# Math Help - Differentiabl Prob.

1. ## Differentiabl Prob.

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

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

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

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