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Show that , if $a ge 9/8$, then $f '(x) ge 0$
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My questions are to do with the solution of the following question:
Let $f(x)=ax-fracx^31+x^2$ , where $a$ is a positive constant.
Show that if $agefrac98$ then $f'(x)ge0$ for all $x$.
Differentiating we get :
$$f'(x)=fraca+(2a-3)x^2+(a-1)x^4(1+x^2)^2$$
Next the following is said/done:
"We can use the inequality $agefrac98$ immediately after finding f'(x) since $a$ appears in f'(x) with a plus sign always:
$$f'(x)=fraca+(2a-3)x^2+(a-1)x^4(1+x^2)^2gefracfrac98+(frac94-3)x^2+(frac98-1)x^4(1+x^2)^2=frac(x^2-3)^28(1+x^2)^2ge0$$
My questions pertain to everything in bold and after the bold statement:
- How is the author using the inequality $agefrac98$ to get to the final result?
- What is the significance of $a$ appearing with a positive sign always?
derivatives inequality
asked Jul 23 at 19:23
stochasticmrfox
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My questions are to do with the solution of the following question:
Let $f(x)=ax-fracx^31+x^2$ , where $a$ is a positive constant.
Show that if $agefrac98$ then $f'(x)ge0$ for all $x$.
Differentiating we get :
$$f'(x)=fraca+(2a-3)x^2+(a-1)x^4(1+x^2)^2$$
Next the following is said/done:
"We can use the inequality $agefrac98$ immediately after finding f'(x) since $a$ appears in f'(x) with a plus sign always:
$$f'(x)=fraca+(2a-3)x^2+(a-1)x^4(1+x^2)^2gefracfrac98+(frac94-3)x^2+(frac98-1)x^4(1+x^2)^2=frac(x^2-3)^28(1+x^2)^2ge0$$
My questions pertain to everything in bold and after the bold statement:
- How is the author using the inequality $agefrac98$ to get to the final result?
- What is the significance of $a$ appearing with a positive sign always?
derivatives inequality
asked Jul 23 at 19:23
stochasticmrfox
216
I think you left out an $x^2$ in the numerator of the next-to-last fraction
– saulspatz
Jul 23 at 19:30
This problem showed $f'(x)geq 0 $ and $f(0)=a>0$, therefore we can say $f(x)>0$ if $ageq frac98$.
– Takahiro Waki
Jul 31 at 14:49
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My questions are to do with the solution of the following question:
Let $f(x)=ax-fracx^31+x^2$ , where $a$ is a positive constant.
Show that if $agefrac98$ then $f'(x)ge0$ for all $x$.
Differentiating we get :
$$f'(x)=fraca+(2a-3)x^2+(a-1)x^4(1+x^2)^2$$
Next the following is said/done:
"We can use the inequality $agefrac98$ immediately after finding f'(x) since $a$ appears in f'(x) with a plus sign always:
$$f'(x)=fraca+(2a-3)x^2+(a-1)x^4(1+x^2)^2gefracfrac98+(frac94-3)x^2+(frac98-1)x^4(1+x^2)^2=frac(x^2-3)^28(1+x^2)^2ge0$$
My questions pertain to everything in bold and after the bold statement:
- How is the author using the inequality $agefrac98$ to get to the final result?
- What is the significance of $a$ appearing with a positive sign always?
derivatives inequality
asked Jul 23 at 19:23
stochasticmrfox
216
My questions are to do with the solution of the following question:
Let $f(x)=ax-fracx^31+x^2$ , where $a$ is a positive constant.
Show that if $agefrac98$ then $f'(x)ge0$ for all $x$.
Differentiating we get :
$$f'(x)=fraca+(2a-3)x^2+(a-1)x^4(1+x^2)^2$$
Next the following is said/done:
"We can use the inequality $agefrac98$ immediately after finding f'(x) since $a$ appears in f'(x) with a plus sign always:
$$f'(x)=fraca+(2a-3)x^2+(a-1)x^4(1+x^2)^2gefracfrac98+(frac94-3)x^2+(frac98-1)x^4(1+x^2)^2=frac(x^2-3)^28(1+x^2)^2ge0$$
My questions pertain to everything in bold and after the bold statement:
- How is the author using the inequality $agefrac98$ to get to the final result?
- What is the significance of $a$ appearing with a positive sign always?
derivatives inequality
asked Jul 23 at 19:23
stochasticmrfox
216
edited Jul 23 at 19:32
asked Jul 23 at 19:23
stochasticmrfox
216
asked Jul 23 at 19:23
stochasticmrfox
216
asked Jul 23 at 19:23
stochasticmrfox
216
216
I think you left out an $x^2$ in the numerator of the next-to-last fraction
– saulspatz
Jul 23 at 19:30
This problem showed $f'(x)geq 0 $ and $f(0)=a>0$, therefore we can say $f(x)>0$ if $ageq frac98$.
– Takahiro Waki
Jul 31 at 14:49
add a comment |Â
I think you left out an $x^2$ in the numerator of the next-to-last fraction
– saulspatz
Jul 23 at 19:30
This problem showed $f'(x)geq 0 $ and $f(0)=a>0$, therefore we can say $f(x)>0$ if $ageq frac98$.
– Takahiro Waki
Jul 31 at 14:49
I think you left out an $x^2$ in the numerator of the next-to-last fraction
– saulspatz
Jul 23 at 19:30
I think you left out an $x^2$ in the numerator of the next-to-last fraction
– saulspatz
Jul 23 at 19:30
This problem showed $f'(x)geq 0 $ and $f(0)=a>0$, therefore we can say $f(x)>0$ if $ageq frac98$.
– Takahiro Waki
Jul 31 at 14:49
This problem showed $f'(x)geq 0 $ and $f(0)=a>0$, therefore we can say $f(x)>0$ if $ageq frac98$.
– Takahiro Waki
Jul 31 at 14:49
add a comment |Â
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First the sign of $f'(x)$ is the sign of its numerator, which is a biquadratic polynomial, so setting $u=x^2$, we obtain a quadratic polynomial:
$$p(u)=(a-1)u^2+(2a-3)u+a.$$
Supposing $ane 1$, so we have a quadratic polynomial, it has a constant sign (or is $0$) if and only if is discriminant is non-positive:
$$Delta=(2a-3)^2-4a(a-1)=9-8ale 0.$$
So the condition is indeed $age frac98$.
Last, since the sign of $p(u)$ is constant, it is also the sign of $p(0)=a$.
answered Jul 23 at 19:36
Bernard
110k635103
You miss various exceptions. $a=1$ and $Δ>0$ is also ok in some case.
– Takahiro Waki
Jul 31 at 13:46
@TakahiroWaki: Yes, I supposed implicitly we had a quadratic equation. I've now added this detail. On the other hand, note I didn't try to find all possible solutions, but only to answer the question and explain why $a>9/8$ is a sufficient condition to have $f'(x)ge 0$.
– Bernard
Jul 31 at 14:07
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First, he's saying that $$agefrac98implies aygefrac9y8$$ for various values of $y$. This is only true when $yge0.$ I'm sure you know that multiplying by a negative number reverses the sense of an inequality.
As for how he's using it, he's just pulling a factor of $frac18$ out of the numerator of the next-to-last fraction, then factoring it.
answered Jul 23 at 19:37
saulspatz
10.5k21323
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1) If $a ge k$ then $a$ times anything positive plus or minus anything else will be greater than $k$ times the same thing plus or minus the thing.
So:
$age frac 98$ means $(2a - 3)x^2 ge (2*frac 98 - 3)x^2$ and $(a-1)x^4 ge (frac 98 - 1)x^4$ etc.
2) If $a ge frac 98$ and if $k > 0$ then $ak > frac 98k$. But if $k < 0$ then $ak < frac 98 k$.
Basically, if $a ge frac 98$ we can remplace $a$ with $frac 98$ and get something less than or equal to the expression with $a$ presuming we never multiply $a$ by anything negative. Which as $x^2, x^4, 2, 1+x^2$ are all non-negative, is a safe assumption.
....
Basically there are two basic axioms of inequalities.
I) If $a ge k$ then $a + c ge k + c$ for all possible (including negative) $c$. and
II) If $a ge k$ and $b > 0$ then $ab ge ak$.
Given those two axioms than
$a ge frac 98 implies 2a-3 ge 2frac 98 - 3; a-1 ge frac 98 - 1implies$
$age 98;(2a-3)x^2> (2frac 98 - 3)x^2; (a-1)x^4 ge (frac 98 - 1)x^4implies$
$a + (2a -3)x^2 + (2a-1)x^4ge frac98+(frac98-3)x^2+(frac98-1)x^4implies$
$fraca+(2a-3)x^2+(a-1)x^4(1+x^2)^2gefracfrac98+(frac94-3)x^2+(frac98-1)x^4(1+x^2)^2$
Okay, we also need the definition
0) $a le b; ble c implies a le c$
To get $a le frac 98; (2a-3)x^2 le (2frac 98 -3)x^2; (a-1)x^4 le (frac 98 -1)x^4implies$
$a + (2a-3)x^2 + (a-1)x^4 ge a + (2a-3)x^2 + (frac 98 - 1)x^4;$
$a + (2a-3)x^2 + (frac 98 - 1)x^4ge a + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4;$
$a + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4ge frac 98 + (2frac 98 - 3)x^2 + (frac 9/8 - 1)x^4;implies$
$a + (2a-3)x^2 + (a-1)x^4gefrac 98 + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4$
answered Jul 23 at 20:09
fleablood
60.3k22575
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This is because since the denominator is always non-negative so must be the numerator. Also the numerator has a quadratic form whose coefficient of the term with highest power must be positive i.e. $$a-1>0\a>1$$ also the $Delta$ must be negative therefore $$(2a-3)^2<4a(a-1)\4a^2-12a+9<4a^2-4a\$$$$Large a>dfrac98$$
answered Jul 23 at 20:36
Mostafa Ayaz
8,5823630
Your answer doesn't tell all cases, too.
– Takahiro Waki
Jul 31 at 13:47
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We must show $(a-1)x^4+(2a-3)x^2+ageq 0$ ,
and can easily assume $ageq 1$.
1: exception case If $a=1$, $f'(x)=-x^2+1$
This eventually become negative.
2: If axis$leq0⇔ -frac2a-3a-1leq0⇔ageqfrac32$ or $a<1 $, this need only $f(0)=ageq0$
3: If axis $>0⇔ 1<a<frac32$, this need $Dleq0$
$D=(2a-3)^2-4a(a-1)=-8a+9leq0$
We get $ageqdfrac98$ by these 3 cases.
answered Jul 31 at 13:41
Takahiro Waki
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5 Answers
5
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5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
First the sign of $f'(x)$ is the sign of its numerator, which is a biquadratic polynomial, so setting $u=x^2$, we obtain a quadratic polynomial:
$$p(u)=(a-1)u^2+(2a-3)u+a.$$
Supposing $ane 1$, so we have a quadratic polynomial, it has a constant sign (or is $0$) if and only if is discriminant is non-positive:
$$Delta=(2a-3)^2-4a(a-1)=9-8ale 0.$$
So the condition is indeed $age frac98$.
Last, since the sign of $p(u)$ is constant, it is also the sign of $p(0)=a$.
answered Jul 23 at 19:36
Bernard
110k635103
You miss various exceptions. $a=1$ and $Δ>0$ is also ok in some case.
– Takahiro Waki
Jul 31 at 13:46
@TakahiroWaki: Yes, I supposed implicitly we had a quadratic equation. I've now added this detail. On the other hand, note I didn't try to find all possible solutions, but only to answer the question and explain why $a>9/8$ is a sufficient condition to have $f'(x)ge 0$.
– Bernard
Jul 31 at 14:07
add a comment |Â
up vote
1
down vote
First the sign of $f'(x)$ is the sign of its numerator, which is a biquadratic polynomial, so setting $u=x^2$, we obtain a quadratic polynomial:
$$p(u)=(a-1)u^2+(2a-3)u+a.$$
Supposing $ane 1$, so we have a quadratic polynomial, it has a constant sign (or is $0$) if and only if is discriminant is non-positive:
$$Delta=(2a-3)^2-4a(a-1)=9-8ale 0.$$
So the condition is indeed $age frac98$.
Last, since the sign of $p(u)$ is constant, it is also the sign of $p(0)=a$.
answered Jul 23 at 19:36
Bernard
110k635103
You miss various exceptions. $a=1$ and $Δ>0$ is also ok in some case.
– Takahiro Waki
Jul 31 at 13:46
@TakahiroWaki: Yes, I supposed implicitly we had a quadratic equation. I've now added this detail. On the other hand, note I didn't try to find all possible solutions, but only to answer the question and explain why $a>9/8$ is a sufficient condition to have $f'(x)ge 0$.
– Bernard
Jul 31 at 14:07
add a comment |Â
up vote
1
down vote
up vote
1
down vote
First the sign of $f'(x)$ is the sign of its numerator, which is a biquadratic polynomial, so setting $u=x^2$, we obtain a quadratic polynomial:
$$p(u)=(a-1)u^2+(2a-3)u+a.$$
Supposing $ane 1$, so we have a quadratic polynomial, it has a constant sign (or is $0$) if and only if is discriminant is non-positive:
$$Delta=(2a-3)^2-4a(a-1)=9-8ale 0.$$
So the condition is indeed $age frac98$.
Last, since the sign of $p(u)$ is constant, it is also the sign of $p(0)=a$.
answered Jul 23 at 19:36
Bernard
110k635103
First the sign of $f'(x)$ is the sign of its numerator, which is a biquadratic polynomial, so setting $u=x^2$, we obtain a quadratic polynomial:
$$p(u)=(a-1)u^2+(2a-3)u+a.$$
Supposing $ane 1$, so we have a quadratic polynomial, it has a constant sign (or is $0$) if and only if is discriminant is non-positive:
$$Delta=(2a-3)^2-4a(a-1)=9-8ale 0.$$
So the condition is indeed $age frac98$.
Last, since the sign of $p(u)$ is constant, it is also the sign of $p(0)=a$.
answered Jul 23 at 19:36
Bernard
110k635103
edited Jul 31 at 13:55
answered Jul 23 at 19:36
Bernard
110k635103
answered Jul 23 at 19:36
Bernard
110k635103
answered Jul 23 at 19:36
Bernard
110k635103
110k635103
You miss various exceptions. $a=1$ and $Δ>0$ is also ok in some case.
– Takahiro Waki
Jul 31 at 13:46
@TakahiroWaki: Yes, I supposed implicitly we had a quadratic equation. I've now added this detail. On the other hand, note I didn't try to find all possible solutions, but only to answer the question and explain why $a>9/8$ is a sufficient condition to have $f'(x)ge 0$.
– Bernard
Jul 31 at 14:07
add a comment |Â
You miss various exceptions. $a=1$ and $Δ>0$ is also ok in some case.
– Takahiro Waki
Jul 31 at 13:46
@TakahiroWaki: Yes, I supposed implicitly we had a quadratic equation. I've now added this detail. On the other hand, note I didn't try to find all possible solutions, but only to answer the question and explain why $a>9/8$ is a sufficient condition to have $f'(x)ge 0$.
– Bernard
Jul 31 at 14:07
You miss various exceptions. $a=1$ and $Δ>0$ is also ok in some case.
– Takahiro Waki
Jul 31 at 13:46
You miss various exceptions. $a=1$ and $Δ>0$ is also ok in some case.
– Takahiro Waki
Jul 31 at 13:46
@TakahiroWaki: Yes, I supposed implicitly we had a quadratic equation. I've now added this detail. On the other hand, note I didn't try to find all possible solutions, but only to answer the question and explain why $a>9/8$ is a sufficient condition to have $f'(x)ge 0$.
– Bernard
Jul 31 at 14:07
@TakahiroWaki: Yes, I supposed implicitly we had a quadratic equation. I've now added this detail. On the other hand, note I didn't try to find all possible solutions, but only to answer the question and explain why $a>9/8$ is a sufficient condition to have $f'(x)ge 0$.
– Bernard
Jul 31 at 14:07
add a comment |Â
up vote
0
down vote
First, he's saying that $$agefrac98implies aygefrac9y8$$ for various values of $y$. This is only true when $yge0.$ I'm sure you know that multiplying by a negative number reverses the sense of an inequality.
As for how he's using it, he's just pulling a factor of $frac18$ out of the numerator of the next-to-last fraction, then factoring it.
answered Jul 23 at 19:37
saulspatz
10.5k21323
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up vote
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down vote
First, he's saying that $$agefrac98implies aygefrac9y8$$ for various values of $y$. This is only true when $yge0.$ I'm sure you know that multiplying by a negative number reverses the sense of an inequality.
As for how he's using it, he's just pulling a factor of $frac18$ out of the numerator of the next-to-last fraction, then factoring it.
answered Jul 23 at 19:37
saulspatz
10.5k21323
add a comment |Â
up vote
0
down vote
up vote
0
down vote
First, he's saying that $$agefrac98implies aygefrac9y8$$ for various values of $y$. This is only true when $yge0.$ I'm sure you know that multiplying by a negative number reverses the sense of an inequality.
As for how he's using it, he's just pulling a factor of $frac18$ out of the numerator of the next-to-last fraction, then factoring it.
answered Jul 23 at 19:37
saulspatz
10.5k21323
First, he's saying that $$agefrac98implies aygefrac9y8$$ for various values of $y$. This is only true when $yge0.$ I'm sure you know that multiplying by a negative number reverses the sense of an inequality.
As for how he's using it, he's just pulling a factor of $frac18$ out of the numerator of the next-to-last fraction, then factoring it.
answered Jul 23 at 19:37
saulspatz
10.5k21323
answered Jul 23 at 19:37
saulspatz
10.5k21323
answered Jul 23 at 19:37
saulspatz
10.5k21323
answered Jul 23 at 19:37
saulspatz
10.5k21323
10.5k21323
add a comment |Â
add a comment |Â
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0
down vote
1) If $a ge k$ then $a$ times anything positive plus or minus anything else will be greater than $k$ times the same thing plus or minus the thing.
So:
$age frac 98$ means $(2a - 3)x^2 ge (2*frac 98 - 3)x^2$ and $(a-1)x^4 ge (frac 98 - 1)x^4$ etc.
2) If $a ge frac 98$ and if $k > 0$ then $ak > frac 98k$. But if $k < 0$ then $ak < frac 98 k$.
Basically, if $a ge frac 98$ we can remplace $a$ with $frac 98$ and get something less than or equal to the expression with $a$ presuming we never multiply $a$ by anything negative. Which as $x^2, x^4, 2, 1+x^2$ are all non-negative, is a safe assumption.
....
Basically there are two basic axioms of inequalities.
I) If $a ge k$ then $a + c ge k + c$ for all possible (including negative) $c$. and
II) If $a ge k$ and $b > 0$ then $ab ge ak$.
Given those two axioms than
$a ge frac 98 implies 2a-3 ge 2frac 98 - 3; a-1 ge frac 98 - 1implies$
$age 98;(2a-3)x^2> (2frac 98 - 3)x^2; (a-1)x^4 ge (frac 98 - 1)x^4implies$
$a + (2a -3)x^2 + (2a-1)x^4ge frac98+(frac98-3)x^2+(frac98-1)x^4implies$
$fraca+(2a-3)x^2+(a-1)x^4(1+x^2)^2gefracfrac98+(frac94-3)x^2+(frac98-1)x^4(1+x^2)^2$
Okay, we also need the definition
0) $a le b; ble c implies a le c$
To get $a le frac 98; (2a-3)x^2 le (2frac 98 -3)x^2; (a-1)x^4 le (frac 98 -1)x^4implies$
$a + (2a-3)x^2 + (a-1)x^4 ge a + (2a-3)x^2 + (frac 98 - 1)x^4;$
$a + (2a-3)x^2 + (frac 98 - 1)x^4ge a + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4;$
$a + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4ge frac 98 + (2frac 98 - 3)x^2 + (frac 9/8 - 1)x^4;implies$
$a + (2a-3)x^2 + (a-1)x^4gefrac 98 + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4$
answered Jul 23 at 20:09
fleablood
60.3k22575
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up vote
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1) If $a ge k$ then $a$ times anything positive plus or minus anything else will be greater than $k$ times the same thing plus or minus the thing.
So:
$age frac 98$ means $(2a - 3)x^2 ge (2*frac 98 - 3)x^2$ and $(a-1)x^4 ge (frac 98 - 1)x^4$ etc.
2) If $a ge frac 98$ and if $k > 0$ then $ak > frac 98k$. But if $k < 0$ then $ak < frac 98 k$.
Basically, if $a ge frac 98$ we can remplace $a$ with $frac 98$ and get something less than or equal to the expression with $a$ presuming we never multiply $a$ by anything negative. Which as $x^2, x^4, 2, 1+x^2$ are all non-negative, is a safe assumption.
....
Basically there are two basic axioms of inequalities.
I) If $a ge k$ then $a + c ge k + c$ for all possible (including negative) $c$. and
II) If $a ge k$ and $b > 0$ then $ab ge ak$.
Given those two axioms than
$a ge frac 98 implies 2a-3 ge 2frac 98 - 3; a-1 ge frac 98 - 1implies$
$age 98;(2a-3)x^2> (2frac 98 - 3)x^2; (a-1)x^4 ge (frac 98 - 1)x^4implies$
$a + (2a -3)x^2 + (2a-1)x^4ge frac98+(frac98-3)x^2+(frac98-1)x^4implies$
$fraca+(2a-3)x^2+(a-1)x^4(1+x^2)^2gefracfrac98+(frac94-3)x^2+(frac98-1)x^4(1+x^2)^2$
Okay, we also need the definition
0) $a le b; ble c implies a le c$
To get $a le frac 98; (2a-3)x^2 le (2frac 98 -3)x^2; (a-1)x^4 le (frac 98 -1)x^4implies$
$a + (2a-3)x^2 + (a-1)x^4 ge a + (2a-3)x^2 + (frac 98 - 1)x^4;$
$a + (2a-3)x^2 + (frac 98 - 1)x^4ge a + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4;$
$a + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4ge frac 98 + (2frac 98 - 3)x^2 + (frac 9/8 - 1)x^4;implies$
$a + (2a-3)x^2 + (a-1)x^4gefrac 98 + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4$
answered Jul 23 at 20:09
fleablood
60.3k22575
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up vote
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up vote
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1) If $a ge k$ then $a$ times anything positive plus or minus anything else will be greater than $k$ times the same thing plus or minus the thing.
So:
$age frac 98$ means $(2a - 3)x^2 ge (2*frac 98 - 3)x^2$ and $(a-1)x^4 ge (frac 98 - 1)x^4$ etc.
2) If $a ge frac 98$ and if $k > 0$ then $ak > frac 98k$. But if $k < 0$ then $ak < frac 98 k$.
Basically, if $a ge frac 98$ we can remplace $a$ with $frac 98$ and get something less than or equal to the expression with $a$ presuming we never multiply $a$ by anything negative. Which as $x^2, x^4, 2, 1+x^2$ are all non-negative, is a safe assumption.
....
Basically there are two basic axioms of inequalities.
I) If $a ge k$ then $a + c ge k + c$ for all possible (including negative) $c$. and
II) If $a ge k$ and $b > 0$ then $ab ge ak$.
Given those two axioms than
$a ge frac 98 implies 2a-3 ge 2frac 98 - 3; a-1 ge frac 98 - 1implies$
$age 98;(2a-3)x^2> (2frac 98 - 3)x^2; (a-1)x^4 ge (frac 98 - 1)x^4implies$
$a + (2a -3)x^2 + (2a-1)x^4ge frac98+(frac98-3)x^2+(frac98-1)x^4implies$
$fraca+(2a-3)x^2+(a-1)x^4(1+x^2)^2gefracfrac98+(frac94-3)x^2+(frac98-1)x^4(1+x^2)^2$
Okay, we also need the definition
0) $a le b; ble c implies a le c$
To get $a le frac 98; (2a-3)x^2 le (2frac 98 -3)x^2; (a-1)x^4 le (frac 98 -1)x^4implies$
$a + (2a-3)x^2 + (a-1)x^4 ge a + (2a-3)x^2 + (frac 98 - 1)x^4;$
$a + (2a-3)x^2 + (frac 98 - 1)x^4ge a + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4;$
$a + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4ge frac 98 + (2frac 98 - 3)x^2 + (frac 9/8 - 1)x^4;implies$
$a + (2a-3)x^2 + (a-1)x^4gefrac 98 + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4$
answered Jul 23 at 20:09
fleablood
60.3k22575
1) If $a ge k$ then $a$ times anything positive plus or minus anything else will be greater than $k$ times the same thing plus or minus the thing.
So:
$age frac 98$ means $(2a - 3)x^2 ge (2*frac 98 - 3)x^2$ and $(a-1)x^4 ge (frac 98 - 1)x^4$ etc.
2) If $a ge frac 98$ and if $k > 0$ then $ak > frac 98k$. But if $k < 0$ then $ak < frac 98 k$.
Basically, if $a ge frac 98$ we can remplace $a$ with $frac 98$ and get something less than or equal to the expression with $a$ presuming we never multiply $a$ by anything negative. Which as $x^2, x^4, 2, 1+x^2$ are all non-negative, is a safe assumption.
....
Basically there are two basic axioms of inequalities.
I) If $a ge k$ then $a + c ge k + c$ for all possible (including negative) $c$. and
II) If $a ge k$ and $b > 0$ then $ab ge ak$.
Given those two axioms than
$a ge frac 98 implies 2a-3 ge 2frac 98 - 3; a-1 ge frac 98 - 1implies$
$age 98;(2a-3)x^2> (2frac 98 - 3)x^2; (a-1)x^4 ge (frac 98 - 1)x^4implies$
$a + (2a -3)x^2 + (2a-1)x^4ge frac98+(frac98-3)x^2+(frac98-1)x^4implies$
$fraca+(2a-3)x^2+(a-1)x^4(1+x^2)^2gefracfrac98+(frac94-3)x^2+(frac98-1)x^4(1+x^2)^2$
Okay, we also need the definition
0) $a le b; ble c implies a le c$
To get $a le frac 98; (2a-3)x^2 le (2frac 98 -3)x^2; (a-1)x^4 le (frac 98 -1)x^4implies$
$a + (2a-3)x^2 + (a-1)x^4 ge a + (2a-3)x^2 + (frac 98 - 1)x^4;$
$a + (2a-3)x^2 + (frac 98 - 1)x^4ge a + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4;$
$a + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4ge frac 98 + (2frac 98 - 3)x^2 + (frac 9/8 - 1)x^4;implies$
$a + (2a-3)x^2 + (a-1)x^4gefrac 98 + (2frac 98 - 3)x^2 + (frac 98 - 1)x^4$
answered Jul 23 at 20:09
fleablood
60.3k22575
edited Jul 23 at 20:16
answered Jul 23 at 20:09
fleablood
60.3k22575
answered Jul 23 at 20:09
fleablood
60.3k22575
answered Jul 23 at 20:09
fleablood
60.3k22575
60.3k22575
add a comment |Â
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up vote
0
down vote
This is because since the denominator is always non-negative so must be the numerator. Also the numerator has a quadratic form whose coefficient of the term with highest power must be positive i.e. $$a-1>0\a>1$$ also the $Delta$ must be negative therefore $$(2a-3)^2<4a(a-1)\4a^2-12a+9<4a^2-4a\$$$$Large a>dfrac98$$
answered Jul 23 at 20:36
Mostafa Ayaz
8,5823630
Your answer doesn't tell all cases, too.
– Takahiro Waki
Jul 31 at 13:47
add a comment |Â
up vote
0
down vote
This is because since the denominator is always non-negative so must be the numerator. Also the numerator has a quadratic form whose coefficient of the term with highest power must be positive i.e. $$a-1>0\a>1$$ also the $Delta$ must be negative therefore $$(2a-3)^2<4a(a-1)\4a^2-12a+9<4a^2-4a\$$$$Large a>dfrac98$$
answered Jul 23 at 20:36
Mostafa Ayaz
8,5823630
Your answer doesn't tell all cases, too.
– Takahiro Waki
Jul 31 at 13:47
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is because since the denominator is always non-negative so must be the numerator. Also the numerator has a quadratic form whose coefficient of the term with highest power must be positive i.e. $$a-1>0\a>1$$ also the $Delta$ must be negative therefore $$(2a-3)^2<4a(a-1)\4a^2-12a+9<4a^2-4a\$$$$Large a>dfrac98$$
answered Jul 23 at 20:36
Mostafa Ayaz
8,5823630
This is because since the denominator is always non-negative so must be the numerator. Also the numerator has a quadratic form whose coefficient of the term with highest power must be positive i.e. $$a-1>0\a>1$$ also the $Delta$ must be negative therefore $$(2a-3)^2<4a(a-1)\4a^2-12a+9<4a^2-4a\$$$$Large a>dfrac98$$
answered Jul 23 at 20:36
Mostafa Ayaz
8,5823630
answered Jul 23 at 20:36
Mostafa Ayaz
8,5823630
answered Jul 23 at 20:36
Mostafa Ayaz
8,5823630
answered Jul 23 at 20:36
Mostafa Ayaz
8,5823630
8,5823630
Your answer doesn't tell all cases, too.
– Takahiro Waki
Jul 31 at 13:47
add a comment |Â
Your answer doesn't tell all cases, too.
– Takahiro Waki
Jul 31 at 13:47
Your answer doesn't tell all cases, too.
– Takahiro Waki
Jul 31 at 13:47
Your answer doesn't tell all cases, too.
– Takahiro Waki
Jul 31 at 13:47
add a comment |Â
up vote
0
down vote
We must show $(a-1)x^4+(2a-3)x^2+ageq 0$ ,
and can easily assume $ageq 1$.
1: exception case If $a=1$, $f'(x)=-x^2+1$
This eventually become negative.
2: If axis$leq0⇔ -frac2a-3a-1leq0⇔ageqfrac32$ or $a<1 $, this need only $f(0)=ageq0$
3: If axis $>0⇔ 1<a<frac32$, this need $Dleq0$
$D=(2a-3)^2-4a(a-1)=-8a+9leq0$
We get $ageqdfrac98$ by these 3 cases.
answered Jul 31 at 13:41
Takahiro Waki
1,984520
add a comment |Â
up vote
0
down vote
We must show $(a-1)x^4+(2a-3)x^2+ageq 0$ ,
and can easily assume $ageq 1$.
1: exception case If $a=1$, $f'(x)=-x^2+1$
This eventually become negative.
2: If axis$leq0⇔ -frac2a-3a-1leq0⇔ageqfrac32$ or $a<1 $, this need only $f(0)=ageq0$
3: If axis $>0⇔ 1<a<frac32$, this need $Dleq0$
$D=(2a-3)^2-4a(a-1)=-8a+9leq0$
We get $ageqdfrac98$ by these 3 cases.
answered Jul 31 at 13:41
Takahiro Waki
1,984520
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We must show $(a-1)x^4+(2a-3)x^2+ageq 0$ ,
and can easily assume $ageq 1$.
1: exception case If $a=1$, $f'(x)=-x^2+1$
This eventually become negative.
2: If axis$leq0⇔ -frac2a-3a-1leq0⇔ageqfrac32$ or $a<1 $, this need only $f(0)=ageq0$
3: If axis $>0⇔ 1<a<frac32$, this need $Dleq0$
$D=(2a-3)^2-4a(a-1)=-8a+9leq0$
We get $ageqdfrac98$ by these 3 cases.
answered Jul 31 at 13:41
Takahiro Waki
1,984520
We must show $(a-1)x^4+(2a-3)x^2+ageq 0$ ,
and can easily assume $ageq 1$.
1: exception case If $a=1$, $f'(x)=-x^2+1$
This eventually become negative.
2: If axis$leq0⇔ -frac2a-3a-1leq0⇔ageqfrac32$ or $a<1 $, this need only $f(0)=ageq0$
3: If axis $>0⇔ 1<a<frac32$, this need $Dleq0$
$D=(2a-3)^2-4a(a-1)=-8a+9leq0$
We get $ageqdfrac98$ by these 3 cases.
answered Jul 31 at 13:41
Takahiro Waki
1,984520
answered Jul 31 at 13:41
Takahiro Waki
1,984520
answered Jul 31 at 13:41
Takahiro Waki
1,984520
answered Jul 31 at 13:41
Takahiro Waki
1,984520
1,984520
add a comment |Â
add a comment |Â
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I think you left out an $x^2$ in the numerator of the next-to-last fraction
– saulspatz
Jul 23 at 19:30
This problem showed $f'(x)geq 0 $ and $f(0)=a>0$, therefore we can say $f(x)>0$ if $ageq frac98$.
– Takahiro Waki
Jul 31 at 14:49