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Let the function \(f\) be continuous and real valued, and satisfy the following conditions:

1. \((f'(x))^2 = f(x)f''(x)\),

2. \(f(0)=1,\)

3. \(f''''(0)=9.\)

 

List out all possible values of \(f'(0).\)

 

Thank you so much ;))))))))))

 Dec 24, 2021
 #1
avatar+117443 
0

I have no idea but I would also like to see someone tackle this one.

 Dec 25, 2021
 #2
avatar+226 
+2

Write (1) as     \(\displaystyle \frac{f ' (x)}{f(x)}=\frac{f ''(x)}{f '(x)}\), and integrate.

\(\displaystyle \log_{e}f(x)=\log_{e}f '(x) + \log_{e}C=\log_{e}\{Cf '(x)\}\) ,

so

\(\displaystyle f(x)=Cf '(x).\)

Now integrate a second time, (same sort of technique), to get an expression for f(x).

 Dec 25, 2021
 #5
avatar+72 
0

Doesn't integrating that simply give us \(f(x) = c\)?

 

In that case, "all possible values of \(f'(0)\)  are simply... \(0\).

 

Thank you!

jsaddern  Dec 26, 2021
 #3
avatar+117443 
+1

Thanks Tiggsy,

I never would have thought of that by myself.

 

It is always great to see you on the forum.   I hope you enjoyed Christmas :)

 

----------------

 

Attn:   jsaddem

I get two possible answers,  do you know what they are?   

Could you interact with us please.

 Dec 25, 2021
edited by Melody  Dec 25, 2021
 #4
avatar+72 
+1

I'd love to interact!

 

I'm not quite sure what the answers are... How would we continue to integrate f(x) = Cf'(x) to achieve the possible values of f'(0)?

 

Thank you both and happy holidays!

jsaddern  Dec 26, 2021
 #6
avatar+117443 
0

If you would love to interact then try interacting with the other questions you have asked (that have been answered)

People want a response from you and they would like you to give them points for their time, effort and/or help.

 

 

I will take some time again with this question.. when I have time,  probably later today.

Tiggsy's answer is really good but I can give you a little more insight.

Melody  Dec 26, 2021
edited by Melody  Dec 26, 2021
 #7
avatar+117443 
+1

 

\((f'(x))^2 = f(x)f''(x)\\~\\ \frac{f'(x)}{f(x)} = \frac{f''(x)}{f'(x)}\\~\\ \int\frac{f'(x)}{f(x)}dx = \int\frac{f''(x)}{f'(x)}dx\\~\\ ln(f(x))+lnC=ln(f'(x))\qquad \text{Where lnC is a constant}\\~\\ ln(Cf(x))=ln(f'(x))\\~\\ Cf(x)=f'(x)\\~\\ C=\frac{f'(x)}{f(x)}\\~\\ \int C\;dx=\int \frac{f'(x)}{f(x)}\;dx\\~\\ Cx+k=ln(f(x))\\~\\ e^{Cx+k}=e^{ln(f(x))}\\~\\ f(x)=e^{Cx+k}\)

 

    Sub f(0)=1  and you get    k=0

 

\(f(x)=e^{Cx}\)

 

  \(​​​​f(x)=e^{Cx}\\ ​​​​f'(x)=Ce^{Cx}\\ ​​​​f''(x)=C^2e^{Cx}\\ ​​​​f'''(x)=C^3e^{Cx}\\ ​​​​f''''(x)=C^4e^{Cx}\\ \quad given\;\;f''''(0)=9\\ \quad \;C^4=9\\ \quad \;C=\pm\sqrt3\\ \)    

\(​​​​f'(x)=\sqrt3\;e^{\sqrt3\;x}\qquad or \qquad f'(x)=-\sqrt3\;e^{-\sqrt3\;x}\\ so\\ ​​​​f'(0)=\sqrt3\qquad or \qquad f'(0)=-\sqrt3\\\)

 

 

 

 

 

LaTex

(f'(x))^2 = f(x)f''(x)\\~\\
\frac{f'(x)}{f(x)} = \frac{f''(x)}{f'(x)}\\~\\
\int\frac{f'(x)}{f(x)}dx = \int\frac{f''(x)}{f'(x)}dx\\~\\
ln(f(x))+lnC=ln(f'(x))\qquad \text{Where lnC is a constant}\\~\\
ln(Cf(x))=ln(f'(x))\\~\\
Cf(x)=f'(x)\\~\\
C=\frac{f'(x)}{f(x)}\\~\\
\int C\;dx=\int \frac{f'(x)}{f(x)}\;dx\\~\\
Cx+k=ln(f(x))\\~\\
e^{Cx+k}=e^{ln(f(x))}\\~\\
f(x)=e^{Cx+k}

 

​​​​f(x)=e^{Cx}\\
​​​​f'(x)=Ce^{Cx}\\
​​​​f''(x)=C^2e^{Cx}\\
​​​​f'''(x)=C^3e^{Cx}\\
​​​​f''''(x)=C^4e^{Cx}\\
\quad given\;\;f''''(0)=9\\
\quad \;C^4=9\\
\quad \;C=\pm\sqrt3\\

 

​​​​f'(x)=\sqrt3\;e^{\sqrt3\;x}\qquad or \qquad f'(x)=-\sqrt3\;e^{-\sqrt3\;x}\\
so\\
​​​​f'(0)=\sqrt3\qquad or \qquad f'(0)=-\sqrt3\\

 Dec 27, 2021
 #8
avatar+72 
+1

Thanks so much! That is indeed a great solution!

 

An alternative that I thought of was:

 

f'(0)^2 = f(0)f''(0) --> Since f(0) = 1, we conclude that f''(0) = f'(0)^2. Now, we can differentiate both sides to get f'(x)f''(x)=f(x)f'''(x). Now, we can differentiate this equation once again! And use the product rule! --> f'(x)f'''(x) + f''(x)^2 = f(x)f''''(x) + f'(x)f'''(x) --> f''(x)^2 = f(x)f''''(x).

 

Plugging in x=0 simply gives us the answer, that is, f''(0)^4 = f'''(0) = 0. Therefore, f''(0) = \(\pm 3\) and further, since f'(0)^2 =f''(0), we get the end result, \(f'(0) = \pm \sqrt{3}\). Yayyyy!

jsaddern  Dec 27, 2021
 #9
avatar+117443 
+1

I am just looking at your solution.   

Now I have finished looking and I really like it.  Thanks.

 

 

from 1 and 2

 

\((f'(0))^2=f(0)f''(0)\\ (f'(0))^2=f''(0)\\~\\ so\\ f(0)=1,\qquad f''''(0)=9\qquad f''(0)=(f'(0))^2 \qquad find\;\; f'(0) \\\)

 

\((f'(x))^2=f(x)f''(x)\\ \text{diff wrt x}\\ 2(f'(x))(f''(x)) =f(x)f'''(x) +f'(x)f''(x)\\ \text{diff again wrt x}\\ 2(f'(x)(f'''(x)) + 2(f''(x)(f''(x)) =f(x)f''''(x) +f'(x)f'''(x) +f'(x)f'''(x) +f''(x)f''(x)\\ f''(x)f''(x) =f(x)f''''(x) \\ sub\;\;x=0\\ f''(0)f''(0) =f(0)f''''(0) \\ (f'(0))^4 =1*9 \\ f'(0)=\pm \sqrt[4]{9}\\ f'(0)=\pm \sqrt{3}\\\)

 

 

 

 

 

 

LaTex:

(f'(0))^2=f(0)f''(0)\\
(f'(0))^2=f''(0)\\~\\
so\\
f(0)=1,\qquad f''''(0)=9\qquad f''(0)=(f'(0))^2 \qquad find\;\; f'(0) \\

(f'(x))^2=f(x)f''(x)\\
\text{diff  wrt x}\\
2(f'(x))(f''(x))    =f(x)f'''(x)    +f'(x)f''(x)\\

\text{diff  again wrt x}\\
2(f'(x)(f'''(x))  + 2(f''(x)(f''(x))  =f(x)f''''(x) +f'(x)f'''(x)   +f'(x)f'''(x)  +f''(x)f''(x)\\
 f''(x)f''(x)  =f(x)f''''(x)    \\
sub\;\;x=0\\
f''(0)f''(0)  =f(0)f''''(0)    \\
(f'(0))^4 =1*9   \\
f'(0)=\pm \sqrt[4]{9}\\
f'(0)=\pm \sqrt{3}\\

Melody  Dec 29, 2021

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