Answer
$f(x)$ is concave up for $-\infty < x < 2$
$f(x)$ is concave down for $2 < x < \infty$
Work Step by Step
$f(x) = -x^3 + 6x^2 - 9x - 1$
1.) Find $f''(x)$
$f'(x) = -3x^2 + 12x - 9$
$f''(x) = -6x + 12$
2.) Find when $f''(x) = 0$
0 = -6x + 12
-12 = -6x
x = 2
3.) Find when $f''(x)$ is positive and negative.
Looking at the timeline below, the graph of $f(x)$ is shown below and on either side of x = 2 is the algebraic sign of $f''(x)$. Since $f''(x)$ is positive for $-\infty < x < 2$ then $f(x)$ is concave up for this interval. Since $f''(x)$ is negative for $2 < x < \infty$ then $f(x)$ is concave down for this interval.
