do this with limit

superconduct · December 31st, 2011, 12:04 AM

Let $f(x)$ be a continuous function for all real values of $x$ such that the following conditions are satisfied:
(I) $f(0) \neq 0$
(II) $f'(0) = 2$
(III) $f(a+b)= f(a)f(b)$ for all real numbers $a$ and $b$ .

Show that $f'(x) = 2f(x)$ for all real values of $x$ .

MarkFL · December 31st, 2011, 01:37 AM

I'm not sure what limits have to do with this, but here is my take on it:

From condition III, we must have:

$f(x)=r^{kx}$ where r and k are non-zero real constants.

$f(a+b)=r^{k(a+b)}=r^{ka}r^{kb}=f(a)f(b)$

From condition II, we then find:

$f'(x)=k\ln(r)r^{kx}$

$f'(0)=k\ln(r)r^{k\cdot0}=k\ln(r)=2\:\therefore \:e^{\frac{2}{k}}=a$ so that we have:

$f(x)=$e^{\frac{2}{k}}$^{kx}=e^{2x}$

This satisfies condition I, and we find:
'
$f'(x)=2e^{2x}=2f(x)$ for all real x.

superconduct · December 31st, 2011, 05:02 AM

oh ur solution is pretty cool, but give some comments with this one:

$f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h}<br /> <br /> = f(x) \lim_{h \to 0} \frac{f(h)-1}{h} = f(x) \lim_{h \to 0} \frac{f'(h)}{1} = 2f(x)$

MarkFL · December 31st, 2011, 10:11 AM

After having some sleep, I feel kind of thick now! Of course, where derivatives are concerned there are limits.

I like your solution, especially the use of L'H�pital's rule.

December 31st, 2011, 12:04 AM	#1
superconduct Member Joined: Dec 2011 Posts: 61 Thanks: 1	do this with limit Let $f(x)$ be a continuous function for all real values of $x$ such that the following conditions are satisfied: (I) $f(0) \neq 0$ (II) $f'(0) = 2$ (III) $f(a+b)= f(a)f(b)$ for all real numbers $a$ and $b$ . Show that $f'(x) = 2f(x)$ for all real values of $x$ .
$superconduct is offline$

December 31st, 2011, 01:37 AM	#2
MarkFL Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs	Re: do this with limit I'm not sure what limits have to do with this, but here is my take on it: From condition III, we must have: $f(x)=r^{kx}$ where r and k are non-zero real constants. $f(a+b)=r^{k(a+b)}=r^{ka}r^{kb}=f(a)f(b)$ From condition II, we then find: $f'(x)=k\ln(r)r^{kx}$ $f'(0)=k\ln(r)r^{k\cdot0}=k\ln(r)=2\:\therefore \:e^{\frac{2}{k}}=a$ so that we have: $f(x)=\(e^{\frac{2}{k}}\)^{kx}=e^{2x}$ This satisfies condition I, and we find: ' $f'(x)=2e^{2x}=2f(x)$ for all real x.
$MarkFL is offline$

December 31st, 2011, 05:02 AM	#3
superconduct Member Joined: Dec 2011 Posts: 61 Thanks: 1	Re: do this with limit oh ur solution is pretty cool, but give some comments with this one: $f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} = \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h}<br /> <br /> = f(x) \lim_{h \to 0} \frac{f(h)-1}{h} = f(x) \lim_{h \to 0} \frac{f'(h)}{1} = 2f(x)$
$superconduct is offline$

December 31st, 2011, 10:11 AM	#4
MarkFL Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs	Re: do this with limit After having some sleep, I feel kind of thick now! Of course, where derivatives are concerned there are limits. I like your solution, especially the use of L'H�pital's rule.
$MarkFL is offline$

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