Question 564698
*[tex \LARGE \frac{\Delta(y)}{\Delta(x)} = \frac{e^{3^2} - e^{(-3)^2)}}{3 - (-3)} = 0]


Hence we want to find all points in [-3,3] such that f'(x) = 0. By the chain rule, we have


*[tex \LARGE f'(x) = 2xe^{x^2}]


This is equal to zero only when x = 0 (since e^(x^2) can never equal zero). There is only one point that satisfies.