[SOLVED] Derivatives of real valued functions

dwsmith

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Mar 2010
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Let f and g be twice differentiable real-valued functions defined on \(\displaystyle \mathbb{R}\). If f'(x)>g'(x) \(\displaystyle \forall x>0\), which of the following must be true for all x>0?
(a) f(x)>g(x)
(b) f''(x)>g''(x)
(c) f(x)-f(0)>g(x)-g(0)
(d) f'(x)-f'(0)>g'(x)-g'(0)
(e) f''(x)-f''(0)>g''(x)-g''(0)

The answer is c but I thought it was d. Can someone show me how to prove it is c?
 
Feb 2010
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Integrate \(\displaystyle f'>g'\) from 0 to x.
 
Feb 2010
422
141
\(\displaystyle f(x)-f(0) = \int_0^x f' > \int_0^x g' = g(x)-g(0)\)...

edit: to see its not (d), take \(\displaystyle f(x)=2x\) and \(\displaystyle g(x)=x\).
 

dwsmith

MHF Hall of Honor
Mar 2010
3,093
582
Florida
Then why wouldn't the derivative of f'(x)>g'(x) also be an answer?
 
Feb 2010
422
141
(a) is not true, take \(\displaystyle f(x) = 2x\) and \(\displaystyle g(x) = x+1\).

(b), (d), and (e) are not true, take \(\displaystyle f(x)=2x\), \(\displaystyle g(x) = x\).
 
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Plato

MHF Helper
Aug 2006
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Let f and g be twice differentiable real-valued functions defined on \(\displaystyle \mathbb{R}\). If f'(x)>g'(x) \(\displaystyle \forall x>0\), which of the following must be true for all x>0?
(c) f(x)-f(0)>g(x)-g(0)
Can someone show me how to prove it is c?
We know the derivative of \(\displaystyle f-g\) is \(\displaystyle f '-g '>0\).
Therefore \(\displaystyle f-g\) is increasing or \(\displaystyle f(x)-g(x)>f(0)-g(0)\).
Thus answer (c).
 
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