2 205021 New Member Joined May 23, 2009 Messages 13 Gender Male HSC 2010 May 5, 2011 #1 Hey guys, I sorta know this, but don't know how to properly explain/express it. So a bit of help would be great! f(x) = 3 - e^(-x) Show that f1(x) > 0 for all x what is lim x->infinity f(x) ?
Hey guys, I sorta know this, but don't know how to properly explain/express it. So a bit of help would be great! f(x) = 3 - e^(-x) Show that f1(x) > 0 for all x what is lim x->infinity f(x) ?
D Deep Blue Member Joined Dec 17, 2010 Messages 150 Gender Male HSC 2012 May 5, 2011 #2 If you mean the derivative of f(x) = 3 - e^(-x) is positive for all then then, f'(x) = -e^(-x) * (-1) = e^(-x) As e^(-x) is greater than zero for all x in its domain, the f'(x) > 0 for all x. Part 2: As x becomes very large (x-> infinity) e^(-x) becomes very small and hence the limit as x goes to infinity is f(x) = 3.
If you mean the derivative of f(x) = 3 - e^(-x) is positive for all then then, f'(x) = -e^(-x) * (-1) = e^(-x) As e^(-x) is greater than zero for all x in its domain, the f'(x) > 0 for all x. Part 2: As x becomes very large (x-> infinity) e^(-x) becomes very small and hence the limit as x goes to infinity is f(x) = 3.
