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

Recall that

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

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