
Concept explainers
In Problems 57-64, use a graphing utility to graph each function over the indicated interval and approximate any
60.

To find: The following values using the given graph,
a. Draw the graph using graphing utility and determine the local maximum and minimum values.
Answer to Problem 56AYU
a. Local maximum point is and local minimum is .
Explanation of Solution
Given:
It is asked to draw the graph using graphing utility and determine the local maximum and minimum values and also find the increasing and decreasing intervals of the given function.
Calculation:
a. By the definition of local maximum, “Let be a function defined on some interval . A function has a local maximum at if there is an open interval containing so that, for all in this open interval, we have . We call a local maximum value of ”, It can be directly concluded from the graph and the definition that the curve has local maximum point at .
The value of the local maximum at is .
Therefore, the local maximum point is .
By the definition of local minimum, “Let be a function defined on some interval . A function has a local minimum at if there is an open interval containing so that, for all in this open interval, we have . We call a local minimum value of ”, It can be directly concluded from the graph and the definition that the curve has local minimum point at .
The value of the local minimum point at is .
Therefore, the local minimum point is .

To find: The following values using the given graph,
b. Increasing and decreasing intervals of the function .
Answer to Problem 56AYU
b. The function is increasing in the intervals and , the function is decreasing in the intervals and . There is no constant interval in the given graph.
Explanation of Solution
Given:
It is asked to draw the graph using graphing utility and determine the local maximum and minimum values and also find the increasing and decreasing intervals of the given function.
Calculation:
b. Increasing intervals, decreasing intervals and constant interval if any.
It can be directly concluded from the graph that the curve is decreasing from to , then increasing from to 0, then decreasing from 0 to and at last increasing from to 2.
Therefore, the function is increasing in the intervals and , the function is decreasing in the intervals and . There is no constant interval in the given graph.
Chapter 2 Solutions
Precalculus
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