Help with the interval question

**450081592** · November 15th 2009, 05:00 PM

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 $\in$ (0,c), then f(x) > g(x) for all x $\in$ (0,c).

**Jose27** · November 15th 2009, 05:22 PM

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

**450081592** · November 15th 2009, 05:56 PM

Originally Posted by Jose27

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

Sorry, We havent cover the integral material yet, we need to prove it by the mean value theorem or increasing or decreasing function properties.

**Jose27** · November 15th 2009, 06:12 PM

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

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