# Calculus / general math related question

1. May 24, 2015

### PsychonautQQ

Let f and g be twice differentiable real-valued functions. If f'(x) > g'(x) for all values of x, which of the following statements must be true?

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 correct answer is C. Can somebody explain to me why the other answers don't work and why C works?

2. May 24, 2015

### pasmith

A counterexample is $f(x) - g(x) = x$. Then $f'(x) - g'(x) = 1 > 0$ for all $x$ yet $f(-1) - g(-1) = -1 < 0$. Hence $f(-1) < g(-1)$.

A counterexample is $f'(x) = 1$ and $g'(x) = \tanh(x)$. Then $f'(x) > g'(x)$ for all $x$, but $f''(x) = 0$ whilst $g''(x) = \mathrm{sech}^2(x) > 0$ for all $x$.

I don't think this one is actually true.

If $f(0) = g(0)$ then this is exactly the same as (A), which was shown to be false. Indeed, the counterexample given for (A) also works here since if $f(x) - g(x) = x$ then $f(0) - g(0) = 0$ so that $f(0) = g(0)$.

What is true is that if $f'(x) > g'(x)$ for all $x$ then $f(x) - f(0) > g(x) - g(0)$ for all positive $x$.

Rearranging gives $f'(x) - g'(x) > f'(0) - g'(0)$.

A counterexample is $f'(x) - g'(x) = \mathrm{sech}(x)$: Then $f'(x) - g'(x) > 0$ for all $x$ but $f'(0) - g'(0) = 1$ is maximal.

Rearranging, $f''(x) - g''(x) > f''(0) - g''(0)$.

A counterexample is $f'(x) - g'(x) = 1 + \tanh(x)$. Then $f'(x) > g'(x)$ for all $x$. But $f''(x) - g''(x) = \mathrm{sech}^2(x)$ so that $f''(0) - g''(0) = 1$ is maximal.