Answer
$f(x)$ is concave up on the interval $-\infty < x < 1$ and concave down on the interval $1 < x < \infty$
Work Step by Step
$f(x) = -x^3 + 3x^2$
1.) Find $f''(x)$
$f'(x) = -3x^2 + 6x$
$f''(x) = -6x + 6$
2.) Determine the intervals for which $f''(x)$ is positive and negative.
Looking at the timeline below, $f''(x)$ is positive on the interval $-\infty < x < 1$ and is negative on the interval $1 < x < \infty$. Therefore, $f(x)$ is concave up on the interval $-\infty < x < 1$ and concave down on the interval $1 < x < \infty$
