Thread: [SOLVED] Derivatives of real valued functions

1. [SOLVED] Derivatives of real valued functions

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?

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

3. Originally Posted by maddas
Integrate $f'>g'$ from 0 to x.
That is just f(x)-f(0) and same for g

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

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

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

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

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

7. 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 $f(x)-g(x)>f(0)-g(0)$.