Science:Math Exam Resources/Courses/MATH110/December 2013/Question 06 (c)/Solution 1

Recall that

$f^{'} (x) = (3 x^{2} + 4 x + 1) \cdot e^{x^{3} + 2 x^{2} + x - 3} .$

From part (a), we know that $e^{x^{3} + 2 x^{2} + x - 3} \neq 0$ . That implies that $f^{'} (x) = 0$ when $3 x^{2} + 4 x + 1 = 0$ . Factoring, we get $(3 x + 1) (x + 1) = 0$ , which means $x = - \frac{1}{3}, - 1$ . That means there are two solutions for when $f^{'} (x) = 0$ .