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Solve the given inequality below in the body.
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Solve the given inequality below in the body.
Rational/Quadratic InequalityProof that $left(sum^n_{k=1}x_kright)left(sum^n_{k=1}y_kright)geq n^2$Is $int_x^{infty}e^{-frac{t^2}{2}} < frac{1}{x}e^{-frac{x^2}{2}}$?Is the inequality solution below legal?Young's inequality or something similarTypical Absolute value inequalityWhich is the proper way to solve the inequality problems?Solve the inequality $sin x > ln x$
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}
$begingroup$
$$frac{1}{|x|-3} le frac 12$$
Let’s consider $|x|=y$
So $$frac{1}{y-3}-frac 12 le 0$$
$$frac{2-y+3}{y-3} le 0$$
$$frac{y-5}{y-3} ge 0$$
$$y in (-infty , 3)cup [5, infty)$$
Now this is where the problem starts. I cannot figure out on how to break the modulus function, so that’s the part where I need help in. Thanks a lot!
inequality
asked 8 hours ago
Aditya Aditya
39910 bronze badges
$endgroup$
add a comment |
$begingroup$
$$frac{1}{|x|-3} le frac 12$$
Let’s consider $|x|=y$
So $$frac{1}{y-3}-frac 12 le 0$$
$$frac{2-y+3}{y-3} le 0$$
$$frac{y-5}{y-3} ge 0$$
$$y in (-infty , 3)cup [5, infty)$$
Now this is where the problem starts. I cannot figure out on how to break the modulus function, so that’s the part where I need help in. Thanks a lot!
inequality
asked 8 hours ago
Aditya Aditya
39910 bronze badges
$endgroup$
add a comment |
$begingroup$
$$frac{1}{|x|-3} le frac 12$$
Let’s consider $|x|=y$
So $$frac{1}{y-3}-frac 12 le 0$$
$$frac{2-y+3}{y-3} le 0$$
$$frac{y-5}{y-3} ge 0$$
$$y in (-infty , 3)cup [5, infty)$$
Now this is where the problem starts. I cannot figure out on how to break the modulus function, so that’s the part where I need help in. Thanks a lot!
inequality
asked 8 hours ago
Aditya Aditya
39910 bronze badges
$endgroup$
$$frac{1}{|x|-3} le frac 12$$
Let’s consider $|x|=y$
So $$frac{1}{y-3}-frac 12 le 0$$
$$frac{2-y+3}{y-3} le 0$$
$$frac{y-5}{y-3} ge 0$$
$$y in (-infty , 3)cup [5, infty)$$
Now this is where the problem starts. I cannot figure out on how to break the modulus function, so that’s the part where I need help in. Thanks a lot!
inequality
inequality
asked 8 hours ago
Aditya Aditya
39910 bronze badges
asked 8 hours ago
Aditya Aditya
39910 bronze badges
asked 8 hours ago
Aditya Aditya
39910 bronze badges
asked 8 hours ago
Aditya Aditya
39910 bronze badges
asked 8 hours ago
Aditya Aditya
39910 bronze badges
39910 bronze badges
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Anyway, in your case, $yge 0$, so you should transform all this to
$$|x|<3;textit{ or }; |x|ge 5iff -3<x<3;textit{ or }; xge 5;textit{ or }; xle -5$$
or, as a set, the union of three intervals:
$$(-infty,-5]cup(-3,3)cup[5,+infty). $$
answered 8 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
$endgroup$
add a comment |
$begingroup$
The equivalences $$lvert Arvert< Biffbegin{cases}Bge 0\ A> -B\ A< Bend{cases}\ lvert Arvertge Biff Ble 0lor begin{cases}B> 0\ Ale -Bend{cases}lorbegin{cases}B>0\ Age Bend{cases}$$
are a good place to start from.
answered 8 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
$endgroup$
$begingroup$
In your first equivalence, the inequalities should be strict.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard Yes, thank you.
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
Even $B$ should be positive, since $|A|ge 0$.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
$endgroup$
– Bernard
8 hours ago
|
show 2 more comments
$begingroup$
For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation
$$frac1{|x|-3}=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.
So the solution set is
$$(-3,3)cup(-infty,-5]cup[5,infty).$$
answered 8 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Anyway, in your case, $yge 0$, so you should transform all this to
$$|x|<3;textit{ or }; |x|ge 5iff -3<x<3;textit{ or }; xge 5;textit{ or }; xle -5$$
or, as a set, the union of three intervals:
$$(-infty,-5]cup(-3,3)cup[5,+infty). $$
answered 8 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
$endgroup$
add a comment |
$begingroup$
Anyway, in your case, $yge 0$, so you should transform all this to
$$|x|<3;textit{ or }; |x|ge 5iff -3<x<3;textit{ or }; xge 5;textit{ or }; xle -5$$
or, as a set, the union of three intervals:
$$(-infty,-5]cup(-3,3)cup[5,+infty). $$
answered 8 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
$endgroup$
add a comment |
$begingroup$
Anyway, in your case, $yge 0$, so you should transform all this to
$$|x|<3;textit{ or }; |x|ge 5iff -3<x<3;textit{ or }; xge 5;textit{ or }; xle -5$$
or, as a set, the union of three intervals:
$$(-infty,-5]cup(-3,3)cup[5,+infty). $$
answered 8 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
$endgroup$
Anyway, in your case, $yge 0$, so you should transform all this to
$$|x|<3;textit{ or }; |x|ge 5iff -3<x<3;textit{ or }; xge 5;textit{ or }; xle -5$$
or, as a set, the union of three intervals:
$$(-infty,-5]cup(-3,3)cup[5,+infty). $$
answered 8 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
answered 8 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
answered 8 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
answered 8 hours ago
BernardBernard
132k7 gold badges43 silver badges126 bronze badges
132k7 gold badges43 silver badges126 bronze badges
add a comment |
add a comment |
$begingroup$
The equivalences $$lvert Arvert< Biffbegin{cases}Bge 0\ A> -B\ A< Bend{cases}\ lvert Arvertge Biff Ble 0lor begin{cases}B> 0\ Ale -Bend{cases}lorbegin{cases}B>0\ Age Bend{cases}$$
are a good place to start from.
answered 8 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
$endgroup$
$begingroup$
In your first equivalence, the inequalities should be strict.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard Yes, thank you.
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
Even $B$ should be positive, since $|A|ge 0$.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
$endgroup$
– Bernard
8 hours ago
|
show 2 more comments
$begingroup$
The equivalences $$lvert Arvert< Biffbegin{cases}Bge 0\ A> -B\ A< Bend{cases}\ lvert Arvertge Biff Ble 0lor begin{cases}B> 0\ Ale -Bend{cases}lorbegin{cases}B>0\ Age Bend{cases}$$
are a good place to start from.
answered 8 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
$endgroup$
$begingroup$
In your first equivalence, the inequalities should be strict.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard Yes, thank you.
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
Even $B$ should be positive, since $|A|ge 0$.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
$endgroup$
– Bernard
8 hours ago
|
show 2 more comments
$begingroup$
The equivalences $$lvert Arvert< Biffbegin{cases}Bge 0\ A> -B\ A< Bend{cases}\ lvert Arvertge Biff Ble 0lor begin{cases}B> 0\ Ale -Bend{cases}lorbegin{cases}B>0\ Age Bend{cases}$$
are a good place to start from.
answered 8 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
$endgroup$
The equivalences $$lvert Arvert< Biffbegin{cases}Bge 0\ A> -B\ A< Bend{cases}\ lvert Arvertge Biff Ble 0lor begin{cases}B> 0\ Ale -Bend{cases}lorbegin{cases}B>0\ Age Bend{cases}$$
are a good place to start from.
answered 8 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
edited 8 hours ago
answered 8 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
answered 8 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
answered 8 hours ago
Gae. S.Gae. S.
2,0648 silver badges17 bronze badges
2,0648 silver badges17 bronze badges
$begingroup$
In your first equivalence, the inequalities should be strict.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard Yes, thank you.
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
Even $B$ should be positive, since $|A|ge 0$.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
$endgroup$
– Bernard
8 hours ago
|
show 2 more comments
$begingroup$
In your first equivalence, the inequalities should be strict.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard Yes, thank you.
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
Even $B$ should be positive, since $|A|ge 0$.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
$endgroup$
– Bernard
8 hours ago
$begingroup$
In your first equivalence, the inequalities should be strict.
$endgroup$
– Bernard
8 hours ago
$begingroup$
In your first equivalence, the inequalities should be strict.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard Yes, thank you.
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
@Bernard Yes, thank you.
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
Even $B$ should be positive, since $|A|ge 0$.
$endgroup$
– Bernard
8 hours ago
$begingroup$
Even $B$ should be positive, since $|A|ge 0$.
$endgroup$
– Bernard
8 hours ago
$begingroup$
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
@Bernard For $B=0$ the other two inequations sort it out (though you may argue that so do they when $B<0$).
$endgroup$
– Gae. S.
8 hours ago
$begingroup$
There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
$endgroup$
– Bernard
8 hours ago
$begingroup$
There remains that $|A|<BRightarrow B>0$, from a strictly logical point of view.
$endgroup$
– Bernard
8 hours ago
|
show 2 more comments
$begingroup$
For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation
$$frac1{|x|-3}=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.
So the solution set is
$$(-3,3)cup(-infty,-5]cup[5,infty).$$
answered 8 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
$endgroup$
add a comment |
$begingroup$
For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation
$$frac1{|x|-3}=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.
So the solution set is
$$(-3,3)cup(-infty,-5]cup[5,infty).$$
answered 8 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
$endgroup$
add a comment |
$begingroup$
For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation
$$frac1{|x|-3}=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.
So the solution set is
$$(-3,3)cup(-infty,-5]cup[5,infty).$$
answered 8 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
$endgroup$
For $|x|<3$, the LHS is negative and the inequation certainly holds. Now assuming $|x|>3$, we can solve the equation
$$frac1{|x|-3}=frac12,$$ which gives $|x|=5$. The inequation holds for $|x|ge5$. Your resolution was right.
So the solution set is
$$(-3,3)cup(-infty,-5]cup[5,infty).$$
answered 8 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
edited 6 hours ago
answered 8 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
answered 8 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
answered 8 hours ago
Yves DaoustYves Daoust
145k10 gold badges89 silver badges248 bronze badges
145k10 gold badges89 silver badges248 bronze badges
add a comment |
add a comment |
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