## Calculus: Early Transcendentals 8th Edition

(a) $f$ is decreasing on $(-1,3)$ and increasing on $(-\infty,-1)$ and $(3,+\infty)$. (b) $(-1,9)$ is the local maximum and $(3,-23)$ is the local minimum of $f$. (c) $f$ is concave downward on $(-\infty,1)$ and concave upward on $(1,+\infty)$ Point $(1,-7)$ is the inflection point of $f$.
$$f(x)=x^3-3x^2-9x+4$$ First, we need to find $f'(x)$ and $f''(x)$. $$f'(x)=3x^2-6x-9$$ $$f'(x)=3(x^2-2x-3)$$ $$f'(x)=3(x+1)(x-3)$$ $$f''(x)=6x-6$$ $$f''(x)=6(x-1)$$ (a) Take $f'(x)=0$ $$3(x+1)(x-3)=0$$ $$x=-1\hspace{.5cm}or\hspace{.5cm}x=3$$ These are the critical points of $f$. To find the trend of increase or decrease of $f$, it's all about the sign of $f'$. A rather useful method here is to build the sign table for $f'$ using the critical points for the intervals as the image below. As we can see, on $(-\infty,-1)$ and $(3,+\infty)$, $f'(x)\gt0$, so $f$ would be increasing there. On $(-1,3)$, $f'(x)\lt0$, so $f$ would be decreasing on this interval. (b) To find local maximum and minimum, we would use the First Derivative Test. Looking at the table again, at $x=-1$, $f'$ changes from positive to negative. So $f$ has a local maximum at $x=-1$. At $x=3$, $f'$ changes from negative to positive. So $f$ has a local minimum at $x=3$. $$f(-1)=(-1)^3-3\times(-1)^2-9\times(-1)+4=-1-3+9+4=9$$ $$f(3)=3^3-3\times3^2-9\times3+4=27-27-27+4=-23$$ Therefore, $(-1,9)$ is the local maximum and $(3,-23)$ the local minimum of $f$. (c) To find out the concavity, we take a close look at the change of signs of $f''(x)$. A table for the signs of $f''(x)$ is also below. Using the Concavity Test and from the table, we see that - On $(-\infty, 1)$, $f''(x)\lt0$, so $f$ is concave downward. - On $(1,+\infty)$, $f''(x)\gt0$, so $f$ is concave upward. Finally, at $x=1$, $f$ changes from concave downward to upward. $$f(1)=1^3-3\times1^2-9\times1+4=-7$$ So, point $(1,-7)$ is the inflection point of $f$. 