January 9th, 2015, 08:05 PM  #1 
Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1  Functions Proof
Given: f(x) and g(x) are continuous and differentiable everywhere f(x) is concave up g(x) is concave down f(x) and g(x) have exactly one intersection point at x = c Prove: f'(c) = g'(c) 
January 10th, 2015, 06:29 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 21,038 Thanks: 2274 
That can't be done (as it's untrue). It's easy to sketch a counterexample.

January 10th, 2015, 10:58 AM  #3 
Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1 
Can you give a counterexample of two functions that satisfy the given conditions?

January 10th, 2015, 11:13 AM  #4  
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond  Quote:
I thought differently. Let the area of above the curve f(x) be called the interior of f(x) and similarly for g(x). If the interior of g(x) lies in the interior of f(x) then f(x) and g(x) must intersect in two places. Since f(x) and g(x) intersect only once they must be tangent, hence f'(c) = g'(c). Perhaps some other conditions are required to give the problem veracity.  
January 10th, 2015, 12:30 PM  #5  
Math Team Joined: Jul 2011 From: Texas Posts: 3,045 Thanks: 1627  Quote:
$\displaystyle \color{blue}{g(x) = 1e^x}$  
January 10th, 2015, 06:41 PM  #6 
Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1 
I just realized that I forgot to include the conditions limit of f(x) as x approaches negative infinity and positive infinity is positive infinity and limit of g(x) as x approaches negative infinity and positive infinity is negative infinity I believe that the statement should be true given these additional parameters. 
January 11th, 2015, 07:18 PM  #7  
Newbie Joined: Jul 2012 From: Houston, Texas Posts: 18 Thanks: 2 Math Focus: Geometry , Modern Algebra  Quote:
 
January 11th, 2015, 07:53 PM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,693 Thanks: 2677 Math Focus: Mainly analysis and algebra 
It is true now. I think you could look at $h(x) = f(x)  g(x)$. It clearly goes to infinity as $x \to \pm \infty$. So we just need to think about how many distinct roots it has and what we can say about $h'(x)$ at those roots. 
January 15th, 2015, 02:26 PM  #9 
Senior Member Joined: Feb 2012 Posts: 628 Thanks: 1 
Observe, then, that h(x) must have a minimum value. (1) If its minimum value is negative then h(x) has two distinct roots, whereas (2) if its minimum value is positive then h(x) has no distinct roots. Therefore, (3) since h(x) must have exactly one distinct root, its minimum value is 0, and the c for which h(c) = 0 also satisfies h'(c) = 0 since c is a critical point. But from the equality h(x) = f(x)  g(x) we get h'(x) = f'(x)  g'(x) and specifically h'(c) = f'(c)  g'(c), which yields 0 = f'(c)  g'(c) or f'(c) = g'(c), and the statement is proven (sort of). You might have to convince yourself of (1), (2), and (3).

January 15th, 2015, 02:58 PM  #10 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,693 Thanks: 2677 Math Focus: Mainly analysis and algebra 
You need to prove that $h(x)$ has only one local minimum. Your discussion of the value of that local (and indeed, global) minimum then applies.


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