How to solve this inequality , when there is a irrational power?Does this inequality hold true, in general?Prove inequality and when does equality hold?How do I solve this rational inequality?How to solve irrational inequality?Help to solve absolute value inequalityIs this inequality satisfied for large values?Inequality $frac1x>frac1y$ when $y>0$ has the same solutions as..inequality with power function
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How to solve this inequality , when there is a irrational power?
Does this inequality hold true, in general?Prove inequality and when does equality hold?How do I solve this rational inequality?How to solve irrational inequality?Help to solve absolute value inequalityIs this inequality satisfied for large values?Inequality $frac1x>frac1y$ when $y>0$ has the same solutions as..inequality with power function
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
$$frac(x-2)(x-3)^sqrt2(x+1)>0$$
I'm confused with $(x-3)^sqrt2$ factor.
When $x>3$ it is positive.
When $x=3$ this inequality does not hold.
When $x<3$, I am confused.
inequality
edited 1 hour ago
Rodrigo de Azevedo
13.5k4 gold badges22 silver badges67 bronze badges
asked 9 hours ago
Angelo MarkAngelo Mark
4,3302 gold badges17 silver badges43 bronze badges
$endgroup$
add a comment |
$begingroup$
$$frac(x-2)(x-3)^sqrt2(x+1)>0$$
I'm confused with $(x-3)^sqrt2$ factor.
When $x>3$ it is positive.
When $x=3$ this inequality does not hold.
When $x<3$, I am confused.
inequality
edited 1 hour ago
Rodrigo de Azevedo
13.5k4 gold badges22 silver badges67 bronze badges
asked 9 hours ago
Angelo MarkAngelo Mark
4,3302 gold badges17 silver badges43 bronze badges
$endgroup$
add a comment |
$begingroup$
$$frac(x-2)(x-3)^sqrt2(x+1)>0$$
I'm confused with $(x-3)^sqrt2$ factor.
When $x>3$ it is positive.
When $x=3$ this inequality does not hold.
When $x<3$, I am confused.
inequality
edited 1 hour ago
Rodrigo de Azevedo
13.5k4 gold badges22 silver badges67 bronze badges
asked 9 hours ago
Angelo MarkAngelo Mark
4,3302 gold badges17 silver badges43 bronze badges
$endgroup$
$$frac(x-2)(x-3)^sqrt2(x+1)>0$$
I'm confused with $(x-3)^sqrt2$ factor.
When $x>3$ it is positive.
When $x=3$ this inequality does not hold.
When $x<3$, I am confused.
inequality
inequality
edited 1 hour ago
Rodrigo de Azevedo
13.5k4 gold badges22 silver badges67 bronze badges
asked 9 hours ago
Angelo MarkAngelo Mark
4,3302 gold badges17 silver badges43 bronze badges
edited 1 hour ago
Rodrigo de Azevedo
13.5k4 gold badges22 silver badges67 bronze badges
asked 9 hours ago
Angelo MarkAngelo Mark
4,3302 gold badges17 silver badges43 bronze badges
edited 1 hour ago
Rodrigo de Azevedo
13.5k4 gold badges22 silver badges67 bronze badges
edited 1 hour ago
Rodrigo de Azevedo
13.5k4 gold badges22 silver badges67 bronze badges
edited 1 hour ago
Rodrigo de Azevedo
13.5k4 gold badges22 silver badges67 bronze badges
13.5k4 gold badges22 silver badges67 bronze badges
asked 9 hours ago
Angelo MarkAngelo Mark
4,3302 gold badges17 silver badges43 bronze badges
asked 9 hours ago
Angelo MarkAngelo Mark
4,3302 gold badges17 silver badges43 bronze badges
asked 9 hours ago
Angelo MarkAngelo Mark
4,3302 gold badges17 silver badges43 bronze badges
4,3302 gold badges17 silver badges43 bronze badges
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Since we have the factor$$(x-3)^sqrt2$$ it must be $$x>3$$ then it is $$(x-3)^sqrt2$$ a real number.
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
88k4 gold badges29 silver badges72 bronze badges
$endgroup$
$begingroup$
So , sir it is never a negative when $x geq 3$ ?
$endgroup$
– Angelo Mark
9 hours ago
$begingroup$
No, it must be $x>3$ since the inequality is strict.
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
$0$ is not greater than zero, so $x=3$ is not a solution.
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
So numbers like $(-2)^sqrt3$ are complex numbers ?
$endgroup$
– Angelo Mark
8 hours ago
add a comment |
$begingroup$
You don't have to worry about $x < 3$ because it is not in the domain. Irrational powers are not defined when the base is negative.
You might remember that for an irrational number $a$, by definition the exponentiation function with exponent $a$ is defined by
$$t^a = e^a ln(t)
$$
and the domain of that function is the positive axis $t > 0$.
So for your function, the domain is $x > 3$.
answered 9 hours ago
Lee MosherLee Mosher
58.8k4 gold badges39 silver badges96 bronze badges
$endgroup$
$begingroup$
$$x=1/2$$ is not defined for instance.
$endgroup$
– Dr. Sonnhard Graubner
9 hours ago
$begingroup$
Thanks, I fixed the clash of variables in my answer.
$endgroup$
– Lee Mosher
8 hours ago
add a comment |
$begingroup$
Hint: For $x<3$, $(x-3)^sqrt 2$ is not defined, so you must impose the condition $xge 3$ from the start. Aftwerwards you may proceed as you did.
answered 9 hours ago
AlexdanutAlexdanut
7581 silver badge13 bronze badges
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since we have the factor$$(x-3)^sqrt2$$ it must be $$x>3$$ then it is $$(x-3)^sqrt2$$ a real number.
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
88k4 gold badges29 silver badges72 bronze badges
$endgroup$
$begingroup$
So , sir it is never a negative when $x geq 3$ ?
$endgroup$
– Angelo Mark
9 hours ago
$begingroup$
No, it must be $x>3$ since the inequality is strict.
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
$0$ is not greater than zero, so $x=3$ is not a solution.
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
So numbers like $(-2)^sqrt3$ are complex numbers ?
$endgroup$
– Angelo Mark
8 hours ago
add a comment |
$begingroup$
Since we have the factor$$(x-3)^sqrt2$$ it must be $$x>3$$ then it is $$(x-3)^sqrt2$$ a real number.
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
88k4 gold badges29 silver badges72 bronze badges
$endgroup$
$begingroup$
So , sir it is never a negative when $x geq 3$ ?
$endgroup$
– Angelo Mark
9 hours ago
$begingroup$
No, it must be $x>3$ since the inequality is strict.
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
$0$ is not greater than zero, so $x=3$ is not a solution.
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
So numbers like $(-2)^sqrt3$ are complex numbers ?
$endgroup$
– Angelo Mark
8 hours ago
add a comment |
$begingroup$
Since we have the factor$$(x-3)^sqrt2$$ it must be $$x>3$$ then it is $$(x-3)^sqrt2$$ a real number.
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
88k4 gold badges29 silver badges72 bronze badges
$endgroup$
Since we have the factor$$(x-3)^sqrt2$$ it must be $$x>3$$ then it is $$(x-3)^sqrt2$$ a real number.
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
88k4 gold badges29 silver badges72 bronze badges
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
88k4 gold badges29 silver badges72 bronze badges
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
88k4 gold badges29 silver badges72 bronze badges
answered 9 hours ago
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
88k4 gold badges29 silver badges72 bronze badges
88k4 gold badges29 silver badges72 bronze badges
$begingroup$
So , sir it is never a negative when $x geq 3$ ?
$endgroup$
– Angelo Mark
9 hours ago
$begingroup$
No, it must be $x>3$ since the inequality is strict.
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
$0$ is not greater than zero, so $x=3$ is not a solution.
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
So numbers like $(-2)^sqrt3$ are complex numbers ?
$endgroup$
– Angelo Mark
8 hours ago
add a comment |
$begingroup$
So , sir it is never a negative when $x geq 3$ ?
$endgroup$
– Angelo Mark
9 hours ago
$begingroup$
No, it must be $x>3$ since the inequality is strict.
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
$0$ is not greater than zero, so $x=3$ is not a solution.
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
So numbers like $(-2)^sqrt3$ are complex numbers ?
$endgroup$
– Angelo Mark
8 hours ago
$begingroup$
So , sir it is never a negative when $x geq 3$ ?
$endgroup$
– Angelo Mark
9 hours ago
$begingroup$
So , sir it is never a negative when $x geq 3$ ?
$endgroup$
– Angelo Mark
9 hours ago
$begingroup$
No, it must be $x>3$ since the inequality is strict.
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
No, it must be $x>3$ since the inequality is strict.
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
$0$ is not greater than zero, so $x=3$ is not a solution.
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
$0$ is not greater than zero, so $x=3$ is not a solution.
$endgroup$
– Dr. Sonnhard Graubner
8 hours ago
$begingroup$
So numbers like $(-2)^sqrt3$ are complex numbers ?
$endgroup$
– Angelo Mark
8 hours ago
$begingroup$
So numbers like $(-2)^sqrt3$ are complex numbers ?
$endgroup$
– Angelo Mark
8 hours ago
add a comment |
$begingroup$
You don't have to worry about $x < 3$ because it is not in the domain. Irrational powers are not defined when the base is negative.
You might remember that for an irrational number $a$, by definition the exponentiation function with exponent $a$ is defined by
$$t^a = e^a ln(t)
$$
and the domain of that function is the positive axis $t > 0$.
So for your function, the domain is $x > 3$.
answered 9 hours ago
Lee MosherLee Mosher
58.8k4 gold badges39 silver badges96 bronze badges
$endgroup$
$begingroup$
$$x=1/2$$ is not defined for instance.
$endgroup$
– Dr. Sonnhard Graubner
9 hours ago
$begingroup$
Thanks, I fixed the clash of variables in my answer.
$endgroup$
– Lee Mosher
8 hours ago
add a comment |
$begingroup$
You don't have to worry about $x < 3$ because it is not in the domain. Irrational powers are not defined when the base is negative.
You might remember that for an irrational number $a$, by definition the exponentiation function with exponent $a$ is defined by
$$t^a = e^a ln(t)
$$
and the domain of that function is the positive axis $t > 0$.
So for your function, the domain is $x > 3$.
answered 9 hours ago
Lee MosherLee Mosher
58.8k4 gold badges39 silver badges96 bronze badges
$endgroup$
$begingroup$
$$x=1/2$$ is not defined for instance.
$endgroup$
– Dr. Sonnhard Graubner
9 hours ago
$begingroup$
Thanks, I fixed the clash of variables in my answer.
$endgroup$
– Lee Mosher
8 hours ago
add a comment |
$begingroup$
You don't have to worry about $x < 3$ because it is not in the domain. Irrational powers are not defined when the base is negative.
You might remember that for an irrational number $a$, by definition the exponentiation function with exponent $a$ is defined by
$$t^a = e^a ln(t)
$$
and the domain of that function is the positive axis $t > 0$.
So for your function, the domain is $x > 3$.
answered 9 hours ago
Lee MosherLee Mosher
58.8k4 gold badges39 silver badges96 bronze badges
$endgroup$
You don't have to worry about $x < 3$ because it is not in the domain. Irrational powers are not defined when the base is negative.
You might remember that for an irrational number $a$, by definition the exponentiation function with exponent $a$ is defined by
$$t^a = e^a ln(t)
$$
and the domain of that function is the positive axis $t > 0$.
So for your function, the domain is $x > 3$.
answered 9 hours ago
Lee MosherLee Mosher
58.8k4 gold badges39 silver badges96 bronze badges
edited 8 hours ago
answered 9 hours ago
Lee MosherLee Mosher
58.8k4 gold badges39 silver badges96 bronze badges
answered 9 hours ago
Lee MosherLee Mosher
58.8k4 gold badges39 silver badges96 bronze badges
answered 9 hours ago
Lee MosherLee Mosher
58.8k4 gold badges39 silver badges96 bronze badges
58.8k4 gold badges39 silver badges96 bronze badges
$begingroup$
$$x=1/2$$ is not defined for instance.
$endgroup$
– Dr. Sonnhard Graubner
9 hours ago
$begingroup$
Thanks, I fixed the clash of variables in my answer.
$endgroup$
– Lee Mosher
8 hours ago
add a comment |
$begingroup$
$$x=1/2$$ is not defined for instance.
$endgroup$
– Dr. Sonnhard Graubner
9 hours ago
$begingroup$
Thanks, I fixed the clash of variables in my answer.
$endgroup$
– Lee Mosher
8 hours ago
$begingroup$
$$x=1/2$$ is not defined for instance.
$endgroup$
– Dr. Sonnhard Graubner
9 hours ago
$begingroup$
$$x=1/2$$ is not defined for instance.
$endgroup$
– Dr. Sonnhard Graubner
9 hours ago
$begingroup$
Thanks, I fixed the clash of variables in my answer.
$endgroup$
– Lee Mosher
8 hours ago
$begingroup$
Thanks, I fixed the clash of variables in my answer.
$endgroup$
– Lee Mosher
8 hours ago
add a comment |
$begingroup$
Hint: For $x<3$, $(x-3)^sqrt 2$ is not defined, so you must impose the condition $xge 3$ from the start. Aftwerwards you may proceed as you did.
answered 9 hours ago
AlexdanutAlexdanut
7581 silver badge13 bronze badges
$endgroup$
add a comment |
$begingroup$
Hint: For $x<3$, $(x-3)^sqrt 2$ is not defined, so you must impose the condition $xge 3$ from the start. Aftwerwards you may proceed as you did.
answered 9 hours ago
AlexdanutAlexdanut
7581 silver badge13 bronze badges
$endgroup$
add a comment |
$begingroup$
Hint: For $x<3$, $(x-3)^sqrt 2$ is not defined, so you must impose the condition $xge 3$ from the start. Aftwerwards you may proceed as you did.
answered 9 hours ago
AlexdanutAlexdanut
7581 silver badge13 bronze badges
$endgroup$
Hint: For $x<3$, $(x-3)^sqrt 2$ is not defined, so you must impose the condition $xge 3$ from the start. Aftwerwards you may proceed as you did.
answered 9 hours ago
AlexdanutAlexdanut
7581 silver badge13 bronze badges
answered 9 hours ago
AlexdanutAlexdanut
7581 silver badge13 bronze badges
answered 9 hours ago
AlexdanutAlexdanut
7581 silver badge13 bronze badges
answered 9 hours ago
AlexdanutAlexdanut
7581 silver badge13 bronze badges
7581 silver badge13 bronze badges
add a comment |
add a comment |
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