Answer
$f'(x) < 0$
$f''(x) > 0$
Work Step by Step
1.) Determine the algebraic sign (Either positive or negative) of $f'(x)$
Since the graph of $f(x)$ is decreasing, $f'(x) < 0$ ($f'(x)$ is less than zero and therefore negative)
2.) Determine the algebraic sign of $f''(x)$
Since the graph of $f(x)$ opens up on the interval, $f(x)$ is concave up for this interval and $f''(x) > 0$ (positive). We can also tell that $f''(x)$ is positive because as $f(x)$ decreases it decreases less and less as x increases ($f(x)$ is decreasing at a decreasing rate) which means even though $f'(x)$ is negative, it is increasing and is approaching zero.
