How do you find the relative extrema of a function?

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Question 3: How do you find the relative extrema of a function?
The strategy for tracking the sign of the derivative is useful for more than determining
where a function is increasing or decreasing. It is also useful for locating the relative
extrema of a function. At a relative extrema, a function changes from increasing to
decreasing or decreasing to increasing. The number lines in the previous question allow
us to see these changes by observing changes in the sign of the derivative of a
function.
When the derivative of a function changes from positive to negative, we know the
function changes from increasing to decreasing. As long as the function is defined at the
critical value where the change occurs, the critical point must be a relative maximum. If
the derivative of a function changes from negative to positive, we know the function
changes from decreasing to increasing. In this case, the critical point is a relative
minimum as long as the function is defined there. If the derivative does not change sign
at a critical value, there is no relative extrema at the corresponding critical point.
The First Derivative Test summarizes these observations and helps us to locate relative
extrema on a function.
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First Derivative Test
Let f be a non-constant function that is defined at a critical
value x  c .
If f  changes from positive to negative at x  c , then a
relative maximum occurs at the critical point  c, f (c)  .
If f  changes from negative to positive at x  c , then a
relative minimum occurs at the critical point  c, f (c)  .
If f  does not change sign at x  c , then there is no relative
extrema at the corresponding critical point.
Example 6
Find the Relative Extrema of a Function
Find the location of the relative extrema of the function
f ( x)  4 x3  21x 2  18 x  5
Solution The first derivative test requires us to construct a number line
for the derivative so that we can identify where the graph is increasing
and decreasing. Using the rules for derivatives, the first derivative of the
function f ( x) is
f ( x)  4
d 3
d
d
d
 x   21  x 2   18  x   5
dx
dx
dx
dx
 4  3x 2   21 2 x   18 1  0
Use Sum / Difference Rule and
the Product with a Constant
Rule.
Use the Power Rule for
Derivatives and the fact that the
derivative of a constant is zero
So the derivative is f ( x)  12 x 2  42 x  18 .
21
We need to use this derivative to find the critical values. Set the
derivative, f ( x)  12 x 2  42 x  18 , equal to zero to find those values.
12 x 2  42 x  18  0
Set the derivative equal to zero
6  2 x 2  7 x  3  0
Factor the greatest common factor from
each term
6  2 x  1 x  3  0
2x 1  0
x 3  0
2x  1
x
x3
1
2
Factor the trinomial
To find where the product is
equal to zero, set each factor
equal to zero and solve for the
variable
In general, critical values may also come from x values where the
derivative is undefined. Since f ( x) is a polynomial, it is defined
everywhere so the derivative is defined everywhere.
Although this derivative could be factored to find the critical values,
most quadratic derivatives are not factorable. In this case, the quadratic
equation yielding the critical values can be solved using the quadratic
formula. This strategy would yield the same critical values as factoring:
2
12
42 x  18
x 

0
a
x
  42  

42  900
24

42  30
24
b
Identify a, b, and c for the quadratic
c
 42   4 12 18
2 12 
2
formula x 
b  b 2  4ac
and put
2a
in the values
Simplify the numerator and denominator
 3, 12
To find the critical values with more complicated derivatives, we may
need to solve the equation using a graph. The solution to the equation
22
12 x 2  42 x  18  0 can also be found by locating the x intercepts on the
derivative f ( x ) .
=0
=0
1
2
3
f ( x)  6  2 x  1 x  3
Like factoring or the quadratic formula, the critical values are located at
x  12 and x  3 . All three strategies yield the same critical values. Keep
in mind that a graph will give approximate values while factoring or the
quadratic formula yield exact values.
If the derivative is not factorable, linear, or quadratic, another method
will need to be used to determine where the derivative is equal to zero.
Even though a graph of the derivative only gives an estimate of the
critical values, it may be the only way to find the critical values if the
derivative is complicated.
With the critical values in hand, label them on a number line so that we
are able to apply the first derivative test.
If we select a test point in each interval and determine the sign of each
factor, we can complete the number line and track the sign of the
derivative.
23
(+)(-)(-) = +
=0
(+)(+)(-) = -
=0
(+)(+)(+) = +
increasing
1
2
decreasing
3
increasing
f ( x)  6  2 x  1 x  3
f ( x )
When a continuous function changes from increasing to decreasing, we
have a relative maximum at the critical value. When a continuous
function changes from decreasing to increasing, we have a relative
minimum at the critical value. In this case, the relative maximum is
located at x  12 and the relative minimum is located at x  3 .
To find the ordered pairs for the relative extrema, we need to substitute
the critical values into the original function f ( x) to find the
corresponding y values:
 12   4  12 
3
 21 12   18  12   5  374
2
Relative Maximum:
f
Relative Minimum:
f  3  4  3  21 3  18  3  5  22
The relative maximum is located at
3
 12 , 374 
2
and the relative minimum is
located at  3, 22  .
24
Example 7
Find the Relative Maximum
Find the location of the relative maximum of the function
g ( x) 
x
ex
Solution The derivative of g ( x) is found with the Quotient Rule for
Derivatives with
ux
v  ex
u  1
v  e x
Put these expressions into the Quotient Rule:
g ( x) 
e x 1  x  e x 
e 
x 2
The Quotient Rule is
d  u  vu  uv

dx  v 
v2
To make it easier to find the critical values, simplify the derivative.
e x 1  x  e x 
e 
x 2

e x 1  x 
x
Factor e from the numerator
e 
x 2
1 x
 x
e
x
Reduce the common e factor in the
numerator and denominator
Now we can find the critical values by determining where this fraction is
equal to zero or undefined.
Any fraction is equal to zero where the numerator is equal to zero. In
this case, this is where 1  x  0 or x  1 . Fractions are undefined where
the denominator is equal to zero. The denominator for this fraction is e x
and is always positive so there are no x values where the fraction is
undefined.
25
To apply the first derivative test, label a number line with this critical
value and test the first derivative on either side of the critical value:



=0
1
increasing



f ( x) 
1 x
ex
decreasing
Since this function is continuous and the derivative changes from
increasing to decreasing at x  1 , the critical point is a relative
maximum. The y value for the relative maximum comes from the
function g ( x) 
x
1
and is g (1)  1 . The ordered pair for the relative
x
e
e
maximum is 1, 1e  .
Example 8
Find the Minimum Average Cost
The total daily cost to produce Q units of a product is given by the
function
C  Q   0.005Q 2  10Q  1000 dollars
The average cost function C  Q  is found by dividing the total daily cost
function C  Q  by the quantity Q.
a. Find the average cost function C  Q  .
Solution The average cost function is defined by C  Q  
C Q 
.
Q
Substitute the total daily cost function into the numerator of this fraction
to yield
26
0.005Q 2  10Q  1000
C Q  
Q
b. Find the quantity that yields the minimum average cost.
Solution The minimum average cost is found by locating the relative
minimum of the average cost function. To use the first derivative test to
find this relative minimum, we need to take the derivative of C  Q  using
the Quotient Rule for Derivatives. The numerator, denominator and their
derivatives are
u  0.005Q 2  10Q  1000
vQ
u  0.010Q  10
v  1
The derivative of the average cost function is
C  Q  
uv
vu 



2
Q  0.010Q  10    0.005Q  10Q  1000  1
2
Q

v2
Since we need to use the derivative to find the critical values, we
simplify the derivative as much as possible:
C  Q  
0.010Q 2  10Q  0.005Q 2  10Q  1000
Q2
0.005Q 2  1000

Q2
Simplify the numerator by carrying
out the multiplication and
subtraction
Combine like terms
Any fraction is equal to zero when the numerator is equal to zero and
undefined where the denominator is equal to zero.
27
Set the numerator equal to 0
0.005Q 2  1000  0
0.005Q 2  1000
Q 2  200000
Q   200000
Q  447
Set the denominator equal to 0
Q2  0
Q0
Since quantities produced must be positive, only Q  447 is a
reasonable critical value for this function.
The number line for this function only includes positive quantities.
Testing on either side of the critical numbers yields the behavior of C  .



0
decreasing
=0
447



C  Q  
0.005Q 2  1000
Q2
increasing
Since the function decreases on the left side of the critical value and
increases on the right side of the critical value, the quantity at
approximately 447 units is a relative minimum.
The minimum average cost is obtained from
C Q  
0.005Q 2  10Q  1000
and is calculated as
Q
28
C


200000 
0.005

200000

2
 10


200000  1000
200000
 14.47
Since the total daily cost is in dollars and we are dividing by the number
of units to get the average cost, the units on the average cost are
dollars per unit. This means that a production level of about 447 units
gives the lowest average cost of 14.47 dollars per unit.
Figure 6 - The relative minimum for the average cost
function.
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