#### Answer

(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 $f$.

#### Work Step by Step

$$f(x)=x^3-3x^2-9x+4$$
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 $f$.