[SOLVED] Derivatives of real valued functions

**dwsmith** · May 9th 2010, 11:53 AM

Let f and g be twice differentiable real-valued functions defined on $\mathbb{R}$ . If f'(x)>g'(x) $\forall x>0$ , 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?

**maddas** · May 9th 2010, 12:23 PM

Integrate $f'>g'$ from 0 to x.

**dwsmith** · May 9th 2010, 12:25 PM

Originally Posted by maddas

Integrate $f'>g'$ from 0 to x.

That is just f(x)-f(0) and same for g

**maddas** · May 9th 2010, 12:27 PM

$f(x)-f(0) = \int_0^x f' > \int_0^x g' = g(x)-g(0)$ ...

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

**dwsmith** · May 9th 2010, 12:29 PM

Then why wouldn't the derivative of f'(x)>g'(x) also be an answer?

**maddas** · May 9th 2010, 12:35 PM

(a) is not true, take $f(x) = 2x$ and $g(x) = x+1$ .

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

**Plato** · May 9th 2010, 12:44 PM

Originally Posted by dwsmith

Let f and g be twice differentiable real-valued functions defined on $\mathbb{R}$ . If f'(x)>g'(x) $\forall x>0$ , 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 $f-g$ is $f '-g '>0$ .
Therefore $f-g$ is increasing or $f(x)-g(x)>f(0)-g(0)$ .
Thus answer (c).

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