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 October 9th, 2011, 02:35 AM #1 Newbie   Joined: Apr 2010 Posts: 24 Thanks: 0 Determining a differentiable function given conditions I am really stuck on this question g is a differentiable function for all pos. values of x such that these 3 conditions are true i. g(1)=0 ii. tangent to the graph at x=1 is inclined at 45 degrees to the positive x-axis iii. d/dx(g(2x)) =g'(x) a) Determine value for g'(2) b) Prove that g(2x) = g(x) + g(2) Any help would be much appreciated
 October 9th, 2011, 03:21 AM #2 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: Determining a differentiable function given conditions [color=#000000]From the second clue, $g'(1)=1$ and from the third $\frac{d}{dx}g(2x)=g'(x)\Rightarrow 2g'(2x)=g'(x)\Rightarrow g'(2x)=\frac{1}{2}g#39;(x)$ for x=1, $g'(2)=\frac{1}{2}g#39;(1)=\frac{1}{2}$. Also, $\frac{d}{dx}\left(g(2x)\right)=g'(x)\Rightarro w \int \frac{d}{dx}\left(g(2x)\right)\;dx=\int g#39;(x)\;dx\Rightarrow g(2x)=g(x)+c$, for x=1 we get $g(2)=g(1)+c\Rightarrow g(2)=c$, so $g(2x)=g(x)+g(2)$.[/color]
October 9th, 2011, 03:43 AM   #3
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Re: Determining a differentiable function given conditions

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 Originally Posted by ZardoZ [color=#000000]From the second clue, $g'(1)=1$ and from the third $\frac{d}{dx}g(2x)=g'(x)\Rightarrow 2g'(2x)=g'(x)\Rightarrow g'(2x)=\frac{1}{2}g#39;(x)$ for x=1, $g'(2)=\frac{1}{2}g#39;(1)=\frac{1}{2}$. Also, $\frac{d}{dx}\left(g(2x)\right)=g'(x)\Rightarro w \int \frac{d}{dx}\left(g(2x)\right)\;dx=\int g#39;(x)\;dx\Rightarrow g(2x)=g(x)+c$, for x=1 we get $g(2)=g(1)+c\Rightarrow g(2)=c$, so $g(2x)=g(x)+g(2)$.[/color]
thanks a lot

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