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 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
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Quote:
 Originally Posted by skipjack That can't be done (as it's untrue). It's easy to sketch a counterexample.

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
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Quote:
 Originally Posted by icemanfan Can you give a counterexample of two functions that satisfy the given conditions?
$\displaystyle \color{red}{f(x) = e^x-1}$

$\displaystyle \color{blue}{g(x) = 1-e^x}$
Attached Images
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 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
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 Originally Posted by icemanfan 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.
Man! You should have included earlier. I spend so much time proving this problem because I thought the given statement was true.

 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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