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)


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) = 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


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