given that f(x) = e^x - x - 1, show that f is an increasing function for x>0
OK i can see that this statement is true... but i just don't know how to write it on paper...
Do you not just differentiate it, and show that is is > 0 on $\displaystyle (1,\infty)$??
So $\displaystyle f'(x) = e^x - 1$
$\displaystyle e^x > 0 $ and $\displaystyle e^x > 1 $ for x > 0
$\displaystyle f'(x) > 0 $ for $\displaystyle x>0$
Therefore
$\displaystyle f(x) $is an increasing function..
Sound right?