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Find the interval of $a$ that makes $x^2e^ax-1gt ln x$ true.

1

$begingroup$

$$x^2e^ax-1gt ln x$$

Then $e^axgt fracln x+1x^2$

Let $f(x)=fracln x+1x^2$, then $f'(x)=frac-2ln x-1x^3$, and $f''(x)=frac6ln x+1x^4$

$-2ln x-1=0Rightarrow x=e^-frac12$

$f''(e^-frac12)=-2e^2 lt0$

Therefore it is the maximum point for $f(x)$ when $x=e^-frac12$.

Then $e^ae^frac12gt f(e^-frac12)=frace2$

Therefore $ae^-frac12gt1-ln2 Rightarrow agtsqrt e(1-ln2)$

But the answer is $agtfrace3$

How do they get $agt frace3$?

calculus inequality

share|cite|improve this question

edited Apr 8 at 8:23

yuanming luo

asked Apr 5 at 16:42

yuanming luoyuanming luo

9711

$endgroup$

This question has an open bounty worth +50
reputation from yuanming luo ending ending at 2019-04-15 08:03:10Z">in 2 days.

One or more of the answers is exemplary and worthy of an additional bounty.

If possible, present the mindset of solving this kind of problem, please.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Where did you get the awnser $frace3$? Graphing on desmos, it seems your awnser should be correct..
  $endgroup$
  – Justin Stevenson
  Apr 5 at 17:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is the question of the practice booklet. I have the answer but do not know how do they get that.
  $endgroup$
  – yuanming luo
  Apr 6 at 1:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or maybe it can be solved in other ways?
  $endgroup$
  – yuanming luo
  Apr 8 at 8:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have any idea of how to do that? Any advice will be appreciated.
  $endgroup$
  – yuanming luo
  Apr 8 at 9:10
  
  

  
  
  
  

add a comment | 

1

$begingroup$

$$x^2e^ax-1gt ln x$$

Then $e^axgt fracln x+1x^2$

Let $f(x)=fracln x+1x^2$, then $f'(x)=frac-2ln x-1x^3$, and $f''(x)=frac6ln x+1x^4$

$-2ln x-1=0Rightarrow x=e^-frac12$

$f''(e^-frac12)=-2e^2 lt0$

Therefore it is the maximum point for $f(x)$ when $x=e^-frac12$.

Then $e^ae^frac12gt f(e^-frac12)=frace2$

Therefore $ae^-frac12gt1-ln2 Rightarrow agtsqrt e(1-ln2)$

But the answer is $agtfrace3$

How do they get $agt frace3$?

calculus inequality

share|cite|improve this question

edited Apr 8 at 8:23

yuanming luo

asked Apr 5 at 16:42

yuanming luoyuanming luo

9711

$endgroup$

This question has an open bounty worth +50
reputation from yuanming luo ending ending at 2019-04-15 08:03:10Z">in 2 days.

One or more of the answers is exemplary and worthy of an additional bounty.

If possible, present the mindset of solving this kind of problem, please.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Where did you get the awnser $frace3$? Graphing on desmos, it seems your awnser should be correct..
  $endgroup$
  – Justin Stevenson
  Apr 5 at 17:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is the question of the practice booklet. I have the answer but do not know how do they get that.
  $endgroup$
  – yuanming luo
  Apr 6 at 1:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Or maybe it can be solved in other ways?
  $endgroup$
  – yuanming luo
  Apr 8 at 8:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you have any idea of how to do that? Any advice will be appreciated.
  $endgroup$
  – yuanming luo
  Apr 8 at 9:10
  
  

  
  
  
  

add a comment | 

$begingroup$

$$x^2e^ax-1gt ln x$$

Then $e^axgt fracln x+1x^2$

Let $f(x)=fracln x+1x^2$, then $f'(x)=frac-2ln x-1x^3$, and $f''(x)=frac6ln x+1x^4$

$-2ln x-1=0Rightarrow x=e^-frac12$

$f''(e^-frac12)=-2e^2 lt0$

Therefore it is the maximum point for $f(x)$ when $x=e^-frac12$.

Then $e^ae^frac12gt f(e^-frac12)=frace2$

Therefore $ae^-frac12gt1-ln2 Rightarrow agtsqrt e(1-ln2)$

But the answer is $agtfrace3$

How do they get $agt frace3$?

calculus inequality

share|cite|improve this question

edited Apr 8 at 8:23

yuanming luo

asked Apr 5 at 16:42

yuanming luoyuanming luo

9711

$endgroup$

share|cite|improve this question

edited Apr 8 at 8:23

yuanming luo

asked Apr 5 at 16:42

yuanming luoyuanming luo

9711

share|cite|improve this question

edited Apr 8 at 8:23

yuanming luo

asked Apr 5 at 16:42

yuanming luoyuanming luo

9711

asked Apr 5 at 16:42

yuanming luoyuanming luo

9711

asked Apr 5 at 16:42

yuanming luoyuanming luo

9711

1 Answer
1

active

oldest

votes

1

$begingroup$

Your way of solution is wrong. If you want
$$e^ax>1+log xover x^2qquadforall>x>0tag1$$
it is not sufficient to find the maximal point $xi$ of the RHS and then to make sure that the inequality is true at this point $xi$. In fact the value $a=sqrte(1-log2)=0.505915$ you have found is too small, and the graphs of the LHS and RHS intersect for this value of $a$.

If we want $(1)$ then we can as well ask for
$$a>log(1+log x)-2log xover x=:g(x)qquadforall>x>0 .$$
When we graph $g$ we obtain the following picture:

There is a unique maximum at $xi:=0.56849$ (can not be computed elementarily) with $$a_*:=g(xi)=0.523596 .$$
If $a>a_*$ then the desired inequality holds for all $x>0$. Your $a=0.505915$ is transcended by $g$ in an $x$-interval of positive length. The value $a':=1over3e>0.9$ is sufficiently large, but obviously too large.

share|cite|improve this answer

answered Apr 8 at 9:21

Christian BlatterChristian Blatter

176k8115328

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So actually the answer booklet is wrong? I use my graphing calculator to do it too... Thanks with that.
  $endgroup$
  – yuanming luo
  Apr 8 at 9:49
  

  
  
  
  

add a comment | 

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1

$begingroup$

Your way of solution is wrong. If you want
$$e^ax>1+log xover x^2qquadforall>x>0tag1$$
it is not sufficient to find the maximal point $xi$ of the RHS and then to make sure that the inequality is true at this point $xi$. In fact the value $a=sqrte(1-log2)=0.505915$ you have found is too small, and the graphs of the LHS and RHS intersect for this value of $a$.

If we want $(1)$ then we can as well ask for
$$a>log(1+log x)-2log xover x=:g(x)qquadforall>x>0 .$$
When we graph $g$ we obtain the following picture:

There is a unique maximum at $xi:=0.56849$ (can not be computed elementarily) with $$a_*:=g(xi)=0.523596 .$$
If $a>a_*$ then the desired inequality holds for all $x>0$. Your $a=0.505915$ is transcended by $g$ in an $x$-interval of positive length. The value $a':=1over3e>0.9$ is sufficiently large, but obviously too large.

share|cite|improve this answer

answered Apr 8 at 9:21

Christian BlatterChristian Blatter

176k8115328

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So actually the answer booklet is wrong? I use my graphing calculator to do it too... Thanks with that.
  $endgroup$
  – yuanming luo
  Apr 8 at 9:49
  

  
  
  
  

add a comment | 

1

$begingroup$

Your way of solution is wrong. If you want
$$e^ax>1+log xover x^2qquadforall>x>0tag1$$
it is not sufficient to find the maximal point $xi$ of the RHS and then to make sure that the inequality is true at this point $xi$. In fact the value $a=sqrte(1-log2)=0.505915$ you have found is too small, and the graphs of the LHS and RHS intersect for this value of $a$.

If we want $(1)$ then we can as well ask for
$$a>log(1+log x)-2log xover x=:g(x)qquadforall>x>0 .$$
When we graph $g$ we obtain the following picture:

There is a unique maximum at $xi:=0.56849$ (can not be computed elementarily) with $$a_*:=g(xi)=0.523596 .$$
If $a>a_*$ then the desired inequality holds for all $x>0$. Your $a=0.505915$ is transcended by $g$ in an $x$-interval of positive length. The value $a':=1over3e>0.9$ is sufficiently large, but obviously too large.

share|cite|improve this answer

answered Apr 8 at 9:21

Christian BlatterChristian Blatter

176k8115328

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So actually the answer booklet is wrong? I use my graphing calculator to do it too... Thanks with that.
  $endgroup$
  – yuanming luo
  Apr 8 at 9:49
  

  
  
  
  

add a comment | 

$begingroup$

Your way of solution is wrong. If you want
$$e^ax>1+log xover x^2qquadforall>x>0tag1$$
it is not sufficient to find the maximal point $xi$ of the RHS and then to make sure that the inequality is true at this point $xi$. In fact the value $a=sqrte(1-log2)=0.505915$ you have found is too small, and the graphs of the LHS and RHS intersect for this value of $a$.

If we want $(1)$ then we can as well ask for
$$a>log(1+log x)-2log xover x=:g(x)qquadforall>x>0 .$$
When we graph $g$ we obtain the following picture:

There is a unique maximum at $xi:=0.56849$ (can not be computed elementarily) with $$a_*:=g(xi)=0.523596 .$$
If $a>a_*$ then the desired inequality holds for all $x>0$. Your $a=0.505915$ is transcended by $g$ in an $x$-interval of positive length. The value $a':=1over3e>0.9$ is sufficiently large, but obviously too large.

share|cite|improve this answer

answered Apr 8 at 9:21

Christian BlatterChristian Blatter

176k8115328

$endgroup$

share|cite|improve this answer

answered Apr 8 at 9:21

Christian BlatterChristian Blatter

176k8115328

answered Apr 8 at 9:21

Christian BlatterChristian Blatter

176k8115328

answered Apr 8 at 9:21

Christian BlatterChristian Blatter

176k8115328