## Calculus: Early Transcendentals 8th Edition

$f$ is increasing on the interval $(3, \infty)$
The graph is increasing when $f'(x) \gt 0$ $f'(x) = (x+1)^2(x-3)^5(x-6)^4$ $(x+1)^2$ is $0$ when $x = -1$ and positive for all other values of $x$. $(x-3)^5$ is $0$ when $x = 3$, positive when $x \gt 3$, and negative when $x \lt 3$ $(x-6)^4$ is $0$ when $x = 6$ and positive for all other values of $x$. Thus $f'(x) \geq 0$ for all $x$ such that $x \gt 3$ Note that $f'(x)= 0$ at $x=6$. However, this is an inflection point, not a point where the graph changes from increasing to decreasing. Therefore, we can say that $f$ is increasing on the interval $(3, \infty)$