Question

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

Short answer

The given sign chart is

Step-by-step solution

Step 1. Given Information. 

The given sign chart is

Step 2. Sketch the graph of f. 

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.

Step 3. The graph of f. 

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