1. ## Differentiabl Prob.

True or False: If $\displaystyle f$ and $\displaystyle g$ are differentiable on $\displaystyle [a,b]$ and $\displaystyle |f'(x)| \le 1 \le |g'(x)|$ $\displaystyle \forall x \in (a,b)$, then $\displaystyle |f(x) - f(a)| \le |g(x) - g(a)|$ $\displaystyle \forall$ $\displaystyle x \in [a,b]$

Since $\displaystyle f,g$ are differentiable on $\displaystyle [a,b]$ they are also continuous on $\displaystyle [a,b]$....All i can derive from this problem so far. Help a brother continue. Thank you.

2. Originally Posted by Endowed
True or False: If $\displaystyle f$ and $\displaystyle g$ are differentiable on $\displaystyle [a,b]$ and $\displaystyle |f'(x)| \le 1 \le |g'(x)|$ $\displaystyle \forall x \in (a,b)$, then $\displaystyle |f(x) - f(a)| \le |g(x) - g(a)|$ $\displaystyle \forall$ $\displaystyle x \in [a,b]$

Since $\displaystyle f,g$ are differentiable on $\displaystyle [a,b]$ they are also continuous on $\displaystyle [a,b]$....All i can derive from this problem so far. Help a brother continue. Thank you.
Suppose that $\displaystyle |f(x)-f(a)|>|g(x)-g(a)|$ for some $\displaystyle x\in(a,b)$ then, clearly $\displaystyle \left|\frac{f(x)-f(a)}{x-a}\right|>\left|\frac{g(x)-g(a)}{x-a}\right|$ and since by assumption $\displaystyle f,g$ are differentiable on $\displaystyle [a,x]$ we know that the mean value theorem guarantees us that $\displaystyle |f'(c)|=\left|\frac{f(x)-f(a)}{x-a}\right|>\left|\frac{g(x)-g(a)}{x-a}\right|=|g'(e)|$ for some $\displaystyle c,e\in(a,x)\subseteq (a,b)$...so..