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Concept explainers
In Problems 57-64, use a graphing utility to graph each function over the indicated interval and approximate any
58.
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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 54AYU
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:
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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 5.
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 2.
Therefore, the local minimum point is .
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To find: The following values using the given graph:
b. Increasing and decreasing intervals of the function .
Answer to Problem 54AYU
b. The function is increasing in the intervals and and the function is decreasing in the interval . 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:
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b. Increasing intervals, decreasing intervals and constant interval if any.
It can be directly concluded from the graph that the curve is increasing from to 0, then decreasing from 0 to 2 and at last increasing from 2 to 3.
Therefore, the function is increasing in the intervals and and the function is decreasing in the interval . There is no constant interval in the given graph.
Chapter 2 Solutions
Precalculus
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