For each set of sign charts in Exercises 53–62, sketch a possible graph of f.

The given sign chart is

To sketch the possible graph off, we will use theorem 3.6 and 3.10.

Theorem 3.6 states that the Derivative Measures Where a Function is Increasing or Decreasing, letf be a function that is differentiable on an intervalI.

(a) If $f\text{'}$is positive in the interior of I, then f is increasing onI.

(b) If $f\text{'}$is negative in the interior of I, then f is decreasing onI.

(c) If $f\text{'}$is zero in the interior of I, then f is constant on I.

Theorem 3.10 states that the Second Derivative Determines Concavity, suppose both f and$f\text{'}$ are differentiable on an interval I.

(a) If $f\text{'}\text{'}$ is positive on I, thenf is concave up on I.

(b) If $f\text{'}\text{'}$is negative onI, thenf is concave down onI.

From the given chart, we conclude that

$f\text{'}$is positive on the intervals $\left(-\infty ,-2\right)and\left(3,\infty \right)$and negative on the intervals $\left(-2,1\right)and\left(1,3\right).$ Thus, f will we increase on the positve intervals and decrease on the negative intervals.

$f\text{'}\text{'}$is positive on the intervals $\left(-1,1\right)and\left(2,\infty \right)$and negative on the intervals $\left(-\infty ,-1\right)and\left(1,2\right).$Thus, fwill be concave up on the positive intervals and concave down on the negative intervals.

The graph has roots at $x=-3,1,4.$A local minimum at $x=3$and inflection points at$x=-1,1,2.$

The graph is