# Thread: Help with the interval question

1. ## Help with the interval question

Assum that f and g are differentiable on the interval (-c, c) and f(0) = g(0).

show that if f'(x) > g'(x) for all x $\displaystyle \in$ (0,c), then f(x) > g(x) for all x $\displaystyle \in$(0,c).

2. By the properties of the integral we have $\displaystyle \int_{0} ^{x} f'(t)dt > \int_{0}^{x} g'(t)dt$ which by the FTC is equivalent to $\displaystyle f(x)-f(0)>g(x)-g(0)$ and since $\displaystyle f(0)=g(0)$ we have $\displaystyle f(x)>g(x)$ for all $\displaystyle x \in (0,c)$

3. Originally Posted by Jose27
By the properties of the integral we have $\displaystyle \int_{0} ^{x} f'(t)dt > \int_{0}^{x} g'(t)dt$ which by the FTC is equivalent to $\displaystyle f(x)-f(0)>g(x)-g(0)$ and since $\displaystyle f(0)=g(0)$ we have $\displaystyle f(x)>g(x)$ for all $\displaystyle x \in (0,c)$
Sorry, We havent cover the integral material yet, we need to prove it by the mean value theorem or increasing or decreasing function properties.

4. Assume there is $\displaystyle d\in (0,c)$ such that $\displaystyle (f-g)(d)=0$ then by Rolle's theorem there exists a point $\displaystyle x_0\in (0,d)$ such that $\displaystyle (f-g)'(x_0)=0$ which is a contradiction since $\displaystyle (f-g)'(x)>0$ in $\displaystyle (0,c)$ and this also implies that in a right neighbourhood of $\displaystyle 0$ we have that $\displaystyle f-g$ is positive. And this is enough because since $\displaystyle f-g$ is continous, if it were negative there would exist a point in which it's zero and we repeat the argument. So $\displaystyle f-g$ is positive in $\displaystyle (0,c)$