Theorums about Differentiable Functions

Paige05 · #1 $Old$ November 4th, 2007, 11:35 AM

I have 3 questions that I'm stuck on.. if you have any idea about any of them please help me out!

Thanks!

1. Use the fact that ln x and e^x are inverse functions to show that the inequalities e^x >/= 1+x and ln x </= x-1 are equivalent for x>0.

2. Prove. If f ' (x) </= 1 for all x and f(0)=0, f(x) </= x for all x>/= 0.

3. Suppose that f ' ' (x) >/= 0 for all x in (a,b). We will show the graph of f lies above the tangent line at (x,f(c)) for any c with a<c<b.
a) Use the increasing function theorum to prove that f'(c)</=f'(x) for c</= x < b and that f'(x) </= f'(c) for a<x</=c.
b) Use (a) and the racetrack principle to conclude that f(c) + f'(c)(x-c) </= f(x) for a<x<b.

Thank you sooooooooooo much

ThePerfectHacker · #2 $Old$ November 4th, 2007, 12:48 PM

Quote:

Originally Posted by Paige05 $View Post$

I have 3 questions that I'm stuck on.. if you have any idea about any of them please help me out!

Thanks!

1. Use the fact that ln x and e^x are inverse functions to show that the inequalities e^x >/= 1+x and ln x </= x-1 are equivalent for x>0.

$e^x \geq 1+x$ is equivalent to saying $e^x - x - 1 \geq 0$ so let $f(x) = e^x - x - 1$ on $(0,\infty)$ since $f'(x) = e^x - 1$ and $f'(x)>0$ it means the function is increasing.