Relative (aka Local) Extrema:
Before I go into the details of what an extrema is, I think a definition of an extrema would be a nice way to start.
Extrema- a maximum or minimum value of a function
A) Draw a number line with the x value(s): <-----x1-------x2-----xn----->
B) Use numbers before and after the calculated x value, plugged into the first derivative equation,
to determine whether the numbers before and after are positive or negative.
C) If the number before is positive and the number after is negative,
the extrema is a local maximum.
D) If the number before is negative and the number after is positive,
the extrema is a local minimum.
E) On the occasion the number before and after is the same sign (positive or negative)
the number is not a local maximum or minimum.
F) The reason the first derivative test is true is because the first derivative is a slope the function.
The point where the slope changes from positive to negative is a maximum because the
function increases as much as it possibly can before its slope equals zero (critical point)
and then decreases after this point because the slope is then negative.
The opposite is true for a minimum.
2. Option 2: The Second Derivative Test
A) The first and most important step of the Second Derivative Test is to calculate
the second derivative.
B) Once the second derivative is calculated, plug in the values of x calculated from setting
the first derivative equal to zero.
C) If the value of the second derivative for the values of x is negative,
the point is a relative maximum because the graph is concave down.
A max must come at the top of a hill, and a hill like this is concave down.
D) If the value of the second derivative for the values of x is positive,
the point is a relative minimum because the graph is concave up.
A max must come at the bottom of a dip, and a dip like this is concave up.
E) On the occasion that the second derivative is equal to zero, the point is neither a relative
maximum or minimum, it is a point of inflection. Inflection points represent a change in
concavity of the graph and do not represent a maximum or a minimum.
Absolute (aka Global) Extrema:
Differences between local and absolute extrema:
Applet demonstrationg optimization: http://www.calculusapplets.com/boxproblem.
Just incase you want a graphical example of local and global exrema (maximums)
Now for some questions!!!!!!!!!! :-)
Multiple Choice Questions:
answers 2, 1, 1
Free Response Questions:
answer: 1: (-4/3, 32/27)
2: (0,0)
3: 2.25
2. A. Determine the constants a and b in order for the function f(x) = x3+ax2+c to have a relative minimum at x = 4 and a point of inflection at x = 1
B. Find the relative maximum of the function found in (A) after plugging in values of a and b, provided that f(0) = 1.
answer:
(A)
f(x) = 3x3+2ax2+b since f has a relative minimum at x=4 then
f '(4) = 3(4)2+2a(4) + b=0
=> 48 + 8a + b=0 (also check if f ''>0, and it is)
Because f has an inflection point at x = 1
=> f ''(1) =0 but f ''(x) =6x+2a
f ''(1)=0 => 6+2a = 0
=> a = -3
Substitute equaiton (2) into equaiton (1) to find b:
48 + 8(-3) + b=0
24+b = 0
b = -24
(B)
When we plug in a = -3 and b = -24 into
f(x) = x3+ax2+c we get
f(x) = x3+3x2+24x+c
But, f(0) = 1
=> 1 = f(0) = 0- 3(0)2-24(0)+c = c
Hence, f(x) = x3+3x2+24x+1
=> f'(x) = 3x2-6x-24
So, f '(x) = 0
=> 3x2-6x-24 = 0
=> x2-2x-8 = 0 (multipying both sides by 1/3)
=> (x-4)(x+2) = 0
=> x = 4 or x = -2
Moreover, f ''(x) = 6x-6
Plugging in the citical points x = 4, x = -2, we see that
=> f ''(4) = 18 > 0 and f '' (-2) = -18 < 0
Hence, f has a relative maximum at x = -2 and a relative minimum at x = 4
So, the realtive maximum of is is
f(-2) = -8 - 12 + 48 + 1
= 29
3. y = f(x) is a function fhwere f ' and f '' exist and have the following characteristics:
x | x<-2 | x=-2 | -2<x<0 | 0<x<2 | x=2 | x>2 |
f ' (x) | + | 0 | - | - | 0 | + |
f '' (x) | - | - | - | + | + | + |
If f(-2) = 8, f(0) = 4 and f(2) = 0, then
(A) Find all inflection points of -2ff
(B) Find all relative minimum and relative maximum values of -2f
(C) Discuss the concavity of -2f
answer:
(A) d2/dx2 (-2f) = -2f ''
From the chart, -2f '' > 0 in (-infinity, 0)
-2f '' < 0 in (0, infinity)
Therefore, -2f has inflection point at x=0
(B)
(i) d/dx (-2f) = -2f ' < 0 in (-infinity, -2) from the chart
-2f '' < 0 in (-2, infinity) from the chart
=> -2f has a relative minimum at x = -2
(ii) By similar argument as in (i), -2f has a relative maximum at x=2 and equals 0
(C)
Again from the table, -2f '' > 0 in (-infinity, 0) and
-2f '' < 0 in (0, infinity)
So, it is concave upward in (-infinity,0) and concave downward in (0,infinity)
Bibliograhy:
Brook, Donald E., Donna M. Smith, and Tefera Worku. Mathematics Calculus AB. 1st. West Piscataway: Research and Education Association, 2000.
http://www.calculusapplets.com/boxproblem.html
http://dictionary.reference.com/search?q=extremum&db=luna
http://img.sparknotes.com/figures/E/eeb0fa64da5c2a1115cdf73c3ef2f0fc/derivapps1.gif