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 ;))))))))))
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).
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 :)
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Attn: jsaddem
I get two possible answers, do you know what they are?
Could you interact with us please.
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!
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.
\((f'(x))^2 = f(x)f''(x)\\~\\ \frac{f'(x)}{f(x)} = \frac{f''(x)}{f'(x)}\\~\\ \int\frac{f'(x)}{f(x)}dx = \int\frac{f''(x)}{f'(x)}dx\\~\\ ln(f(x))+lnC=ln(f'(x))\qquad \text{Where lnC is a constant}\\~\\ ln(Cf(x))=ln(f'(x))\\~\\ Cf(x)=f'(x)\\~\\ C=\frac{f'(x)}{f(x)}\\~\\ \int C\;dx=\int \frac{f'(x)}{f(x)}\;dx\\~\\ Cx+k=ln(f(x))\\~\\ e^{Cx+k}=e^{ln(f(x))}\\~\\ f(x)=e^{Cx+k}\)
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\\
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!
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}\\