Answer
a)
$f$ increasing on $(-\infty,1\cup(2,\infty)$
$f$ decreasing on $(1,2)$
b)
$f(1) = 2$ is a local maximum value and
$f(2) = 1$ is a local minimum value.
c)
$f$ is concave upward on $\left(\frac{3}{2},\infty\right)$
$f$ is concave downward on $\left(-\infty,\frac{3}{2}\right)$
There is an inflection point at $\left(\frac{3}{2},\frac{3}{2}\right)$
Work Step by Step
a)
$f(x) = 2x^{3}-9x^{2}+12x-3$
$f'(x) = 6x^{2}-18x+12 = 6(x-2)(x-1)$
$f$ increasing on $(-\infty,1)\cup(2,\infty)$
$f$ decreasing on $(1,2)$
b)
$f$ changes from increasing to decreasing at $x = 1$ and from decreasing to increasing at $x = 2$. Thus,
$f(1) = 2$ is a local maximum value and
$f(2) = 1$ is a local minimum value.
c)
$f''(x) = 12x-18 = 6(2x-3)$
$f''(x) \gt 0$ for $x \gt \frac{3}{2}$
$f''(x) \lt 0$ for $x \lt \frac{3}{2}$
Thus
$f$ is concave upward on $\left(\frac{3}{2},\infty\right)$.
$f$ is concave downward on $\left(-\infty,\frac{3}{2}\right)$.
There is an inflection point at $\left(\frac{3}{2},\frac{3}{2}\right)$.
