Quick Question (1 Viewer)

[205021]

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) ?

 

[Deep Blue]

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.