How do you find all intervals where the function #f(x)=e^(x^2)# is increasing?

Question

How do you find all intervals where the function #f(x)=e^(x^2)#  is increasing?

Emily Ammerman · Accepted Answer

Investigate the sign of #f'(x)#.

On intervals on which #f'(x)# is positive (#>0#), #f(x)# is increasing. #f(x)=e^(x^2)# #f'(x)=2xe^(x^2)# Because #f'(x)# is never undefined, it could possibly change sign only at #x# values where #f'(x) =0# #2xe^(x^2) = 0# if and only if #2x = 0#, so #x = 0# # or #e^(x^2) = 0# but #e^n# is never #0# for any #n#. Since #2e^(x^2) > 0# for all #x#, the sign of #f'(x)# is the same as the sign of #x# which is (of course) negative for #x<0# and positive for #x>0#. So #f# is increasing on the interval #(0, oo)#. (In my experience the usual practice is to state open intervals on which a function is increasing. It is also true that this function is increasing on the closed interval: #[0,oo)#.)

Adam Adkins · Answer

undefined To find all intervals where the function $ f(x) = e^{x^2} $ is increasing, we need to examine its derivative.

The derivative of $ f(x) = e^{x^2} $ with respect to $ x $ can be found using the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

So, if we let $ u = x^2 $, then $ \frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} $.

Thus, $ \frac{df}{dx} = e^{x^2} \cdot 2x $.

To find where $ f(x) $ is increasing, we need to find where $ f'(x) $ is positive. Since $ e^{x^2} $ is always positive, the sign of $ f'(x) $ is determined solely by the sign of $ 2x $.

When $ 2x > 0 $, which means $ x > 0 $, $ f'(x) > 0 $, and thus $ f(x) $ is increasing.

So, the function $ f(x) = e^{x^2} $ is increasing on the interval $ (0, \infty) $.