Mixing
Find the set of the real numbers $x$ satisfying the given inequality $frac{1}{x-4}<frac{5}{x+1}$
$begingroup$
Find the set of the real numbers $x$ satisfying the given inequality and sketch them on the real line:
$frac{1}{x-4}<frac{5}{x+1}$
We know that $frac{1}{x-4}<frac{5}{x+1} iff frac{1}{x-4}-frac{5}{x+1}<0$
$iff frac{x+1-5x+20}{(x-4)(x+1)}<0$
$iff frac{21-4x}{(x-4)(x+1)}<0$
$frac{21-4x}{(x-4)(x+1)}$ will be negative, if
begin{description}
$21-4x<0$ and $(x-4)(x+1)>0$.
or,
$21-4x>0$ and $(x-4)(x+1)<0$
The first case,
$21-4x<0 iff 4x>21 iff x>frac{21}{4}$
$(x-4)(x+1)>0$, if $(x-4)$ and $(x+1)$ have the same sign, this means that:
$x-4<0$ and $x+1<0$, then $x<4$ and $x<-1$
Thus, $forall x in (-infty,4)cap (-infty,-1)=(-infty,-1)$, $(x-4)(x+1)>0$
$x-4>0$ and $x+1>0$, then $x>4$ and $x>-1$.
Thus, $forall x in (4,infty)cap (-1,infty)=(4,infty)$, $(x-4)(x+1)>0$
SO, $forall x in (-infty,-1) cup (4,infty)$, $(x-4)(x+1)>0$
Therefore, $forall x in (frac{21}{4},infty) cap [(-infty,-1) cup (4,infty)]=(frac{21}{4},infty)$, $frac{1}{x-4}<frac{5}{x+1}$.
The second case,
$21-4x>0 iff 4x<21 iff x<frac{21}{4}$
$(x-4)(x+1)<0$, if $(x-4)$ and $(x+1)$ have a different signs, this means that:
$x-4<0$ and $x+1>0$, then $x<4$ and $x>-1$
Thus, $forall x in (-1,4)$, $(x-4)(x+1)<0$.
$x-4>0$ and $x+1<0$, then $x>4$ and $x<-1$, which is impossible.
Therefore, $forall x in (-infty,-frac{21}{4})cup (-1,4)=(-1,4)$, $frac{1}{x-4}<frac{5}{x+1}$.
As a result, the solution set is $(frac{21}{4},infty) cup (-1,4)$.
Is that true, please?
real-analysis proof-verification inequality
edited Feb 18 at 4:16
Michael Rozenberg
108k1895200
asked Jan 26 at 21:25
DimaDima
827516
$endgroup$
add a comment |
$begingroup$
Find the set of the real numbers $x$ satisfying the given inequality and sketch them on the real line:
$frac{1}{x-4}<frac{5}{x+1}$
We know that $frac{1}{x-4}<frac{5}{x+1} iff frac{1}{x-4}-frac{5}{x+1}<0$
$iff frac{x+1-5x+20}{(x-4)(x+1)}<0$
$iff frac{21-4x}{(x-4)(x+1)}<0$
$frac{21-4x}{(x-4)(x+1)}$ will be negative, if
begin{description}
$21-4x<0$ and $(x-4)(x+1)>0$.
or,
$21-4x>0$ and $(x-4)(x+1)<0$
The first case,
$21-4x<0 iff 4x>21 iff x>frac{21}{4}$
$(x-4)(x+1)>0$, if $(x-4)$ and $(x+1)$ have the same sign, this means that:
$x-4<0$ and $x+1<0$, then $x<4$ and $x<-1$
Thus, $forall x in (-infty,4)cap (-infty,-1)=(-infty,-1)$, $(x-4)(x+1)>0$
$x-4>0$ and $x+1>0$, then $x>4$ and $x>-1$.
Thus, $forall x in (4,infty)cap (-1,infty)=(4,infty)$, $(x-4)(x+1)>0$
SO, $forall x in (-infty,-1) cup (4,infty)$, $(x-4)(x+1)>0$
Therefore, $forall x in (frac{21}{4},infty) cap [(-infty,-1) cup (4,infty)]=(frac{21}{4},infty)$, $frac{1}{x-4}<frac{5}{x+1}$.
The second case,
$21-4x>0 iff 4x<21 iff x<frac{21}{4}$
$(x-4)(x+1)<0$, if $(x-4)$ and $(x+1)$ have a different signs, this means that:
$x-4<0$ and $x+1>0$, then $x<4$ and $x>-1$
Thus, $forall x in (-1,4)$, $(x-4)(x+1)<0$.
$x-4>0$ and $x+1<0$, then $x>4$ and $x<-1$, which is impossible.
Therefore, $forall x in (-infty,-frac{21}{4})cup (-1,4)=(-1,4)$, $frac{1}{x-4}<frac{5}{x+1}$.
As a result, the solution set is $(frac{21}{4},infty) cup (-1,4)$.
Is that true, please?
real-analysis proof-verification inequality
edited Feb 18 at 4:16
Michael Rozenberg
108k1895200
asked Jan 26 at 21:25
DimaDima
827516
$endgroup$
1
$begingroup$
You have a typo in the third to last line. It should say $forall x in (-infty, frac{21}{4}) cup cdots$. Other than that you are correct.
$endgroup$
– kccu
Jan 26 at 21:30
$begingroup$
@kccu Thank you so much.
$endgroup$
– Dima
Jan 26 at 21:36
add a comment |
$begingroup$
Find the set of the real numbers $x$ satisfying the given inequality and sketch them on the real line:
$frac{1}{x-4}<frac{5}{x+1}$
We know that $frac{1}{x-4}<frac{5}{x+1} iff frac{1}{x-4}-frac{5}{x+1}<0$
$iff frac{x+1-5x+20}{(x-4)(x+1)}<0$
$iff frac{21-4x}{(x-4)(x+1)}<0$
$frac{21-4x}{(x-4)(x+1)}$ will be negative, if
begin{description}
$21-4x<0$ and $(x-4)(x+1)>0$.
or,
$21-4x>0$ and $(x-4)(x+1)<0$
The first case,
$21-4x<0 iff 4x>21 iff x>frac{21}{4}$
$(x-4)(x+1)>0$, if $(x-4)$ and $(x+1)$ have the same sign, this means that:
$x-4<0$ and $x+1<0$, then $x<4$ and $x<-1$
Thus, $forall x in (-infty,4)cap (-infty,-1)=(-infty,-1)$, $(x-4)(x+1)>0$
$x-4>0$ and $x+1>0$, then $x>4$ and $x>-1$.
Thus, $forall x in (4,infty)cap (-1,infty)=(4,infty)$, $(x-4)(x+1)>0$
SO, $forall x in (-infty,-1) cup (4,infty)$, $(x-4)(x+1)>0$
Therefore, $forall x in (frac{21}{4},infty) cap [(-infty,-1) cup (4,infty)]=(frac{21}{4},infty)$, $frac{1}{x-4}<frac{5}{x+1}$.
The second case,
$21-4x>0 iff 4x<21 iff x<frac{21}{4}$
$(x-4)(x+1)<0$, if $(x-4)$ and $(x+1)$ have a different signs, this means that:
$x-4<0$ and $x+1>0$, then $x<4$ and $x>-1$
Thus, $forall x in (-1,4)$, $(x-4)(x+1)<0$.
$x-4>0$ and $x+1<0$, then $x>4$ and $x<-1$, which is impossible.
Therefore, $forall x in (-infty,-frac{21}{4})cup (-1,4)=(-1,4)$, $frac{1}{x-4}<frac{5}{x+1}$.
As a result, the solution set is $(frac{21}{4},infty) cup (-1,4)$.
Is that true, please?
real-analysis proof-verification inequality
edited Feb 18 at 4:16
Michael Rozenberg
108k1895200
asked Jan 26 at 21:25
DimaDima
827516
$endgroup$
Find the set of the real numbers $x$ satisfying the given inequality and sketch them on the real line:
$frac{1}{x-4}<frac{5}{x+1}$
We know that $frac{1}{x-4}<frac{5}{x+1} iff frac{1}{x-4}-frac{5}{x+1}<0$
$iff frac{x+1-5x+20}{(x-4)(x+1)}<0$
$iff frac{21-4x}{(x-4)(x+1)}<0$
$frac{21-4x}{(x-4)(x+1)}$ will be negative, if
begin{description}
$21-4x<0$ and $(x-4)(x+1)>0$.
or,
$21-4x>0$ and $(x-4)(x+1)<0$
The first case,
$21-4x<0 iff 4x>21 iff x>frac{21}{4}$
$(x-4)(x+1)>0$, if $(x-4)$ and $(x+1)$ have the same sign, this means that:
$x-4<0$ and $x+1<0$, then $x<4$ and $x<-1$
Thus, $forall x in (-infty,4)cap (-infty,-1)=(-infty,-1)$, $(x-4)(x+1)>0$
$x-4>0$ and $x+1>0$, then $x>4$ and $x>-1$.
Thus, $forall x in (4,infty)cap (-1,infty)=(4,infty)$, $(x-4)(x+1)>0$
SO, $forall x in (-infty,-1) cup (4,infty)$, $(x-4)(x+1)>0$
Therefore, $forall x in (frac{21}{4},infty) cap [(-infty,-1) cup (4,infty)]=(frac{21}{4},infty)$, $frac{1}{x-4}<frac{5}{x+1}$.
The second case,
$21-4x>0 iff 4x<21 iff x<frac{21}{4}$
$(x-4)(x+1)<0$, if $(x-4)$ and $(x+1)$ have a different signs, this means that:
$x-4<0$ and $x+1>0$, then $x<4$ and $x>-1$
Thus, $forall x in (-1,4)$, $(x-4)(x+1)<0$.
$x-4>0$ and $x+1<0$, then $x>4$ and $x<-1$, which is impossible.
Therefore, $forall x in (-infty,-frac{21}{4})cup (-1,4)=(-1,4)$, $frac{1}{x-4}<frac{5}{x+1}$.
As a result, the solution set is $(frac{21}{4},infty) cup (-1,4)$.
Is that true, please?
real-analysis proof-verification inequality
real-analysis proof-verification inequality
edited Feb 18 at 4:16
Michael Rozenberg
108k1895200
asked Jan 26 at 21:25
DimaDima
827516
edited Feb 18 at 4:16
Michael Rozenberg
108k1895200
asked Jan 26 at 21:25
DimaDima
827516
edited Feb 18 at 4:16
Michael Rozenberg
108k1895200
edited Feb 18 at 4:16
Michael Rozenberg
108k1895200
edited Feb 18 at 4:16
Michael Rozenberg
108k1895200
108k1895200
asked Jan 26 at 21:25
DimaDima
827516
asked Jan 26 at 21:25
DimaDima
827516
asked Jan 26 at 21:25
DimaDima
827516
827516
1
$begingroup$
You have a typo in the third to last line. It should say $forall x in (-infty, frac{21}{4}) cup cdots$. Other than that you are correct.
$endgroup$
– kccu
Jan 26 at 21:30
$begingroup$
@kccu Thank you so much.
$endgroup$
– Dima
Jan 26 at 21:36
add a comment |
1
$begingroup$
You have a typo in the third to last line. It should say $forall x in (-infty, frac{21}{4}) cup cdots$. Other than that you are correct.
$endgroup$
– kccu
Jan 26 at 21:30
$begingroup$
@kccu Thank you so much.
$endgroup$
– Dima
Jan 26 at 21:36
1
1
$begingroup$
You have a typo in the third to last line. It should say $forall x in (-infty, frac{21}{4}) cup cdots$. Other than that you are correct.
$endgroup$
– kccu
Jan 26 at 21:30
$begingroup$
You have a typo in the third to last line. It should say $forall x in (-infty, frac{21}{4}) cup cdots$. Other than that you are correct.
$endgroup$
– kccu
Jan 26 at 21:30
$begingroup$
@kccu Thank you so much.
$endgroup$
– Dima
Jan 26 at 21:36
$begingroup$
@kccu Thank you so much.
$endgroup$
– Dima
Jan 26 at 21:36
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, you are right, but after writing $$frac{4x-21}{(x-4)(x+1)}>0$$ we get the answer immediately by the intervals method:
$$(-1,4)cupleft(frac{21}{4},+inftyright).$$
The intervals method here it's the following.
We need to draw a $x$ axis and to put points $-1$, $4$ and $frac{21}{4}$.
Now, easy easy to see that the sing of $frac{4x-21}{(x-4)(x+1)}$ for $x>frac{21}{4}$ is $+$
and since degree of our points are odd (they are equal to $1$,
we see the the sing of the expression is changed.
Id est, we got the following sings on segments $(-infty,-1),$ $(-1,4),$ $left(4,frac{21}{4}right)$ and $left(frac{21}{4},+inftyright)$.
$$-,+,-,+$$
and we can write the answer.
answered Jan 26 at 21:34
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
What's the intervals method?
$endgroup$
– Dr. Mathva
Jan 27 at 12:30
$begingroup$
@Dr. Mathva I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 27 at 12:48
add a comment |
$begingroup$
Here's a shorter way, with less computations:
- If $x<-1$, both sides of the inequality are negative, so that
$$frac{1}{x-4}<frac{5}{x+1}iff x-4>frac{x+1}5iff5x-20>x+1iff x>frac{21}5,$$
which is impossible since $x<-4$. - If $-1<x<4$, the l.h.s. $<0$ whereas the r.h.s. is $>0$, so the inequality is satisfied for all $xin (-1,4)$.
- If $x>0$, both sides are positive, so, as in the first case, we obtain $x>smash{dfrac{21}5}$, condition which is not incompatible with this case.
To sum it up the solutions are
$$(-1,4)cupBigl(frac{21}5,+inftyBigr).$$
answered Jan 26 at 22:00
BernardBernard
123k741117
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Yes, you are right, but after writing $$frac{4x-21}{(x-4)(x+1)}>0$$ we get the answer immediately by the intervals method:
$$(-1,4)cupleft(frac{21}{4},+inftyright).$$
The intervals method here it's the following.
We need to draw a $x$ axis and to put points $-1$, $4$ and $frac{21}{4}$.
Now, easy easy to see that the sing of $frac{4x-21}{(x-4)(x+1)}$ for $x>frac{21}{4}$ is $+$
and since degree of our points are odd (they are equal to $1$,
we see the the sing of the expression is changed.
Id est, we got the following sings on segments $(-infty,-1),$ $(-1,4),$ $left(4,frac{21}{4}right)$ and $left(frac{21}{4},+inftyright)$.
$$-,+,-,+$$
and we can write the answer.
answered Jan 26 at 21:34
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
What's the intervals method?
$endgroup$
– Dr. Mathva
Jan 27 at 12:30
$begingroup$
@Dr. Mathva I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 27 at 12:48
add a comment |
$begingroup$
Yes, you are right, but after writing $$frac{4x-21}{(x-4)(x+1)}>0$$ we get the answer immediately by the intervals method:
$$(-1,4)cupleft(frac{21}{4},+inftyright).$$
The intervals method here it's the following.
We need to draw a $x$ axis and to put points $-1$, $4$ and $frac{21}{4}$.
Now, easy easy to see that the sing of $frac{4x-21}{(x-4)(x+1)}$ for $x>frac{21}{4}$ is $+$
and since degree of our points are odd (they are equal to $1$,
we see the the sing of the expression is changed.
Id est, we got the following sings on segments $(-infty,-1),$ $(-1,4),$ $left(4,frac{21}{4}right)$ and $left(frac{21}{4},+inftyright)$.
$$-,+,-,+$$
and we can write the answer.
answered Jan 26 at 21:34
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
What's the intervals method?
$endgroup$
– Dr. Mathva
Jan 27 at 12:30
$begingroup$
@Dr. Mathva I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 27 at 12:48
add a comment |
$begingroup$
Yes, you are right, but after writing $$frac{4x-21}{(x-4)(x+1)}>0$$ we get the answer immediately by the intervals method:
$$(-1,4)cupleft(frac{21}{4},+inftyright).$$
The intervals method here it's the following.
We need to draw a $x$ axis and to put points $-1$, $4$ and $frac{21}{4}$.
Now, easy easy to see that the sing of $frac{4x-21}{(x-4)(x+1)}$ for $x>frac{21}{4}$ is $+$
and since degree of our points are odd (they are equal to $1$,
we see the the sing of the expression is changed.
Id est, we got the following sings on segments $(-infty,-1),$ $(-1,4),$ $left(4,frac{21}{4}right)$ and $left(frac{21}{4},+inftyright)$.
$$-,+,-,+$$
and we can write the answer.
answered Jan 26 at 21:34
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
Yes, you are right, but after writing $$frac{4x-21}{(x-4)(x+1)}>0$$ we get the answer immediately by the intervals method:
$$(-1,4)cupleft(frac{21}{4},+inftyright).$$
The intervals method here it's the following.
We need to draw a $x$ axis and to put points $-1$, $4$ and $frac{21}{4}$.
Now, easy easy to see that the sing of $frac{4x-21}{(x-4)(x+1)}$ for $x>frac{21}{4}$ is $+$
and since degree of our points are odd (they are equal to $1$,
we see the the sing of the expression is changed.
Id est, we got the following sings on segments $(-infty,-1),$ $(-1,4),$ $left(4,frac{21}{4}right)$ and $left(frac{21}{4},+inftyright)$.
$$-,+,-,+$$
and we can write the answer.
answered Jan 26 at 21:34
Michael RozenbergMichael Rozenberg
108k1895200
edited Jan 27 at 12:47
answered Jan 26 at 21:34
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 26 at 21:34
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 26 at 21:34
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
$begingroup$
What's the intervals method?
$endgroup$
– Dr. Mathva
Jan 27 at 12:30
$begingroup$
@Dr. Mathva I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 27 at 12:48
add a comment |
$begingroup$
What's the intervals method?
$endgroup$
– Dr. Mathva
Jan 27 at 12:30
$begingroup$
@Dr. Mathva I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 27 at 12:48
$begingroup$
What's the intervals method?
$endgroup$
– Dr. Mathva
Jan 27 at 12:30
$begingroup$
What's the intervals method?
$endgroup$
– Dr. Mathva
Jan 27 at 12:30
$begingroup$
@Dr. Mathva I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 27 at 12:48
$begingroup$
@Dr. Mathva I added something. See now.
$endgroup$
– Michael Rozenberg
Jan 27 at 12:48
add a comment |
$begingroup$
Here's a shorter way, with less computations:
- If $x<-1$, both sides of the inequality are negative, so that
$$frac{1}{x-4}<frac{5}{x+1}iff x-4>frac{x+1}5iff5x-20>x+1iff x>frac{21}5,$$
which is impossible since $x<-4$. - If $-1<x<4$, the l.h.s. $<0$ whereas the r.h.s. is $>0$, so the inequality is satisfied for all $xin (-1,4)$.
- If $x>0$, both sides are positive, so, as in the first case, we obtain $x>smash{dfrac{21}5}$, condition which is not incompatible with this case.
To sum it up the solutions are
$$(-1,4)cupBigl(frac{21}5,+inftyBigr).$$
answered Jan 26 at 22:00
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Here's a shorter way, with less computations:
- If $x<-1$, both sides of the inequality are negative, so that
$$frac{1}{x-4}<frac{5}{x+1}iff x-4>frac{x+1}5iff5x-20>x+1iff x>frac{21}5,$$
which is impossible since $x<-4$. - If $-1<x<4$, the l.h.s. $<0$ whereas the r.h.s. is $>0$, so the inequality is satisfied for all $xin (-1,4)$.
- If $x>0$, both sides are positive, so, as in the first case, we obtain $x>smash{dfrac{21}5}$, condition which is not incompatible with this case.
To sum it up the solutions are
$$(-1,4)cupBigl(frac{21}5,+inftyBigr).$$
answered Jan 26 at 22:00
BernardBernard
123k741117
$endgroup$
add a comment |
$begingroup$
Here's a shorter way, with less computations:
- If $x<-1$, both sides of the inequality are negative, so that
$$frac{1}{x-4}<frac{5}{x+1}iff x-4>frac{x+1}5iff5x-20>x+1iff x>frac{21}5,$$
which is impossible since $x<-4$. - If $-1<x<4$, the l.h.s. $<0$ whereas the r.h.s. is $>0$, so the inequality is satisfied for all $xin (-1,4)$.
- If $x>0$, both sides are positive, so, as in the first case, we obtain $x>smash{dfrac{21}5}$, condition which is not incompatible with this case.
To sum it up the solutions are
$$(-1,4)cupBigl(frac{21}5,+inftyBigr).$$
answered Jan 26 at 22:00
BernardBernard
123k741117
$endgroup$
Here's a shorter way, with less computations:
- If $x<-1$, both sides of the inequality are negative, so that
$$frac{1}{x-4}<frac{5}{x+1}iff x-4>frac{x+1}5iff5x-20>x+1iff x>frac{21}5,$$
which is impossible since $x<-4$. - If $-1<x<4$, the l.h.s. $<0$ whereas the r.h.s. is $>0$, so the inequality is satisfied for all $xin (-1,4)$.
- If $x>0$, both sides are positive, so, as in the first case, we obtain $x>smash{dfrac{21}5}$, condition which is not incompatible with this case.
To sum it up the solutions are
$$(-1,4)cupBigl(frac{21}5,+inftyBigr).$$
answered Jan 26 at 22:00
BernardBernard
123k741117
edited Feb 18 at 9:30
answered Jan 26 at 22:00
BernardBernard
123k741117
answered Jan 26 at 22:00
BernardBernard
123k741117
answered Jan 26 at 22:00
BernardBernard
123k741117
123k741117
add a comment |
add a comment |
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$begingroup$
You have a typo in the third to last line. It should say $forall x in (-infty, frac{21}{4}) cup cdots$. Other than that you are correct.
$endgroup$
– kccu
Jan 26 at 21:30
$begingroup$
@kccu Thank you so much.
$endgroup$
– Dima
Jan 26 at 21:36