Given:- Function f(x) = e^2x

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)

Algorithm:-

(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) = e^2x

⇒

⇒ f’(x) = 2e^2x

For f(x) to be increasing, we must have

⇒ f’(x) > 0

⇒ 2e^2x > 0

⇒ e^2x > 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