My Math Forum Combination of knowledge!

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 March 15th, 2011, 09:11 AM #1 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 Combination of knowledge! [color=#000000]1. Let A(1,0), B(0,1) and ?(1,1) three points on the 2D plane. For a random point M, which lies on the line segment AB without it being point A or B, prove that $\lambda_{M\Gamma}\cdot \lambda_{OM}=1$ (? is the rate of direction and O is the beginning of the axis) 2. If a function f is continuous on the interval [0,1] and differentiable on (0,1) with f(0)=0 and f(1)=1 prove that there exist $\xi_{1}$ and $\xi_{2}$ with $\xi_{1}\neq \xi_{2}$ such that $f'(\xi_{1})\cdot f'(\xi_{2})=1$.[/color]
 March 17th, 2011, 09:14 PM #2 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Combination of knowledge! Intermediate value theorem, mean value theorem? No-one else going to touch this? ("? is the rate of direction" means "? is the slope.")
 March 24th, 2011, 04:53 PM #3 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: Combination of knowledge! [color=#000000]Aswoods is right, noone is going to touch this?[/color]
 March 25th, 2011, 02:54 AM #4 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Combination of knowledge! 1. We have |OB|=1, |A?|=1, |OA|=1 and |B?|=1 (and data provided by ZardoZ) The line segment is described by y=1-x Let M have coördinates (x,y) such that it has coördinates (x,1-x). Let's find slope $\lambda_{M\Gamma}$ It is $\frac{(1-x)-0}{x-0}=\frac{1-x}{x}$ for $x\ne 0$ Now, we need slope $\lambda_{OM}$ as well. It is $\frac{1-(1-x)}{(1)-(x)}=\frac{x}{1-x}$ for $x\ne 1$ Now, $\lambda_{M\Gamma} \cdot \lambda_{OM}=\frac{1-x}{x} \cdot \frac{x}{1-x}=1$ for $x \ne 0,1$ 2. Not sure for this one. Maybe, split in 3 cases. C1: f'(0)<1. Then there exists f'(x)>1 (for otherwise, $f(1) < 1$)? such that $f'(0) \cdot f'(x)=f'(\xi_1) \cdot f#39;(\xi_2)=1$ C2: f'(0)=1. Then there exists f'(x)=1 such that $f'(0) \cdot f'(x)=f'(\xi_1) \cdot f#39;(\xi_2)=1$ C3: f'(0)>1. Then there exists f'(x)=1 (for otherwise, $f(1) > 1$)?such that $f'(0) \cdot f'(x)=f'(\xi_1) \cdot f#39;(\xi_2)=1$ Maybe, some-one is inspired now.
 March 25th, 2011, 06:01 AM #5 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 Re: Combination of knowledge! Let g(x) = 1-x. Then f(0)=0, g(0)=1, but f(1)=1, g(1)=0. Use the intermediate value theorem to prove that the curves intersect for some x=a, then apply the mean value theorem to f over the intervals [0,a] and [a,1].
 March 26th, 2011, 10:22 AM #6 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: Combination of knowledge! [color=#000000]Well done![/color]

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