Transcript DP Studies Y2

DP STUDIES Y2
Chapter 21 Application of Differential Calculus
CONTENTS
A. Increasing and decreasing functions
B. Stationary points
C. Rates of change
D. Optimization
A: INCREASING AND DECREASING FUNCTIONS
Quick reminder
of notations for
inequalities an
intervals.
A: INCREASING
AND
DECREASING FUNCTIONS
A: INCREASING
AND
DECREASING FUNCTIONS
We can determine intervals where a curve is increasing
or decreasing by considering f'(x) on the interval in
question. For most functions that we deal with in this
course:
A: INCREASING
AND
DECREASING FUNCTIONS
Many functions are either increasing or decreasing for
all x є R . We say these functions are monotone
increasing or monotone decreasing.
A: INCREASING
AND
DECREASING FUNCTIONS
for a strictly increasing function, an increase in x
produces an increase in y
“Monotone increasing is an uphill battle all the way.”
A: INCREASING
AND
DECREASING FUNCTIONS
for a strictly decreasing function, an increase in x
produces a decrease in y.
“It’s all downhill with monotone decreasing.”
EXAMPLE 1
Find the intervals where f(x) is:
a.
Increasing
b.
decreasing
A: INCREASING
AND
DECREASING FUNCTIONS
a. The increasing part is as x
increase, y also increase.
x  1 and 2  x
b. The decreasing part is as x
increase, y decrease.
1  x  2
SIGN DIAGRAMS
Sign diagrams for the derivative are extremely useful for
determining intervals where a function is increasing or
decreasing.
derivative = slope
If the derivative is negative, the function is decreasing.
If the derivative is positive, the function is increasing.
The critical values for f’(x) are the values of x for which
f’(x) = 0 or f’(x) is undefined.
When f’(x) = 0, the critical values are shown on a number
line using tick marks.
When f’(x) is undefined, the critical values are shown with a
vertical dotted line.
SIGN DIAGRAMS
turning point (derivative =0)
SIGN DIAGRAMS
SIGN DIAGRAMS
Example 2:
Find the intervals where f(x) = 2x3 + 3x2 – 12x – 5 is
increasing or decreasing.
SIGN DIAGRAMS
Solution to example 2:
f(x) = 2x3 + 3x2 – 12x – 5
f(x) = 6x2 – 6x – 12
Set f(x) = 0 and solve by factoring or
quadratic formula
6(x2 – x – 2) = 0
6(x – 2)(x + 1) = 0 => x = 2, x = -1
C: STATIONARY POINTS
A stationary point of a function is a point where f’ (x) =
0.
It could be a local maximum, local minimum, or
stationary inflection.
EXAMPLE 3
Find and classify all stationary points of
f(x) = x3 – 3x2 – 9x + 5
f ( x)  x 3  3x 2  9 x  5
f ' ( x)  3x 2  6 x  9
0  3( x 2  2 x  3)
0  3( x  3)( x  1)
x  3, x  1
Stationary points at x = 3 and x = -1
Determine the interval of increasing and
decreasing by
choosing x-values less than -1, between -1
and 3, and
greater 3.
x < -1: (-2 – 3)(-2 + 1) = (–)(–)=+
-1 < x < 3: (0 – 3)(0 + 1) = (–)(+)=–
x > 3: (4 – 3)(4 + 1) = (+)(+) = +
+
–
+
increasing -1 decreasing 3 increasing
local maximum
local minimum
+
increasing
+
–
-1
decreasing
local maximum
3
increasing
local minimum
Example 4:
Find the greatest and least value of y = x3 – 6x2 +
5 on the interval -2 < x < 5.
1.
2.
3.
4.
Take the derivative of the function.
Find the stationary points.
Draw the sign diagram.
Determine the local max. and local min. within the
given interval.
Step 1 : take the derivative
y  x3  6 x 2  5
y '  3 x 2  12 x
Step 2 : set  0 and solve for x
y '  3 x  12 x
2
0  3 x  12 x
2
0  3 x( x  4)
x  0, and x  4
3: Draw the sign diagram
increasing
0  3 x ( x  4)
choose x  0
3(1)(1  4)  positive
choose 0  x  4
3(1)(1  4)  negative
choose x  4
3(5)(5  4)  positive
+
–
+
0
decreasing
local maximum
4
increasing
local minimum
CHECK VALUES
Check y-values at the stationary points and the endpoints.
Stationary point, x = 0
y = 03 – 6(0)2 + 5 = 5
Stationary point, x = 4
y = 43 – 6(4)2 + 5 = -27
Left endpoint, x = -2
y = 23 – 6(-2)2 + 5 = -27
Right endpoint, x = 5
y = 53 – 6(5)2 + 5 = -20
The greatest value, 5, occurs at x = 0.
The least value, -27, occurs at x = 4 and at x = -2
C. RATE OF CHANGE
Example 5:
C. RATE OF CHANGE
Solution to example 5:
C. RATE OF CHANGE
C. RATE OF CHANGE
Example 6
C. RATE OF CHANGE
D. OPTIMIZATION
There are many problems for which we need to find the
maximum or minimum value of a function.
The solution is often referred to as the optimum solution
and the process is called optimization.
D. OPTIMIZATION
We can find optimum solutions
in several ways:
using technology to graph the
function and search
for the
maximum or minimum value
using analytical methods such
as the formula x =-b/2a for the
vertex of a parabola
using differential calculus to
locate the turning points of a
function.
These last two methods are
useful especially when exact
solutions are required.
WARNING!!
The maximum or minimum value does not always occur
when the first derivative is zero. It is essential to also
examine the values of the function at the endpoint(s) of
the interval under consideration for global maxima and
minima.
D. OPTIMIZATION
D. OPTIMIZATION
Example 7:
A 4 liter container must have a square base, vertical
sides, and an open top. Find the most economical
shape which minimizes the surface area of material
needed.
D. OPTIMIZATION
D. OPTIMIZATION
D. OPTIMIZATION
Example 8:
A square sheet of metal 12 cm x 12 cm has smaller
squares cut from its corners as shown.
What sized square should be cut out so that when
the sheet is bent into an open box it will hold the
maximum amount of liquid?
V’ = 12x2 – 96x + 144
V’ = 12(x2 – 8x + 12)
V’ = 12(x – 6)(x – 2)
Set the derivative to 0 and solve
x–6=0

x=6
x–2=0

x=2
If we put the x-values back into the original
equation we get
V = x(12 – 2x)2
at x = 6, V = 6(12 – 2(6)2) = 0
at x = 2, V = 2(12 – 2(2)2) = 8
Therefore, the only value for x is 2. The square we
are cutting out is a 2cm x 2cm.
Example 9:
Solution to example 9: