## Calculus 8th Edition

Consider $f:R \to R$ such that $f(x)=|x^{2}+x|$ Since, $x^{2}+x\gt 0$ for all $x∈(-\infty,-1)∪(0,+\infty)$ we have that $f(x)=|x^{2}+x|=x^{2}+x$ for all $x∈(-\infty,-1)∪(0,+\infty)$ Hence, $f$ is differniable on $(-\infty,-1)∪(0,+\infty)$ and $f'(x)=2x+1$ for all $x∈(-\infty,-1)∪(0,+\infty)$ Take $x=-3$ $f'(-3)=2(-3)+1=-6+1=-5\ne 5=|2(-3)+1|$ Hence, the given statement is false.