9367 54 proof.
Posted: Mon Dec 19, 2011 12:17 pm
Hi,
How do we prove the answer for 9367 Q54?
If f and g are real-valued differentiable functions and if f'(x)>=g'(x) for all x in the closed interval [0,1], which of the following must be true?
(C) f(1)-g(1)>=f(0)-g(0)
Can we prove it by
$$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}\geq\lim_{x\to 0}\frac{g(x)-g(0)}{x-0}$$
$$\therefore f(x)-f(0)\geq g(x)-g(0)$$, for all x in [0,1]
$$\therefore f(1)-f(0)\geq g(1)-g(0)$$
Thanks.
How do we prove the answer for 9367 Q54?
If f and g are real-valued differentiable functions and if f'(x)>=g'(x) for all x in the closed interval [0,1], which of the following must be true?
(C) f(1)-g(1)>=f(0)-g(0)
Can we prove it by
$$\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}\geq\lim_{x\to 0}\frac{g(x)-g(0)}{x-0}$$
$$\therefore f(x)-f(0)\geq g(x)-g(0)$$, for all x in [0,1]
$$\therefore f(1)-f(0)\geq g(1)-g(0)$$
Thanks.