Applications of derivatives

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Applications of Derivatives 1. Critical Number: A number is known as a critical number of a function if exists or does not exist Working Rule Step 1: Find and factor it. Step 2: Set and solve for x. These x values are the candidates for critical numbers. Step 3: The values from step 2 will be the critical numbers if exist at these points. Solved Examples Find the critical numbers of the function Solution: We have Step 1: Find and factor it. Step 2: Set Step 3: exists. Thus is the critical number of the function Find the critical numbers of the function Solution: Step 1: Find and factor it.

Transcript of Applications of derivatives

Page 1: Applications of derivatives

  Applications of Derivatives

1. Critical Number: A number is known as a critical number of a function if exists or does not exist

Working RuleStep 1: Find and factor it.Step 2: Set and solve for x. These x values are the candidates for critical numbers.Step 3: The values from step 2 will be the critical numbers if exist at these points.

Solved Examples

Find the critical numbers of the function Solution: We have Step 1: Find and factor it. Step 2: Set

Step 3: exists.

Thus is the critical number of the function

Find the critical numbers of the function Solution: Step 1: Find and factor it. Step 2: Set

Step 3: and both exist.

Thus are the critical numbers of the function.

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Find the critical numbers of the function

Solution: Step 1: Find and factor it.

Step 2: Set

Step 3: exists.Thus is the critical number of the function. Find the critical numbers of the function

Solution: Step 1: Find and factor it.

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Step 2: Set

Also does not exist for

Candidates for critical numbers are

Step 3: and exist.

Thus are the critical numbers of the function.

Find the critical numbers of the function Solution: Step 1: Find and factor it.

Step 2: Set

Step 3: exists.

Thus is the critical number of the function.

Find the critical numbers of the function

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Solution: Step 1: Find and factor it. Step 2: Set

Step 3: exists.

Thus is the critical number of the function.

Find the critical numbers of the function

Solution: Step 1: Find and factor it.

Step 2: Set Also dose not exist at Candidates for critical numbers are

Step 3: and exist but does not exist.Thus are the critical numbers of the function.

2. Intervals of Increasing and Decreasing Functions

A function is increasing on an interval if for any x1 and x2 in the interval then

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A function is decreasing on an interval if for any x1 and x2 in the interval then

How does this relate to derivatives?  Recall that the derivative is the limit 

                            

If  x1 <  x2, then the denominator will be positive.  Now if

, then the numerator will be positive, hence the derivative will be positive.

, then the numerator will be negative and the derivative will be negative.

A function is called an increasing function in an interval if is differentiable on the interval .

for all values of .

A function is called a decreasing function in an interval if is differentiable on the interval .

for all values of .

Working RuleStep 1: Find and factor it.Step 2: Set and solve for x. These x values will divide real number line into intervals I.

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Step 3: If for all , then the function is increasing on I. If for all , then the function is decreasing on I.

Solved Examples

Determine the intervals on which the function is increasing or decreasing.Solution: Set Here the intervals are .

Intervals Test Point ConclusionFunction is decreasing

Function is increasing

3. Local (Relative) Minimum and Local Maximum Values

The First Derivative Test

Suppose c is a critical number of a continuous function f. (a) If changes from to at c, then the function f has a local

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maximum or relative maximum at a point c.(b) If changes from to at c, then the function f has a local minimum or relative minimum at a point c.(c) If does not change the sign at c, then the function f has no local minimum or local maximum at c.

Find the local maximum and minimum values of the function .Solution:

Set Since and exist, therefore are the critical numbers of the function.These points divide the domain of the function into three intervals: .

Intervals Test Point Conclusion

Since changes from to at 3, so the function has a local minimum at x = 3.

Local minimum value is

4. Absolute Minimum and Absolute Maximum Values Closed Interval Method

To find the absolute maximum and minimum values of a continuous function f on a closed interval [a, b]:

1. Find the values of f at the critical numbers of f in (a, b).2. Find the values of f at the end points of the interval.3. The largest of the values from step 1 and 2 is the absolute maximum value;

the smallest of the values from step 1 and 2 is the absolute minimum value.

Find the absolute and local maximum minimum values of the function , Solution: Set

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is the only critical number.

Now Absolute maximum value = 5Absolute minimum value = -7

Concavity

A function is called concave upward on an interval I if is an increasing function on I and is called concave downward on I if is decreasing.

A point where a curve changes its direction of concavity is called an inflection point.

Concavity Test

o If for all x in the interval I, then the graph of is concave upward on I.o If for all x in the interval I, then the graph of is concave downward on I.

Second Derivative TestSuppose is continuous near c.

o If and , then the graph of has a local minimum at c.o If and , then the graph of has a local maximum at c.

Please remember

o Concave down – does not hold watero Concave up – does hold water

Test for Concavity

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Mr. Happy Mr. Sad Mr. Quiet Concave up Concave down No ConcavitySecond Derivative is Positive Second Derivative is Negative Second Derivative is Zero      Example: Discuss the curve with respect to concavity, points of inflection and local extrema.Solution:

Set Since and exist, therefore are the critical numbers of the function.For second derivative test, we need to evaluate at these critical numbers.We have

and Since and , therefore f has a local minimum at and local minimum value is .Since , the second derivative test gives no information about the critical number 0.We also notice from the table that the function has no local min or local max for 0< x< 3.Intervals Test Point Conclusion

The function is decreasing

The function is decreasing. No change

Set These points divide the domain of into three intervals:

. Intervals Test Point Conclusion

Concave upwardConcave downward

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Concave upwardAt x = 0 and 2, the curve changes its direction of concavity.

, Here points of inflection are (0, 0) and (2,-16).