Show that f(x) = e2x is increasing on R.
Given:- Function f(x) = e2x
Theorem:- Let f be a differentiable real function defined on an open interval (a,b).
(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)
(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)
(i) Obtain the function and put it equal to f(x)
(ii) Find f’(x)
(iii) Put f’(x) > 0 and solve this inequation.
For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.
Here we have,
f(x) = e2x
⇒ f’(x) = 2e2x
For f(x) to be increasing, we must have
⇒ f’(x) > 0
⇒ 2e2x > 0
⇒ e2x > 0
since, the value of e lies between 2 and 3
so, whatever be the power of e (i.e x in domain R) will be greater than zero.
Thus f(x) is increasing on interval R