Mixing
Proof $x^3+ax^2+bx+c-e^x=0$ has at least one solution
$begingroup$
I need to show that the following equation has at least one solution: $$x^3+ax^2+bx+c=e^x$$ where $a,b,c in mathbb{R}$.
What I did:
I defined the real-valued function $f(x)=x^3+ax^2+bx+c-e^x$
and showed that for $c=1$ we get $x=0$ as a solution, and that for $c>1$ we get a solution as $f(0)>0$ and $lim_{x to infty} f(x)=- infty$
but Im stuck with the case of $c<1$.
Any suggestions? or other ways of proving this statement. any help will be much appreciated
calculus proof-verification continuity
edited Jan 1 at 10:40
José Alejandro Aburto Araneda
825110
asked Jan 1 at 9:06
user114138user114138
504211
$endgroup$
add a comment |
$begingroup$
I need to show that the following equation has at least one solution: $$x^3+ax^2+bx+c=e^x$$ where $a,b,c in mathbb{R}$.
What I did:
I defined the real-valued function $f(x)=x^3+ax^2+bx+c-e^x$
and showed that for $c=1$ we get $x=0$ as a solution, and that for $c>1$ we get a solution as $f(0)>0$ and $lim_{x to infty} f(x)=- infty$
but Im stuck with the case of $c<1$.
Any suggestions? or other ways of proving this statement. any help will be much appreciated
calculus proof-verification continuity
edited Jan 1 at 10:40
José Alejandro Aburto Araneda
825110
asked Jan 1 at 9:06
user114138user114138
504211
$endgroup$
$begingroup$
You could always find $f'''(x)=6-e^x$ and work backwards.
$endgroup$
– TheSimpliFire
Jan 1 at 9:12
$begingroup$
could you elaborate a bit further pls?
$endgroup$
– user114138
Jan 1 at 9:13
$begingroup$
If you know $f'''(x)$ then you can use its properties (for example, set to zero) to imply properties about $f''(x)$ (for example, increasing or decreasing). Do the same for $f'(x)$ and finally $f(x)$.
$endgroup$
– TheSimpliFire
Jan 1 at 9:15
1
$begingroup$
Do you know it is true?
$endgroup$
– mikado
Jan 1 at 9:16
3
$begingroup$
As written, the claim is false. $x^3 - x^2 - mathrm{e}^x$ has maximum value $-0.8559 dots$, so is never zero. Do you have additional constraints not present in your Question?
$endgroup$
– Eric Towers
Jan 1 at 9:21
add a comment |
$begingroup$
I need to show that the following equation has at least one solution: $$x^3+ax^2+bx+c=e^x$$ where $a,b,c in mathbb{R}$.
What I did:
I defined the real-valued function $f(x)=x^3+ax^2+bx+c-e^x$
and showed that for $c=1$ we get $x=0$ as a solution, and that for $c>1$ we get a solution as $f(0)>0$ and $lim_{x to infty} f(x)=- infty$
but Im stuck with the case of $c<1$.
Any suggestions? or other ways of proving this statement. any help will be much appreciated
calculus proof-verification continuity
edited Jan 1 at 10:40
José Alejandro Aburto Araneda
825110
asked Jan 1 at 9:06
user114138user114138
504211
$endgroup$
I need to show that the following equation has at least one solution: $$x^3+ax^2+bx+c=e^x$$ where $a,b,c in mathbb{R}$.
What I did:
I defined the real-valued function $f(x)=x^3+ax^2+bx+c-e^x$
and showed that for $c=1$ we get $x=0$ as a solution, and that for $c>1$ we get a solution as $f(0)>0$ and $lim_{x to infty} f(x)=- infty$
but Im stuck with the case of $c<1$.
Any suggestions? or other ways of proving this statement. any help will be much appreciated
calculus proof-verification continuity
calculus proof-verification continuity
edited Jan 1 at 10:40
José Alejandro Aburto Araneda
825110
asked Jan 1 at 9:06
user114138user114138
504211
edited Jan 1 at 10:40
José Alejandro Aburto Araneda
825110
asked Jan 1 at 9:06
user114138user114138
504211
edited Jan 1 at 10:40
José Alejandro Aburto Araneda
825110
edited Jan 1 at 10:40
José Alejandro Aburto Araneda
825110
edited Jan 1 at 10:40
José Alejandro Aburto Araneda
825110
825110
asked Jan 1 at 9:06
user114138user114138
504211
asked Jan 1 at 9:06
user114138user114138
504211
asked Jan 1 at 9:06
user114138user114138
504211
504211
$begingroup$
You could always find $f'''(x)=6-e^x$ and work backwards.
$endgroup$
– TheSimpliFire
Jan 1 at 9:12
$begingroup$
could you elaborate a bit further pls?
$endgroup$
– user114138
Jan 1 at 9:13
$begingroup$
If you know $f'''(x)$ then you can use its properties (for example, set to zero) to imply properties about $f''(x)$ (for example, increasing or decreasing). Do the same for $f'(x)$ and finally $f(x)$.
$endgroup$
– TheSimpliFire
Jan 1 at 9:15
1
$begingroup$
Do you know it is true?
$endgroup$
– mikado
Jan 1 at 9:16
3
$begingroup$
As written, the claim is false. $x^3 - x^2 - mathrm{e}^x$ has maximum value $-0.8559 dots$, so is never zero. Do you have additional constraints not present in your Question?
$endgroup$
– Eric Towers
Jan 1 at 9:21
add a comment |
$begingroup$
You could always find $f'''(x)=6-e^x$ and work backwards.
$endgroup$
– TheSimpliFire
Jan 1 at 9:12
$begingroup$
could you elaborate a bit further pls?
$endgroup$
– user114138
Jan 1 at 9:13
$begingroup$
If you know $f'''(x)$ then you can use its properties (for example, set to zero) to imply properties about $f''(x)$ (for example, increasing or decreasing). Do the same for $f'(x)$ and finally $f(x)$.
$endgroup$
– TheSimpliFire
Jan 1 at 9:15
1
$begingroup$
Do you know it is true?
$endgroup$
– mikado
Jan 1 at 9:16
3
$begingroup$
As written, the claim is false. $x^3 - x^2 - mathrm{e}^x$ has maximum value $-0.8559 dots$, so is never zero. Do you have additional constraints not present in your Question?
$endgroup$
– Eric Towers
Jan 1 at 9:21
$begingroup$
You could always find $f'''(x)=6-e^x$ and work backwards.
$endgroup$
– TheSimpliFire
Jan 1 at 9:12
$begingroup$
You could always find $f'''(x)=6-e^x$ and work backwards.
$endgroup$
– TheSimpliFire
Jan 1 at 9:12
$begingroup$
could you elaborate a bit further pls?
$endgroup$
– user114138
Jan 1 at 9:13
$begingroup$
could you elaborate a bit further pls?
$endgroup$
– user114138
Jan 1 at 9:13
$begingroup$
If you know $f'''(x)$ then you can use its properties (for example, set to zero) to imply properties about $f''(x)$ (for example, increasing or decreasing). Do the same for $f'(x)$ and finally $f(x)$.
$endgroup$
– TheSimpliFire
Jan 1 at 9:15
$begingroup$
If you know $f'''(x)$ then you can use its properties (for example, set to zero) to imply properties about $f''(x)$ (for example, increasing or decreasing). Do the same for $f'(x)$ and finally $f(x)$.
$endgroup$
– TheSimpliFire
Jan 1 at 9:15
1
1
$begingroup$
Do you know it is true?
$endgroup$
– mikado
Jan 1 at 9:16
$begingroup$
Do you know it is true?
$endgroup$
– mikado
Jan 1 at 9:16
3
3
$begingroup$
As written, the claim is false. $x^3 - x^2 - mathrm{e}^x$ has maximum value $-0.8559 dots$, so is never zero. Do you have additional constraints not present in your Question?
$endgroup$
– Eric Towers
Jan 1 at 9:21
$begingroup$
As written, the claim is false. $x^3 - x^2 - mathrm{e}^x$ has maximum value $-0.8559 dots$, so is never zero. Do you have additional constraints not present in your Question?
$endgroup$
– Eric Towers
Jan 1 at 9:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is false, in general : note that ${x^3 over 6}< exp x$.
Indeed if $x<0$ this is clear as the rhs is $<0$, and for $x>0$ this follows the Taylor series of the exponential $exp x = 1+x+ x^2/2+x^6/6+ sum _{n>3} x^n/n!$.
Then, for all $x$, $x^3< exp (x+ ln 6)$.
With $y= x+ln 6$; we get that for all $y$, $(y-ln 6)^3 < exp y$, a counter example
answered Jan 1 at 9:22
ThomasThomas
3,889510
$endgroup$
$begingroup$
I used $y$ : $(y-ln 6)^3$ has the required form.
$endgroup$
– Thomas
Jan 1 at 17:32
$begingroup$
Yes it is $(y-ln 6)^3=y^3- 3ln 6 y^2+3{ln 6 } ^2y-{ ln 6}^3$
$endgroup$
– Thomas
Jan 2 at 18:04
add a comment |
$begingroup$
The result is false. Let be $f(x)=e^x-x^3+Cx^2$. It is a convex function for some $Cinmathbb{R}$, since $f(x)''=e^x-6x+C$, let $h(x)=e^x-6x$, it has minimum on $log(6)$, and $h(log(6))=6(1-log(6))$, now take $C=6log(6)$, and you have that $f(x)''=6>0$. So that, as $f$ is strictrly convex it has at most one minimum, let $m=inf f$, the function $g(x)=f(x)+|m|+1$ is of the form $e^x-p(x)$, where $p(x)$ is a real monic polynomial (leader coefficient equal to $1$) of degree 3, so that your result is not true in general.
answered Jan 1 at 15:11
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is false, in general : note that ${x^3 over 6}< exp x$.
Indeed if $x<0$ this is clear as the rhs is $<0$, and for $x>0$ this follows the Taylor series of the exponential $exp x = 1+x+ x^2/2+x^6/6+ sum _{n>3} x^n/n!$.
Then, for all $x$, $x^3< exp (x+ ln 6)$.
With $y= x+ln 6$; we get that for all $y$, $(y-ln 6)^3 < exp y$, a counter example
answered Jan 1 at 9:22
ThomasThomas
3,889510
$endgroup$
$begingroup$
I used $y$ : $(y-ln 6)^3$ has the required form.
$endgroup$
– Thomas
Jan 1 at 17:32
$begingroup$
Yes it is $(y-ln 6)^3=y^3- 3ln 6 y^2+3{ln 6 } ^2y-{ ln 6}^3$
$endgroup$
– Thomas
Jan 2 at 18:04
add a comment |
$begingroup$
It is false, in general : note that ${x^3 over 6}< exp x$.
Indeed if $x<0$ this is clear as the rhs is $<0$, and for $x>0$ this follows the Taylor series of the exponential $exp x = 1+x+ x^2/2+x^6/6+ sum _{n>3} x^n/n!$.
Then, for all $x$, $x^3< exp (x+ ln 6)$.
With $y= x+ln 6$; we get that for all $y$, $(y-ln 6)^3 < exp y$, a counter example
answered Jan 1 at 9:22
ThomasThomas
3,889510
$endgroup$
$begingroup$
I used $y$ : $(y-ln 6)^3$ has the required form.
$endgroup$
– Thomas
Jan 1 at 17:32
$begingroup$
Yes it is $(y-ln 6)^3=y^3- 3ln 6 y^2+3{ln 6 } ^2y-{ ln 6}^3$
$endgroup$
– Thomas
Jan 2 at 18:04
add a comment |
$begingroup$
It is false, in general : note that ${x^3 over 6}< exp x$.
Indeed if $x<0$ this is clear as the rhs is $<0$, and for $x>0$ this follows the Taylor series of the exponential $exp x = 1+x+ x^2/2+x^6/6+ sum _{n>3} x^n/n!$.
Then, for all $x$, $x^3< exp (x+ ln 6)$.
With $y= x+ln 6$; we get that for all $y$, $(y-ln 6)^3 < exp y$, a counter example
answered Jan 1 at 9:22
ThomasThomas
3,889510
$endgroup$
It is false, in general : note that ${x^3 over 6}< exp x$.
Indeed if $x<0$ this is clear as the rhs is $<0$, and for $x>0$ this follows the Taylor series of the exponential $exp x = 1+x+ x^2/2+x^6/6+ sum _{n>3} x^n/n!$.
Then, for all $x$, $x^3< exp (x+ ln 6)$.
With $y= x+ln 6$; we get that for all $y$, $(y-ln 6)^3 < exp y$, a counter example
answered Jan 1 at 9:22
ThomasThomas
3,889510
answered Jan 1 at 9:22
ThomasThomas
3,889510
answered Jan 1 at 9:22
ThomasThomas
3,889510
answered Jan 1 at 9:22
ThomasThomas
3,889510
3,889510
$begingroup$
I used $y$ : $(y-ln 6)^3$ has the required form.
$endgroup$
– Thomas
Jan 1 at 17:32
$begingroup$
Yes it is $(y-ln 6)^3=y^3- 3ln 6 y^2+3{ln 6 } ^2y-{ ln 6}^3$
$endgroup$
– Thomas
Jan 2 at 18:04
add a comment |
$begingroup$
I used $y$ : $(y-ln 6)^3$ has the required form.
$endgroup$
– Thomas
Jan 1 at 17:32
$begingroup$
Yes it is $(y-ln 6)^3=y^3- 3ln 6 y^2+3{ln 6 } ^2y-{ ln 6}^3$
$endgroup$
– Thomas
Jan 2 at 18:04
$begingroup$
I used $y$ : $(y-ln 6)^3$ has the required form.
$endgroup$
– Thomas
Jan 1 at 17:32
$begingroup$
I used $y$ : $(y-ln 6)^3$ has the required form.
$endgroup$
– Thomas
Jan 1 at 17:32
$begingroup$
Yes it is $(y-ln 6)^3=y^3- 3ln 6 y^2+3{ln 6 } ^2y-{ ln 6}^3$
$endgroup$
– Thomas
Jan 2 at 18:04
$begingroup$
Yes it is $(y-ln 6)^3=y^3- 3ln 6 y^2+3{ln 6 } ^2y-{ ln 6}^3$
$endgroup$
– Thomas
Jan 2 at 18:04
add a comment |
$begingroup$
The result is false. Let be $f(x)=e^x-x^3+Cx^2$. It is a convex function for some $Cinmathbb{R}$, since $f(x)''=e^x-6x+C$, let $h(x)=e^x-6x$, it has minimum on $log(6)$, and $h(log(6))=6(1-log(6))$, now take $C=6log(6)$, and you have that $f(x)''=6>0$. So that, as $f$ is strictrly convex it has at most one minimum, let $m=inf f$, the function $g(x)=f(x)+|m|+1$ is of the form $e^x-p(x)$, where $p(x)$ is a real monic polynomial (leader coefficient equal to $1$) of degree 3, so that your result is not true in general.
answered Jan 1 at 15:11
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
$endgroup$
add a comment |
$begingroup$
The result is false. Let be $f(x)=e^x-x^3+Cx^2$. It is a convex function for some $Cinmathbb{R}$, since $f(x)''=e^x-6x+C$, let $h(x)=e^x-6x$, it has minimum on $log(6)$, and $h(log(6))=6(1-log(6))$, now take $C=6log(6)$, and you have that $f(x)''=6>0$. So that, as $f$ is strictrly convex it has at most one minimum, let $m=inf f$, the function $g(x)=f(x)+|m|+1$ is of the form $e^x-p(x)$, where $p(x)$ is a real monic polynomial (leader coefficient equal to $1$) of degree 3, so that your result is not true in general.
answered Jan 1 at 15:11
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
$endgroup$
add a comment |
$begingroup$
The result is false. Let be $f(x)=e^x-x^3+Cx^2$. It is a convex function for some $Cinmathbb{R}$, since $f(x)''=e^x-6x+C$, let $h(x)=e^x-6x$, it has minimum on $log(6)$, and $h(log(6))=6(1-log(6))$, now take $C=6log(6)$, and you have that $f(x)''=6>0$. So that, as $f$ is strictrly convex it has at most one minimum, let $m=inf f$, the function $g(x)=f(x)+|m|+1$ is of the form $e^x-p(x)$, where $p(x)$ is a real monic polynomial (leader coefficient equal to $1$) of degree 3, so that your result is not true in general.
answered Jan 1 at 15:11
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
$endgroup$
The result is false. Let be $f(x)=e^x-x^3+Cx^2$. It is a convex function for some $Cinmathbb{R}$, since $f(x)''=e^x-6x+C$, let $h(x)=e^x-6x$, it has minimum on $log(6)$, and $h(log(6))=6(1-log(6))$, now take $C=6log(6)$, and you have that $f(x)''=6>0$. So that, as $f$ is strictrly convex it has at most one minimum, let $m=inf f$, the function $g(x)=f(x)+|m|+1$ is of the form $e^x-p(x)$, where $p(x)$ is a real monic polynomial (leader coefficient equal to $1$) of degree 3, so that your result is not true in general.
answered Jan 1 at 15:11
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
answered Jan 1 at 15:11
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
answered Jan 1 at 15:11
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
answered Jan 1 at 15:11
José Alejandro Aburto AranedaJosé Alejandro Aburto Araneda
825110
825110
add a comment |
add a comment |
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$begingroup$
You could always find $f'''(x)=6-e^x$ and work backwards.
$endgroup$
– TheSimpliFire
Jan 1 at 9:12
$begingroup$
could you elaborate a bit further pls?
$endgroup$
– user114138
Jan 1 at 9:13
$begingroup$
If you know $f'''(x)$ then you can use its properties (for example, set to zero) to imply properties about $f''(x)$ (for example, increasing or decreasing). Do the same for $f'(x)$ and finally $f(x)$.
$endgroup$
– TheSimpliFire
Jan 1 at 9:15
1
$begingroup$
Do you know it is true?
$endgroup$
– mikado
Jan 1 at 9:16
3
$begingroup$
As written, the claim is false. $x^3 - x^2 - mathrm{e}^x$ has maximum value $-0.8559 dots$, so is never zero. Do you have additional constraints not present in your Question?
$endgroup$
– Eric Towers
Jan 1 at 9:21