#### Answer

(a) $f$ increases on $(-\infty, 1)$ and $(2,\infty)$ and decreases on $(1,2)$.
(b) $(1,2)$ is the local maximum and $(2,1)$ is the local minimum of $f$.
(c) $f$ is concave upward on $(\frac{3}{2},+\infty)$ and concave downward on $(-\infty, \frac{3}{2})$
Point $(\frac{3}{2},\frac{3}{2})$ is the inflection point of $f$.

#### Work Step by Step

$$f(x)=2x^3-9x^2+12x-3$$
First, we need to find $f'(x)$ and $f''(x)$.
$$f'(x)=6x^2-18x+12$$ $$f'(x)=6(x^2-3x+2)$$ $$f'(x)=6(x-1)(x-2)$$
$$f''(x)=12x-18$$ $$f''(x)=6(2x-3)$$
(a)
Take $f'(x)=0$ $$6(x-1)(x-2)=0$$ $$x=1\hspace{.5cm}or\hspace{.5cm}x=2$$
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 $(2,+\infty)$, $f'(x)\gt0$, so $f$ would be increasing there.
On $(1,2)$, $f'(x)\lt0$, 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 $x=2$, $f'$ changes from negative to positive. So $f$ has a local minimum at $x=2$.
$$f(1)=2\times1^3-9\times1^2+12\times1-3=2-9+12-3=2$$
$$f(2)=2\times2^3-9\times2^2+12\times2-3=16-36+24-3=1$$
Therefore, $(1,2)$ is the local maximum and $(2,1)$ 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 $(-\infty, \frac{3}{2})$, $f''(x)\lt0$, so $f$ is concave downward.
- On $(\frac{3}{2},+\infty)$, $f''(x)\gt0$, so $f$ is concave upward.
Finally, at $x=\frac{3}{2}$, $f$ changes from concave downward to upward.
$$f(\frac{3}{2})=2\times(\frac{3}{2})^3-9\times(\frac{3}{2})^2+12\times\frac{3}{2}-3=\frac{27}{4}-\frac{81}{4}+18-3=\frac{3}{2}$$
So, point $(\frac{3}{2},\frac{3}{2})$ is the inflection point of $f$.