my answer, you might accept it.
– Babelfish
Aug 7 at 14:33
 |Â
show 1 more comment
up vote
0
down vote
Alternative Solution
By the Triangle Inequality,
$$|x-1|+|x-2|=|x-1|+|2-x|geq big|(x-1)+(2-x)big|=1,.$$
The inequality becomes an equality if and only if $2-x=0$, or $x-1=lambda(2-x)$ for some $lambdageq0$ (which gives, by the way, $x=frac2lambda+1lambda+1in[1,2)$). It follows immediately that $[1,2]$ is the solution set for $xinmathbbC$ such that $|x-1|+|x-2|=1$.
answered Jul 24 at 14:13
Batominovski
23.2k22777
add a comment |Â
up vote
0
down vote
You are absolutely right. Just you should've plotted a strip whose endpoints are in $1,2$ in other word you should find the intersection points of the function with the following set$$S=1le xle 2$$
answered Jul 24 at 14:52
Mostafa Ayaz
8,5373630
add a comment |Â
up vote
-1
down vote
By direct way we need to distinguish three cases
$x<1 implies |x-1| + |x-2| = 1-x+2-x=1 implies -2x=-2implies x=1$
$1le x<2 implies |x-1| + |x-2| = x-1+2-x=1 implies 1=1$
$xge 2 implies |x-1| + |x-2| = x-1+x-2=1 implies 2x=4implies x=2$
therefore $1le xle 2$.
answered Jul 24 at 14:32
gimusi
65.1k73583
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Case 1 tells you, that the only $x$ in $[2,infty)$ satisfying the equation is $x=2$.
Case 2 tells you, that all $x$ in $[1,2)$ satisfy the equation.
Case 3 tells you that no $x$ in $(-infty,1)$ satisfy the equation.
So the union of those, $[1,2]$ is the set of all $x$ satisfying the equation.
In other words: Let $x in mathbbR$ satisfy the equation. Then either C1) $xin [2,infty)$ or C2) $xin [1,2)$ or C3) $xin (-infty,1)$. You tackle all three cases, but $x$ only has to fulfill one of the three cases.
Apparently, desmos graphing calculator shows a wrong plot. See two other examples. The green curve is plottet the right way, the red one is obviously false.
answered Jul 24 at 14:14
Babelfish
418112
Thanks that makes sense! But why do you suggest the graphing calculator shows only two values of x instead of showing the entire region between 1 and 2 inclusive?
– William
Jul 24 at 14:18
I don't really know, wolframalpha shows a meaningful plot. What graphing calculator is this?
– Babelfish
Jul 24 at 14:20
desmos graphing calculator
– William
Jul 24 at 14:21
That's strange indeed. Even funnier: If you put in "$left|x-1right|+left|x-2right|=y$", the result is correct. $yleft(left|x-1right|+left|x-2right|-1right)=0$ also obviously has the wrong plot.
– Babelfish
Jul 24 at 14:28
@William If you are happy with my answer, you might accept it.
– Babelfish
Aug 7 at 14:33
 |Â
show 1 more comment
up vote
3
down vote
accepted
Case 1 tells you, that the only $x$ in $[2,infty)$ satisfying the equation is $x=2$.
Case 2 tells you, that all $x$ in $[1,2)$ satisfy the equation.
Case 3 tells you that no $x$ in $(-infty,1)$ satisfy the equation.
So the union of those, $[1,2]$ is the set of all $x$ satisfying the equation.
In other words: Let $x in mathbbR$ satisfy the equation. Then either C1) $xin [2,infty)$ or C2) $xin [1,2)$ or C3) $xin (-infty,1)$. You tackle all three cases, but $x$ only has to fulfill one of the three cases.
Apparently, desmos graphing calculator shows a wrong plot. See two other examples. The green curve is plottet the right way, the red one is obviously false.
answered Jul 24 at 14:14
Babelfish
418112
Thanks that makes sense! But why do you suggest the graphing calculator shows only two values of x instead of showing the entire region between 1 and 2 inclusive?
– William
Jul 24 at 14:18
I don't really know, wolframalpha shows a meaningful plot. What graphing calculator is this?
– Babelfish
Jul 24 at 14:20
desmos graphing calculator
– William
Jul 24 at 14:21
That's strange indeed. Even funnier: If you put in "$left|x-1right|+left|x-2right|=y$", the result is correct. $yleft(left|x-1right|+left|x-2right|-1right)=0$ also obviously has the wrong plot.
– Babelfish
Jul 24 at 14:28
@William If you are happy with my answer, you might accept it.
– Babelfish
Aug 7 at 14:33
 |Â
show 1 more comment
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Case 1 tells you, that the only $x$ in $[2,infty)$ satisfying the equation is $x=2$.
Case 2 tells you, that all $x$ in $[1,2)$ satisfy the equation.
Case 3 tells you that no $x$ in $(-infty,1)$ satisfy the equation.
So the union of those, $[1,2]$ is the set of all $x$ satisfying the equation.
In other words: Let $x in mathbbR$ satisfy the equation. Then either C1) $xin [2,infty)$ or C2) $xin [1,2)$ or C3) $xin (-infty,1)$. You tackle all three cases, but $x$ only has to fulfill one of the three cases.
Apparently, desmos graphing calculator shows a wrong plot. See two other examples. The green curve is plottet the right way, the red one is obviously false.
answered Jul 24 at 14:14
Babelfish
418112
Case 1 tells you, that the only $x$ in $[2,infty)$ satisfying the equation is $x=2$.
Case 2 tells you, that all $x$ in $[1,2)$ satisfy the equation.
Case 3 tells you that no $x$ in $(-infty,1)$ satisfy the equation.
So the union of those, $[1,2]$ is the set of all $x$ satisfying the equation.
In other words: Let $x in mathbbR$ satisfy the equation. Then either C1) $xin [2,infty)$ or C2) $xin [1,2)$ or C3) $xin (-infty,1)$. You tackle all three cases, but $x$ only has to fulfill one of the three cases.
Apparently, desmos graphing calculator shows a wrong plot. See two other examples. The green curve is plottet the right way, the red one is obviously false.
answered Jul 24 at 14:14
Babelfish
418112
edited Jul 24 at 14:40
answered Jul 24 at 14:14
Babelfish
418112
answered Jul 24 at 14:14
Babelfish
418112
answered Jul 24 at 14:14
Babelfish
418112
418112
Thanks that makes sense! But why do you suggest the graphing calculator shows only two values of x instead of showing the entire region between 1 and 2 inclusive?
– William
Jul 24 at 14:18
I don't really know, wolframalpha shows a meaningful plot. What graphing calculator is this?
– Babelfish
Jul 24 at 14:20
desmos graphing calculator
– William
Jul 24 at 14:21
That's strange indeed. Even funnier: If you put in "$left|x-1right|+left|x-2right|=y$", the result is correct. $yleft(left|x-1right|+left|x-2right|-1right)=0$ also obviously has the wrong plot.
– Babelfish
Jul 24 at 14:28
@William If you are happy with my answer, you might accept it.
– Babelfish
Aug 7 at 14:33
 |Â
show 1 more comment
Thanks that makes sense! But why do you suggest the graphing calculator shows only two values of x instead of showing the entire region between 1 and 2 inclusive?
– William
Jul 24 at 14:18
I don't really know, wolframalpha shows a meaningful plot. What graphing calculator is this?
– Babelfish
Jul 24 at 14:20
desmos graphing calculator
– William
Jul 24 at 14:21
That's strange indeed. Even funnier: If you put in "$left|x-1right|+left|x-2right|=y$", the result is correct. $yleft(left|x-1right|+left|x-2right|-1right)=0$ also obviously has the wrong plot.
– Babelfish
Jul 24 at 14:28
@William If you are happy with my answer, you might accept it.
– Babelfish
Aug 7 at 14:33
Thanks that makes sense! But why do you suggest the graphing calculator shows only two values of x instead of showing the entire region between 1 and 2 inclusive?
– William
Jul 24 at 14:18
Thanks that makes sense! But why do you suggest the graphing calculator shows only two values of x instead of showing the entire region between 1 and 2 inclusive?
– William
Jul 24 at 14:18
I don't really know, wolframalpha shows a meaningful plot. What graphing calculator is this?
– Babelfish
Jul 24 at 14:20
I don't really know, wolframalpha shows a meaningful plot. What graphing calculator is this?
– Babelfish
Jul 24 at 14:20
desmos graphing calculator
– William
Jul 24 at 14:21
desmos graphing calculator
– William
Jul 24 at 14:21
That's strange indeed. Even funnier: If you put in "$left|x-1right|+left|x-2right|=y$", the result is correct. $yleft(left|x-1right|+left|x-2right|-1right)=0$ also obviously has the wrong plot.
– Babelfish
Jul 24 at 14:28
That's strange indeed. Even funnier: If you put in "$left|x-1right|+left|x-2right|=y$", the result is correct. $yleft(left|x-1right|+left|x-2right|-1right)=0$ also obviously has the wrong plot.
– Babelfish
Jul 24 at 14:28
@William If you are happy with my answer, you might accept it.
– Babelfish
Aug 7 at 14:33
@William If you are happy with my answer, you might accept it.
– Babelfish
Aug 7 at 14:33
 |Â
show 1 more comment
up vote
0
down vote
Alternative Solution
By the Triangle Inequality,
$$|x-1|+|x-2|=|x-1|+|2-x|geq big|(x-1)+(2-x)big|=1,.$$
The inequality becomes an equality if and only if $2-x=0$, or $x-1=lambda(2-x)$ for some $lambdageq0$ (which gives, by the way, $x=frac2lambda+1lambda+1in[1,2)$). It follows immediately that $[1,2]$ is the solution set for $xinmathbbC$ such that $|x-1|+|x-2|=1$.
answered Jul 24 at 14:13
Batominovski
23.2k22777
add a comment |Â
up vote
0
down vote
Alternative Solution
By the Triangle Inequality,
$$|x-1|+|x-2|=|x-1|+|2-x|geq big|(x-1)+(2-x)big|=1,.$$
The inequality becomes an equality if and only if $2-x=0$, or $x-1=lambda(2-x)$ for some $lambdageq0$ (which gives, by the way, $x=frac2lambda+1lambda+1in[1,2)$). It follows immediately that $[1,2]$ is the solution set for $xinmathbbC$ such that $|x-1|+|x-2|=1$.
answered Jul 24 at 14:13
Batominovski
23.2k22777
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Alternative Solution
By the Triangle Inequality,
$$|x-1|+|x-2|=|x-1|+|2-x|geq big|(x-1)+(2-x)big|=1,.$$
The inequality becomes an equality if and only if $2-x=0$, or $x-1=lambda(2-x)$ for some $lambdageq0$ (which gives, by the way, $x=frac2lambda+1lambda+1in[1,2)$). It follows immediately that $[1,2]$ is the solution set for $xinmathbbC$ such that $|x-1|+|x-2|=1$.
answered Jul 24 at 14:13
Batominovski
23.2k22777
Alternative Solution
By the Triangle Inequality,
$$|x-1|+|x-2|=|x-1|+|2-x|geq big|(x-1)+(2-x)big|=1,.$$
The inequality becomes an equality if and only if $2-x=0$, or $x-1=lambda(2-x)$ for some $lambdageq0$ (which gives, by the way, $x=frac2lambda+1lambda+1in[1,2)$). It follows immediately that $[1,2]$ is the solution set for $xinmathbbC$ such that $|x-1|+|x-2|=1$.
answered Jul 24 at 14:13
Batominovski
23.2k22777
answered Jul 24 at 14:13
Batominovski
23.2k22777
answered Jul 24 at 14:13
Batominovski
23.2k22777
answered Jul 24 at 14:13
Batominovski
23.2k22777
23.2k22777
add a comment |Â
add a comment |Â
up vote
0
down vote
You are absolutely right. Just you should've plotted a strip whose endpoints are in $1,2$ in other word you should find the intersection points of the function with the following set$$S=1le xle 2$$
answered Jul 24 at 14:52
Mostafa Ayaz
8,5373630
add a comment |Â
up vote
0
down vote
You are absolutely right. Just you should've plotted a strip whose endpoints are in $1,2$ in other word you should find the intersection points of the function with the following set$$S=1le xle 2$$
answered Jul 24 at 14:52
Mostafa Ayaz
8,5373630
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You are absolutely right. Just you should've plotted a strip whose endpoints are in $1,2$ in other word you should find the intersection points of the function with the following set$$S=1le xle 2$$
answered Jul 24 at 14:52
Mostafa Ayaz
8,5373630
You are absolutely right. Just you should've plotted a strip whose endpoints are in $1,2$ in other word you should find the intersection points of the function with the following set$$S=1le xle 2$$
answered Jul 24 at 14:52
Mostafa Ayaz
8,5373630
answered Jul 24 at 14:52
Mostafa Ayaz
8,5373630
answered Jul 24 at 14:52
Mostafa Ayaz
8,5373630
answered Jul 24 at 14:52
Mostafa Ayaz
8,5373630
8,5373630
add a comment |Â
add a comment |Â
up vote
-1
down vote
By direct way we need to distinguish three cases
$x<1 implies |x-1| + |x-2| = 1-x+2-x=1 implies -2x=-2implies x=1$
$1le x<2 implies |x-1| + |x-2| = x-1+2-x=1 implies 1=1$
$xge 2 implies |x-1| + |x-2| = x-1+x-2=1 implies 2x=4implies x=2$
therefore $1le xle 2$.
answered Jul 24 at 14:32
gimusi
65.1k73583
add a comment |Â
up vote
-1
down vote
By direct way we need to distinguish three cases
$x<1 implies |x-1| + |x-2| = 1-x+2-x=1 implies -2x=-2implies x=1$
$1le x<2 implies |x-1| + |x-2| = x-1+2-x=1 implies 1=1$
$xge 2 implies |x-1| + |x-2| = x-1+x-2=1 implies 2x=4implies x=2$
therefore $1le xle 2$.
answered Jul 24 at 14:32
gimusi
65.1k73583
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
By direct way we need to distinguish three cases
$x<1 implies |x-1| + |x-2| = 1-x+2-x=1 implies -2x=-2implies x=1$
$1le x<2 implies |x-1| + |x-2| = x-1+2-x=1 implies 1=1$
$xge 2 implies |x-1| + |x-2| = x-1+x-2=1 implies 2x=4implies x=2$
therefore $1le xle 2$.
answered Jul 24 at 14:32
gimusi
65.1k73583
By direct way we need to distinguish three cases
$x<1 implies |x-1| + |x-2| = 1-x+2-x=1 implies -2x=-2implies x=1$
$1le x<2 implies |x-1| + |x-2| = x-1+2-x=1 implies 1=1$
$xge 2 implies |x-1| + |x-2| = x-1+x-2=1 implies 2x=4implies x=2$
therefore $1le xle 2$.
answered Jul 24 at 14:32
gimusi
65.1k73583
answered Jul 24 at 14:32
gimusi
65.1k73583
answered Jul 24 at 14:32
gimusi
65.1k73583
answered Jul 24 at 14:32
gimusi
65.1k73583
65.1k73583
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2
You are supposed to take the union (not the intersection) of the solution sets in the three cases: $2cup[1,2)cupemptyset=[1,2]$. And it is a bad notation to write $x=emptyset$. It would be weird, but look better if you write $xinemptyset$ (but it's best to just say that there is no solution in the third case).
– Batominovski
Jul 24 at 14:10
You should take the union of all three cases.
– xarles
Jul 24 at 14:11
@Batominovski why union though? Shouldn't the x values be those which satisfy all the three cases?
– William
Jul 24 at 14:12
1
@William No, because you are examining three different cases with three different restrictions on $x$. For example, the equation $x^2+3x+2$ has solutions $xin -1cup -2$, not $xin -1cap -2=varnothing$.
– Crosby
Jul 24 at 14:13
@Crosby Owwww!!!! Right ,I get it, sort of, also explain the graph too please! Why is it showing only 2 values of x and not the entire region between 1 and 2 inclusive?
– William
Jul 24 at 14:16