# [SOLVED] Derivatives of real valued functions

#### dwsmith

MHF Hall of Honor
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?

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

#### dwsmith

MHF Hall of Honor
Integrate $$\displaystyle f'>g'$$ from 0 to x.
That is just f(x)-f(0) and same for g

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

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

#### dwsmith

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

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

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

• dwsmith

#### Plato

MHF Helper
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 $$\displaystyle f(x)-g(x)>f(0)-g(0)$$.
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