If you know how to derive you can calculate f'(x) and find the variations of f and show that f has a minimum for x=0
show that the function f(x) = e^x-x-1 has an absolute value and deduce that
e^x>= x+1 for all x inside R.
for the first part of the question, i tried
f'(x)= e^x -1 = 0 => x = ln1 = 0 ==> x=0, y=0
but how we know that it is an absolute min since the range of the function is not given??? then how to solve the second part of the question??? plz help me...