Mixing
Method to solve $|x| + |2-x| leq x+ 1$
$begingroup$
Even if it seems really easy, I'm struggling to solve $$|x| +|2-x |leq x+1.$$
The book says that $ x in [1,3] $.
I first rewrote as $x+(2-x)leq x+1$
with $xgeq 0$ and $-x-(2-x)leq x+1$ with $x<0$. Then I solved.
For the first, I got $1leq x$ and for the 2nd, $-3leq x$
$$xin ]-3;0[cup[1;+infty[ $$
It is maybe a very stupid question, but I can't see what I did wrong?
Thanks for your help.
algebra-precalculus inequality absolute-value
edited Jan 20 at 6:09
Martin Sleziak
44.8k10119272
asked May 17 '17 at 17:42
PoujhPoujh
616516
$endgroup$
add a comment |
$begingroup$
Even if it seems really easy, I'm struggling to solve $$|x| +|2-x |leq x+1.$$
The book says that $ x in [1,3] $.
I first rewrote as $x+(2-x)leq x+1$
with $xgeq 0$ and $-x-(2-x)leq x+1$ with $x<0$. Then I solved.
For the first, I got $1leq x$ and for the 2nd, $-3leq x$
$$xin ]-3;0[cup[1;+infty[ $$
It is maybe a very stupid question, but I can't see what I did wrong?
Thanks for your help.
algebra-precalculus inequality absolute-value
edited Jan 20 at 6:09
Martin Sleziak
44.8k10119272
asked May 17 '17 at 17:42
PoujhPoujh
616516
$endgroup$
2
$begingroup$
Consider three cases: (i) $xle0$, (ii) $0<xle2$, (iii) $x>2$.
$endgroup$
– George Law
May 17 '17 at 17:47
$begingroup$
look at where the graph is below the x-axis
$endgroup$
– user66081
May 17 '17 at 17:50
1
$begingroup$
I will just mention that geometrical meaning of $|x|+|x-2|$ is the sum of distances from $0$ and $2$. So if you move along real line, it should be clear that it is equal to two between $0$ and $2$, So between these two points it is equal to two - it is the length of the line segment from $0$ to $2$. And the graph of $|x|+|x-2|$ is increasing to the right of this interval, decreasing to the left of this interval. You can find a few past question with graphs of similar functions.
$endgroup$
– Martin Sleziak
Jan 20 at 6:14
1
$begingroup$
One of the answers here has a picture with a similar argument as I mentioned above: How to solve equations involving modulus function of the type $|x+1| - |1-x|=2 $ and $ |x-1|=|x|+a$? Some other posts which have graphs: How to solve this absolute value inequality? $ |x| + |x - 2| gt 5 $](https://math.stackexchange.com/q/1472709), [Solve the following $|x-1| + |x+1|= x-3$ and How could we solve $x$, in $|x+1|-|1-x|=2$?
$endgroup$
– Martin Sleziak
Jan 20 at 6:18
add a comment |
$begingroup$
Even if it seems really easy, I'm struggling to solve $$|x| +|2-x |leq x+1.$$
The book says that $ x in [1,3] $.
I first rewrote as $x+(2-x)leq x+1$
with $xgeq 0$ and $-x-(2-x)leq x+1$ with $x<0$. Then I solved.
For the first, I got $1leq x$ and for the 2nd, $-3leq x$
$$xin ]-3;0[cup[1;+infty[ $$
It is maybe a very stupid question, but I can't see what I did wrong?
Thanks for your help.
algebra-precalculus inequality absolute-value
edited Jan 20 at 6:09
Martin Sleziak
44.8k10119272
asked May 17 '17 at 17:42
PoujhPoujh
616516
$endgroup$
Even if it seems really easy, I'm struggling to solve $$|x| +|2-x |leq x+1.$$
The book says that $ x in [1,3] $.
I first rewrote as $x+(2-x)leq x+1$
with $xgeq 0$ and $-x-(2-x)leq x+1$ with $x<0$. Then I solved.
For the first, I got $1leq x$ and for the 2nd, $-3leq x$
$$xin ]-3;0[cup[1;+infty[ $$
It is maybe a very stupid question, but I can't see what I did wrong?
Thanks for your help.
algebra-precalculus inequality absolute-value
algebra-precalculus inequality absolute-value
edited Jan 20 at 6:09
Martin Sleziak
44.8k10119272
asked May 17 '17 at 17:42
PoujhPoujh
616516
edited Jan 20 at 6:09
Martin Sleziak
44.8k10119272
asked May 17 '17 at 17:42
PoujhPoujh
616516
edited Jan 20 at 6:09
Martin Sleziak
44.8k10119272
edited Jan 20 at 6:09
Martin Sleziak
44.8k10119272
edited Jan 20 at 6:09
Martin Sleziak
44.8k10119272
44.8k10119272
asked May 17 '17 at 17:42
PoujhPoujh
616516
asked May 17 '17 at 17:42
PoujhPoujh
616516
asked May 17 '17 at 17:42
PoujhPoujh
616516
616516
2
$begingroup$
Consider three cases: (i) $xle0$, (ii) $0<xle2$, (iii) $x>2$.
$endgroup$
– George Law
May 17 '17 at 17:47
$begingroup$
look at where the graph is below the x-axis
$endgroup$
– user66081
May 17 '17 at 17:50
1
$begingroup$
I will just mention that geometrical meaning of $|x|+|x-2|$ is the sum of distances from $0$ and $2$. So if you move along real line, it should be clear that it is equal to two between $0$ and $2$, So between these two points it is equal to two - it is the length of the line segment from $0$ to $2$. And the graph of $|x|+|x-2|$ is increasing to the right of this interval, decreasing to the left of this interval. You can find a few past question with graphs of similar functions.
$endgroup$
– Martin Sleziak
Jan 20 at 6:14
1
$begingroup$
One of the answers here has a picture with a similar argument as I mentioned above: How to solve equations involving modulus function of the type $|x+1| - |1-x|=2 $ and $ |x-1|=|x|+a$? Some other posts which have graphs: How to solve this absolute value inequality? $ |x| + |x - 2| gt 5 $](https://math.stackexchange.com/q/1472709), [Solve the following $|x-1| + |x+1|= x-3$ and How could we solve $x$, in $|x+1|-|1-x|=2$?
$endgroup$
– Martin Sleziak
Jan 20 at 6:18
add a comment |
2
$begingroup$
Consider three cases: (i) $xle0$, (ii) $0<xle2$, (iii) $x>2$.
$endgroup$
– George Law
May 17 '17 at 17:47
$begingroup$
look at where the graph is below the x-axis
$endgroup$
– user66081
May 17 '17 at 17:50
1
$begingroup$
I will just mention that geometrical meaning of $|x|+|x-2|$ is the sum of distances from $0$ and $2$. So if you move along real line, it should be clear that it is equal to two between $0$ and $2$, So between these two points it is equal to two - it is the length of the line segment from $0$ to $2$. And the graph of $|x|+|x-2|$ is increasing to the right of this interval, decreasing to the left of this interval. You can find a few past question with graphs of similar functions.
$endgroup$
– Martin Sleziak
Jan 20 at 6:14
1
$begingroup$
One of the answers here has a picture with a similar argument as I mentioned above: How to solve equations involving modulus function of the type $|x+1| - |1-x|=2 $ and $ |x-1|=|x|+a$? Some other posts which have graphs: How to solve this absolute value inequality? $ |x| + |x - 2| gt 5 $](https://math.stackexchange.com/q/1472709), [Solve the following $|x-1| + |x+1|= x-3$ and How could we solve $x$, in $|x+1|-|1-x|=2$?
$endgroup$
– Martin Sleziak
Jan 20 at 6:18
2
2
$begingroup$
Consider three cases: (i) $xle0$, (ii) $0<xle2$, (iii) $x>2$.
$endgroup$
– George Law
May 17 '17 at 17:47
$begingroup$
Consider three cases: (i) $xle0$, (ii) $0<xle2$, (iii) $x>2$.
$endgroup$
– George Law
May 17 '17 at 17:47
$begingroup$
look at where the graph is below the x-axis
$endgroup$
– user66081
May 17 '17 at 17:50
$begingroup$
look at where the graph is below the x-axis
$endgroup$
– user66081
May 17 '17 at 17:50
1
1
$begingroup$
I will just mention that geometrical meaning of $|x|+|x-2|$ is the sum of distances from $0$ and $2$. So if you move along real line, it should be clear that it is equal to two between $0$ and $2$, So between these two points it is equal to two - it is the length of the line segment from $0$ to $2$. And the graph of $|x|+|x-2|$ is increasing to the right of this interval, decreasing to the left of this interval. You can find a few past question with graphs of similar functions.
$endgroup$
– Martin Sleziak
Jan 20 at 6:14
$begingroup$
I will just mention that geometrical meaning of $|x|+|x-2|$ is the sum of distances from $0$ and $2$. So if you move along real line, it should be clear that it is equal to two between $0$ and $2$, So between these two points it is equal to two - it is the length of the line segment from $0$ to $2$. And the graph of $|x|+|x-2|$ is increasing to the right of this interval, decreasing to the left of this interval. You can find a few past question with graphs of similar functions.
$endgroup$
– Martin Sleziak
Jan 20 at 6:14
1
1
$begingroup$
One of the answers here has a picture with a similar argument as I mentioned above: How to solve equations involving modulus function of the type $|x+1| - |1-x|=2 $ and $ |x-1|=|x|+a$? Some other posts which have graphs: How to solve this absolute value inequality? $ |x| + |x - 2| gt 5 $](https://math.stackexchange.com/q/1472709), [Solve the following $|x-1| + |x+1|= x-3$ and How could we solve $x$, in $|x+1|-|1-x|=2$?
$endgroup$
– Martin Sleziak
Jan 20 at 6:18
$begingroup$
One of the answers here has a picture with a similar argument as I mentioned above: How to solve equations involving modulus function of the type $|x+1| - |1-x|=2 $ and $ |x-1|=|x|+a$? Some other posts which have graphs: How to solve this absolute value inequality? $ |x| + |x - 2| gt 5 $](https://math.stackexchange.com/q/1472709), [Solve the following $|x-1| + |x+1|= x-3$ and How could we solve $x$, in $|x+1|-|1-x|=2$?
$endgroup$
– Martin Sleziak
Jan 20 at 6:18
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
There are 3 cases:
a) $xleq 0$ (quite useless, because it generates a bright lower bound):
$$|x| +|2-x |leq x+1implies -x+ 2-xleq x+1$$
$$-3xleq -1$$
$$3xgeq1$$
$$xgeqfrac{1}{3}$$
b) $0<xleq 2$:
$$|x| +|2-x |leq x+1implies x+2-xleq x+1$$
$$xgeq 1$$
c) $x > 2$:
$$|x| +|2-x |leq x+1implies x+x-2leq x+1$$
$$2xleq x+3implies xleq 3$$
Now, you know the lower bound of $x$, from case b, $x_{min}=1$, and you know the upper bound of $x$ from case c, $x_{max}=3$. Therefore, $1leq xleq 3implies xin [1;3]$
answered May 17 '17 at 18:06
Don't be a x-triple dotDon't be a x-triple dot
845321
$endgroup$
add a comment |
$begingroup$
Note that $|x| = x$ if $x ge 0$ and $|x| = -x$ if $x < 0$.
Similarly, $|2-x| = 2-x$ if $2 -x ge 0$ (i.e., if $x le 2$), and $|2-x| = x-2$ if $2-x < 0$ (i.e., if $x > 2$).
One way is to consider separately the three cases:
$x > 2$ (and so $x ge 0$ also), so $|x| = x$ and $|2-x| = x-2$.
The inequality becomes $x + x - 2 le x + 1$.
$0 le x le 2$, so $|x| = x$ and $|2-x| = 2-x$.
The inequality becomes $x + 2-x le x + 1$.
$x < 0$ (and so $x le 2$ also), so $|x| = -x$ and $|2-x| = 2-x$.
The inequality becomes $-x+2-x le x + 1$.
Technically there is a fourth combination of $x < 0$ and $x ge 2$, but clearly this case can't exist since no number can be $ < 0$ and $ge 2$ at the same time.
Solve the inequalities in each case, and make sure the answer you get makes sense within the restriction of the corresponding case.
For example, for the second case, solving the inequality will give us $x ge 1$. But remember that this is the case where $0 le x le 2$. Thus from the second case we only get $1 le x le 2$.
Solve the first and third cases, put it all together, and you'll get $x in [1,3]$. Let me know if you require further assistance.
answered May 17 '17 at 17:52
tilpertilper
12.9k11145
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add a comment |
$begingroup$
we can write $$|x|+|x-2|le x+1$$
the first case: $$xgeq 2$$ then we have to solve $$x+x-2le x+1$$
second: $$0le x<2$$ then we have $$x-x+2le x+1$$
last case:
$$x<0$$ then we get $$-x-x+2le x+1$$
Can you finish?
answered May 17 '17 at 17:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
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add a comment |
$begingroup$
A change of sign is due to continuity only possible where
$$|x| +|2-x |= x+1iff |2- x|=x+1-|x|.$$
Squaring the equation yields
$$2|x|(x-1)=(x-1)(x+3).$$
Since $x=1$ is a solution we may happily square
$$2|x|=x+3$$
again leading to
$$3(x-3)(x+1)=0.$$
Now consider the truth of
$$|x| +|2-x |leq x+1$$
in the intervals given by the zeroes $-1$, $1$, and $3$.
answered May 17 '17 at 18:45
Michael HoppeMichael Hoppe
11.1k31837
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add a comment |
$begingroup$
The mistake you have made is that,you have taken
$|2-x| = 2-x forall xge 0$, which is not true.
For $x gt 2$, $|2-x|$ is $ = x-2$.
Hence, you must take intervals as three cases, as George Law points out,
$ (i)xle 0$,
$(ii) 0lt xle2$
$(iii) x>2$
answered May 17 '17 at 17:56
Ananth KamathAnanth Kamath
1,525824
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add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are 3 cases:
a) $xleq 0$ (quite useless, because it generates a bright lower bound):
$$|x| +|2-x |leq x+1implies -x+ 2-xleq x+1$$
$$-3xleq -1$$
$$3xgeq1$$
$$xgeqfrac{1}{3}$$
b) $0<xleq 2$:
$$|x| +|2-x |leq x+1implies x+2-xleq x+1$$
$$xgeq 1$$
c) $x > 2$:
$$|x| +|2-x |leq x+1implies x+x-2leq x+1$$
$$2xleq x+3implies xleq 3$$
Now, you know the lower bound of $x$, from case b, $x_{min}=1$, and you know the upper bound of $x$ from case c, $x_{max}=3$. Therefore, $1leq xleq 3implies xin [1;3]$
answered May 17 '17 at 18:06
Don't be a x-triple dotDon't be a x-triple dot
845321
$endgroup$
add a comment |
$begingroup$
There are 3 cases:
a) $xleq 0$ (quite useless, because it generates a bright lower bound):
$$|x| +|2-x |leq x+1implies -x+ 2-xleq x+1$$
$$-3xleq -1$$
$$3xgeq1$$
$$xgeqfrac{1}{3}$$
b) $0<xleq 2$:
$$|x| +|2-x |leq x+1implies x+2-xleq x+1$$
$$xgeq 1$$
c) $x > 2$:
$$|x| +|2-x |leq x+1implies x+x-2leq x+1$$
$$2xleq x+3implies xleq 3$$
Now, you know the lower bound of $x$, from case b, $x_{min}=1$, and you know the upper bound of $x$ from case c, $x_{max}=3$. Therefore, $1leq xleq 3implies xin [1;3]$
answered May 17 '17 at 18:06
Don't be a x-triple dotDon't be a x-triple dot
845321
$endgroup$
add a comment |
$begingroup$
There are 3 cases:
a) $xleq 0$ (quite useless, because it generates a bright lower bound):
$$|x| +|2-x |leq x+1implies -x+ 2-xleq x+1$$
$$-3xleq -1$$
$$3xgeq1$$
$$xgeqfrac{1}{3}$$
b) $0<xleq 2$:
$$|x| +|2-x |leq x+1implies x+2-xleq x+1$$
$$xgeq 1$$
c) $x > 2$:
$$|x| +|2-x |leq x+1implies x+x-2leq x+1$$
$$2xleq x+3implies xleq 3$$
Now, you know the lower bound of $x$, from case b, $x_{min}=1$, and you know the upper bound of $x$ from case c, $x_{max}=3$. Therefore, $1leq xleq 3implies xin [1;3]$
answered May 17 '17 at 18:06
Don't be a x-triple dotDon't be a x-triple dot
845321
$endgroup$
There are 3 cases:
a) $xleq 0$ (quite useless, because it generates a bright lower bound):
$$|x| +|2-x |leq x+1implies -x+ 2-xleq x+1$$
$$-3xleq -1$$
$$3xgeq1$$
$$xgeqfrac{1}{3}$$
b) $0<xleq 2$:
$$|x| +|2-x |leq x+1implies x+2-xleq x+1$$
$$xgeq 1$$
c) $x > 2$:
$$|x| +|2-x |leq x+1implies x+x-2leq x+1$$
$$2xleq x+3implies xleq 3$$
Now, you know the lower bound of $x$, from case b, $x_{min}=1$, and you know the upper bound of $x$ from case c, $x_{max}=3$. Therefore, $1leq xleq 3implies xin [1;3]$
answered May 17 '17 at 18:06
Don't be a x-triple dotDon't be a x-triple dot
845321
edited May 17 '17 at 18:23
answered May 17 '17 at 18:06
Don't be a x-triple dotDon't be a x-triple dot
845321
answered May 17 '17 at 18:06
Don't be a x-triple dotDon't be a x-triple dot
845321
answered May 17 '17 at 18:06
Don't be a x-triple dotDon't be a x-triple dot
845321
845321
add a comment |
add a comment |
$begingroup$
Note that $|x| = x$ if $x ge 0$ and $|x| = -x$ if $x < 0$.
Similarly, $|2-x| = 2-x$ if $2 -x ge 0$ (i.e., if $x le 2$), and $|2-x| = x-2$ if $2-x < 0$ (i.e., if $x > 2$).
One way is to consider separately the three cases:
$x > 2$ (and so $x ge 0$ also), so $|x| = x$ and $|2-x| = x-2$.
The inequality becomes $x + x - 2 le x + 1$.
$0 le x le 2$, so $|x| = x$ and $|2-x| = 2-x$.
The inequality becomes $x + 2-x le x + 1$.
$x < 0$ (and so $x le 2$ also), so $|x| = -x$ and $|2-x| = 2-x$.
The inequality becomes $-x+2-x le x + 1$.
Technically there is a fourth combination of $x < 0$ and $x ge 2$, but clearly this case can't exist since no number can be $ < 0$ and $ge 2$ at the same time.
Solve the inequalities in each case, and make sure the answer you get makes sense within the restriction of the corresponding case.
For example, for the second case, solving the inequality will give us $x ge 1$. But remember that this is the case where $0 le x le 2$. Thus from the second case we only get $1 le x le 2$.
Solve the first and third cases, put it all together, and you'll get $x in [1,3]$. Let me know if you require further assistance.
answered May 17 '17 at 17:52
tilpertilper
12.9k11145
$endgroup$
add a comment |
$begingroup$
Note that $|x| = x$ if $x ge 0$ and $|x| = -x$ if $x < 0$.
Similarly, $|2-x| = 2-x$ if $2 -x ge 0$ (i.e., if $x le 2$), and $|2-x| = x-2$ if $2-x < 0$ (i.e., if $x > 2$).
One way is to consider separately the three cases:
$x > 2$ (and so $x ge 0$ also), so $|x| = x$ and $|2-x| = x-2$.
The inequality becomes $x + x - 2 le x + 1$.
$0 le x le 2$, so $|x| = x$ and $|2-x| = 2-x$.
The inequality becomes $x + 2-x le x + 1$.
$x < 0$ (and so $x le 2$ also), so $|x| = -x$ and $|2-x| = 2-x$.
The inequality becomes $-x+2-x le x + 1$.
Technically there is a fourth combination of $x < 0$ and $x ge 2$, but clearly this case can't exist since no number can be $ < 0$ and $ge 2$ at the same time.
Solve the inequalities in each case, and make sure the answer you get makes sense within the restriction of the corresponding case.
For example, for the second case, solving the inequality will give us $x ge 1$. But remember that this is the case where $0 le x le 2$. Thus from the second case we only get $1 le x le 2$.
Solve the first and third cases, put it all together, and you'll get $x in [1,3]$. Let me know if you require further assistance.
answered May 17 '17 at 17:52
tilpertilper
12.9k11145
$endgroup$
add a comment |
$begingroup$
Note that $|x| = x$ if $x ge 0$ and $|x| = -x$ if $x < 0$.
Similarly, $|2-x| = 2-x$ if $2 -x ge 0$ (i.e., if $x le 2$), and $|2-x| = x-2$ if $2-x < 0$ (i.e., if $x > 2$).
One way is to consider separately the three cases:
$x > 2$ (and so $x ge 0$ also), so $|x| = x$ and $|2-x| = x-2$.
The inequality becomes $x + x - 2 le x + 1$.
$0 le x le 2$, so $|x| = x$ and $|2-x| = 2-x$.
The inequality becomes $x + 2-x le x + 1$.
$x < 0$ (and so $x le 2$ also), so $|x| = -x$ and $|2-x| = 2-x$.
The inequality becomes $-x+2-x le x + 1$.
Technically there is a fourth combination of $x < 0$ and $x ge 2$, but clearly this case can't exist since no number can be $ < 0$ and $ge 2$ at the same time.
Solve the inequalities in each case, and make sure the answer you get makes sense within the restriction of the corresponding case.
For example, for the second case, solving the inequality will give us $x ge 1$. But remember that this is the case where $0 le x le 2$. Thus from the second case we only get $1 le x le 2$.
Solve the first and third cases, put it all together, and you'll get $x in [1,3]$. Let me know if you require further assistance.
answered May 17 '17 at 17:52
tilpertilper
12.9k11145
$endgroup$
Note that $|x| = x$ if $x ge 0$ and $|x| = -x$ if $x < 0$.
Similarly, $|2-x| = 2-x$ if $2 -x ge 0$ (i.e., if $x le 2$), and $|2-x| = x-2$ if $2-x < 0$ (i.e., if $x > 2$).
One way is to consider separately the three cases:
$x > 2$ (and so $x ge 0$ also), so $|x| = x$ and $|2-x| = x-2$.
The inequality becomes $x + x - 2 le x + 1$.
$0 le x le 2$, so $|x| = x$ and $|2-x| = 2-x$.
The inequality becomes $x + 2-x le x + 1$.
$x < 0$ (and so $x le 2$ also), so $|x| = -x$ and $|2-x| = 2-x$.
The inequality becomes $-x+2-x le x + 1$.
Technically there is a fourth combination of $x < 0$ and $x ge 2$, but clearly this case can't exist since no number can be $ < 0$ and $ge 2$ at the same time.
Solve the inequalities in each case, and make sure the answer you get makes sense within the restriction of the corresponding case.
For example, for the second case, solving the inequality will give us $x ge 1$. But remember that this is the case where $0 le x le 2$. Thus from the second case we only get $1 le x le 2$.
Solve the first and third cases, put it all together, and you'll get $x in [1,3]$. Let me know if you require further assistance.
answered May 17 '17 at 17:52
tilpertilper
12.9k11145
answered May 17 '17 at 17:52
tilpertilper
12.9k11145
answered May 17 '17 at 17:52
tilpertilper
12.9k11145
answered May 17 '17 at 17:52
tilpertilper
12.9k11145
12.9k11145
add a comment |
add a comment |
$begingroup$
we can write $$|x|+|x-2|le x+1$$
the first case: $$xgeq 2$$ then we have to solve $$x+x-2le x+1$$
second: $$0le x<2$$ then we have $$x-x+2le x+1$$
last case:
$$x<0$$ then we get $$-x-x+2le x+1$$
Can you finish?
answered May 17 '17 at 17:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
add a comment |
$begingroup$
we can write $$|x|+|x-2|le x+1$$
the first case: $$xgeq 2$$ then we have to solve $$x+x-2le x+1$$
second: $$0le x<2$$ then we have $$x-x+2le x+1$$
last case:
$$x<0$$ then we get $$-x-x+2le x+1$$
Can you finish?
answered May 17 '17 at 17:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
add a comment |
$begingroup$
we can write $$|x|+|x-2|le x+1$$
the first case: $$xgeq 2$$ then we have to solve $$x+x-2le x+1$$
second: $$0le x<2$$ then we have $$x-x+2le x+1$$
last case:
$$x<0$$ then we get $$-x-x+2le x+1$$
Can you finish?
answered May 17 '17 at 17:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
$endgroup$
we can write $$|x|+|x-2|le x+1$$
the first case: $$xgeq 2$$ then we have to solve $$x+x-2le x+1$$
second: $$0le x<2$$ then we have $$x-x+2le x+1$$
last case:
$$x<0$$ then we get $$-x-x+2le x+1$$
Can you finish?
answered May 17 '17 at 17:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
answered May 17 '17 at 17:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
answered May 17 '17 at 17:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
answered May 17 '17 at 17:51
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.8k42866
76.8k42866
add a comment |
add a comment |
$begingroup$
A change of sign is due to continuity only possible where
$$|x| +|2-x |= x+1iff |2- x|=x+1-|x|.$$
Squaring the equation yields
$$2|x|(x-1)=(x-1)(x+3).$$
Since $x=1$ is a solution we may happily square
$$2|x|=x+3$$
again leading to
$$3(x-3)(x+1)=0.$$
Now consider the truth of
$$|x| +|2-x |leq x+1$$
in the intervals given by the zeroes $-1$, $1$, and $3$.
answered May 17 '17 at 18:45
Michael HoppeMichael Hoppe
11.1k31837
$endgroup$
add a comment |
$begingroup$
A change of sign is due to continuity only possible where
$$|x| +|2-x |= x+1iff |2- x|=x+1-|x|.$$
Squaring the equation yields
$$2|x|(x-1)=(x-1)(x+3).$$
Since $x=1$ is a solution we may happily square
$$2|x|=x+3$$
again leading to
$$3(x-3)(x+1)=0.$$
Now consider the truth of
$$|x| +|2-x |leq x+1$$
in the intervals given by the zeroes $-1$, $1$, and $3$.
answered May 17 '17 at 18:45
Michael HoppeMichael Hoppe
11.1k31837
$endgroup$
add a comment |
$begingroup$
A change of sign is due to continuity only possible where
$$|x| +|2-x |= x+1iff |2- x|=x+1-|x|.$$
Squaring the equation yields
$$2|x|(x-1)=(x-1)(x+3).$$
Since $x=1$ is a solution we may happily square
$$2|x|=x+3$$
again leading to
$$3(x-3)(x+1)=0.$$
Now consider the truth of
$$|x| +|2-x |leq x+1$$
in the intervals given by the zeroes $-1$, $1$, and $3$.
answered May 17 '17 at 18:45
Michael HoppeMichael Hoppe
11.1k31837
$endgroup$
A change of sign is due to continuity only possible where
$$|x| +|2-x |= x+1iff |2- x|=x+1-|x|.$$
Squaring the equation yields
$$2|x|(x-1)=(x-1)(x+3).$$
Since $x=1$ is a solution we may happily square
$$2|x|=x+3$$
again leading to
$$3(x-3)(x+1)=0.$$
Now consider the truth of
$$|x| +|2-x |leq x+1$$
in the intervals given by the zeroes $-1$, $1$, and $3$.
answered May 17 '17 at 18:45
Michael HoppeMichael Hoppe
11.1k31837
answered May 17 '17 at 18:45
Michael HoppeMichael Hoppe
11.1k31837
answered May 17 '17 at 18:45
Michael HoppeMichael Hoppe
11.1k31837
answered May 17 '17 at 18:45
Michael HoppeMichael Hoppe
11.1k31837
11.1k31837
add a comment |
add a comment |
$begingroup$
The mistake you have made is that,you have taken
$|2-x| = 2-x forall xge 0$, which is not true.
For $x gt 2$, $|2-x|$ is $ = x-2$.
Hence, you must take intervals as three cases, as George Law points out,
$ (i)xle 0$,
$(ii) 0lt xle2$
$(iii) x>2$
answered May 17 '17 at 17:56
Ananth KamathAnanth Kamath
1,525824
$endgroup$
add a comment |
$begingroup$
The mistake you have made is that,you have taken
$|2-x| = 2-x forall xge 0$, which is not true.
For $x gt 2$, $|2-x|$ is $ = x-2$.
Hence, you must take intervals as three cases, as George Law points out,
$ (i)xle 0$,
$(ii) 0lt xle2$
$(iii) x>2$
answered May 17 '17 at 17:56
Ananth KamathAnanth Kamath
1,525824
$endgroup$
add a comment |
$begingroup$
The mistake you have made is that,you have taken
$|2-x| = 2-x forall xge 0$, which is not true.
For $x gt 2$, $|2-x|$ is $ = x-2$.
Hence, you must take intervals as three cases, as George Law points out,
$ (i)xle 0$,
$(ii) 0lt xle2$
$(iii) x>2$
answered May 17 '17 at 17:56
Ananth KamathAnanth Kamath
1,525824
$endgroup$
The mistake you have made is that,you have taken
$|2-x| = 2-x forall xge 0$, which is not true.
For $x gt 2$, $|2-x|$ is $ = x-2$.
Hence, you must take intervals as three cases, as George Law points out,
$ (i)xle 0$,
$(ii) 0lt xle2$
$(iii) x>2$
answered May 17 '17 at 17:56
Ananth KamathAnanth Kamath
1,525824
edited Jan 20 at 5:48
answered May 17 '17 at 17:56
Ananth KamathAnanth Kamath
1,525824
answered May 17 '17 at 17:56
Ananth KamathAnanth Kamath
1,525824
answered May 17 '17 at 17:56
Ananth KamathAnanth Kamath
1,525824
1,525824
add a comment |
add a comment |
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2
$begingroup$
Consider three cases: (i) $xle0$, (ii) $0<xle2$, (iii) $x>2$.
$endgroup$
– George Law
May 17 '17 at 17:47
$begingroup$
look at where the graph is below the x-axis
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– user66081
May 17 '17 at 17:50
1
$begingroup$
I will just mention that geometrical meaning of $|x|+|x-2|$ is the sum of distances from $0$ and $2$. So if you move along real line, it should be clear that it is equal to two between $0$ and $2$, So between these two points it is equal to two - it is the length of the line segment from $0$ to $2$. And the graph of $|x|+|x-2|$ is increasing to the right of this interval, decreasing to the left of this interval. You can find a few past question with graphs of similar functions.
$endgroup$
– Martin Sleziak
Jan 20 at 6:14
1
$begingroup$
One of the answers here has a picture with a similar argument as I mentioned above: How to solve equations involving modulus function of the type $|x+1| - |1-x|=2 $ and $ |x-1|=|x|+a$? Some other posts which have graphs: How to solve this absolute value inequality? $ |x| + |x - 2| gt 5 $](https://math.stackexchange.com/q/1472709), [Solve the following $|x-1| + |x+1|= x-3$ and How could we solve $x$, in $|x+1|-|1-x|=2$?
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– Martin Sleziak
Jan 20 at 6:18