+72 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 ;))))))))))
+396 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).
+118608 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.
+72 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!
+118608 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.
+118608
\((f'(x))^2 = f(x)f''(x)\\~\\ \frac{f'(x)}{f(x)} = \frac{f''(x)}{f'(x)}\\~\\ \int\frac{f'(x)}{f(x)}dx = \int\frac{f''(x)}{f'(x)}dx\\~\\ ln(f(x))+lnC=ln(f'(x))\qquad \text{Where lnC is a constant}\\~\\ ln(Cf(x))=ln(f'(x))\\~\\ Cf(x)=f'(x)\\~\\ C=\frac{f'(x)}{f(x)}\\~\\ \int C\;dx=\int \frac{f'(x)}{f(x)}\;dx\\~\\ Cx+k=ln(f(x))\\~\\ e^{Cx+k}=e^{ln(f(x))}\\~\\ f(x)=e^{Cx+k}\)
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\\
+72 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!
+118608 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}\\
