score on last try: 0 of 2 pts. see details for more.\nat least one scored part is incorrect. jump to first…

score on last try: 0 of 2 pts. see details for more.\nat least one scored part is incorrect. jump to first changeable incorrect part.\nyou can retry this question below\nthe function graphed above has:\npositive derivative on the open interval(s)\nnegative derivative on the open interval(s)\nif an answer contains more than one interval, write your answer as a union of intervals.\nadd work

score on last try: 0 of 2 pts. see details for more.\nat least one scored part is incorrect. jump to first changeable incorrect part.\nyou can retry this question below\nthe function graphed above has:\npositive derivative on the open interval(s)\nnegative derivative on the open interval(s)\nif an answer contains more than one interval, write your answer as a union of intervals.\nadd work

Answer

Explanation:

Step1: Recall derivative - slope relationship

The derivative of a function at a point is the slope of the tangent line at that point. A positive - slope means a positive derivative and a negative - slope means a negative derivative.

Step2: Identify positive - slope intervals

Looking at the graph, the function has a positive slope (is increasing) on the open interval where the graph is going up as we move from left to right. From the graph, the function is increasing on the interval $(-3,0)$.

Step3: Identify negative - slope intervals

The function has a negative slope (is decreasing) on the open intervals where the graph is going down as we move from left to right. From the graph, the function is decreasing on the intervals $(-\infty,-3)\cup(0,\infty)$.

Answer:

Positive derivative on the open interval(s): $(-3,0)$ Negative derivative on the open interval(s): $(-\infty,-3)\cup(0,\infty)$