Derivatives in graphing-dfs

Derivatives in
Graphing and
Applications
Calculus-I
Chapter 4

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4.1 Analysis of Functions- I (PG:232)
 Increase, Decrease and Concavity

Remember-
 To find Increasing/Decreasing functions, always
Read from Left to Right-

4.2 Analysis of Functions-II (pg:244)
 Relative Extrema;
 Graphing Polynomial

Slope of a Tangent line dy/dx
Rate of change and slope

 Secant line = Average Rate of Change = Slope


 A tangent line is a straight line that touches a
function at only one point. (See above.) The
tangent line represents the instantaneous rate
of change of the function at that one point. The
slope of the tangent line at a point on the
function is equal to the derivative of the function
at the same point (See below.)


 In this first animation we see the secant line
become the tangent line i.e we go from the
Average Rate of Change to the Instantaneous
Rate of Change by letting the interval over
which the Average Rate of Change is
measured go to zero.

 Tangent Line = Instantaneous Rate of Change =
Derivative

Finding SLOPE of a line
 Have you ever heard of Mr. SLOPE GUY?

Mnemonics-
Example: To find slopes
 For horizontal lines:
 Slope=0
 Z for zero
 For vertical lines:
 Slope is undefined
 N for No slope

Memorizing the formulas:
 Use- Pascal’s Triangle

Increasing and Decreasing Functions
 Definition: 4.1.1 (pg: 233)
.sin)(),()( 2121 functiongincreaanisxfthenxfxfimpliesxxIf 
.sin)(),()( 2121 functiongdecreaaisxfthenxfxfimpliesxxIf 

Increasing and Decreasing Functions
 Theorem 4.1.2 (pg:233)
 Let f be a function that is continuous on a closed
interval
[a, b] and differentiable on the open interval (a,
b).
(a) If f’(x) > 0 for every x in (a, b), then f is
increasing on (a,b)
(b) If f’(x) < 0 for every x in (a, b), then f is
decreasing on (a,b)
(c) If f’(x) = 0 for every x in (a, b), then f is constant
on (a,b)

Example
Use the graph below to make a conjecture about the
intervals on which f is increasing or decreasing.
21243)( 234
 xxxxf
is decreasing if 2f x  
is increasing if -2 0f x 
is decreasing if 0 1f x 
is increasing if 1f x 

Use increasing/decreasing test to verify your conjecture.
21243)( 234
 xxxxf
xxxxf 241212)( 23/

)1)(2(12)(/
 xxxxf
-2 0 1
- - - - - + + + - - - - + + +
Increasing:
Decreasing:

Continuous function
 Find the intervals on which the function f(x) is
increasing and where it is decreasing.

 Find the intervals on which the function f(x) is
increasing and where it is decreasing.

4.2 CRITICAL POINTS (pg:245)
 Critical Values: The x-values at which the f ‘(x)=0 or
f ‘(x) fails to exist.
 Note: 1. The x-values at which the f ‘(x)=0 are called
Stationary Points.
 Note: The critical values are the points where the
graph will switch from increasing to decreasing or vice
versa.
 Find the critical points for the following functions:
Examples:
1.f(x) = 3x2 – 6x + 3 2. f(x) = x3/2 – 3x + 7
x = 1 x = 4

Finding points where horizontal
tangents to a curve occur

4.2 Exp: 3 (pg:245)
 Find all critical points of
3/23/5
153)( xxxf 

Qs: 7 -14 Ex: 4.2 (pg: 252)
 Locate the critical points and identify which critical
points are stationary points.
 Qs: 8
 Qs: 9
x12x3)x(f 4

3x
1x
)x(f 2




Key Topics
 Relative maximum: the highest value for f(x) at that
particular “peak” in the graph
 Relative minimum: the lowest value for f(x) at that
particular “valley” in the graph
Relative maximum
Relative maximum
Relative minimum Relative minimum
Relative minimum

Key Topics
 How to determine whether it is a relative maximum or
a relative minimum at a focal point:
Step 1: Find the points of the graph to determine the
intervals on which f(x) is increasing or decreasing
Step 2: Choose an x-value in each interval to determine
whether the function is increasing or decreasing within that
interval
Step 3: If f(x) switches from increasing to decreasing at a
critical point, there is a relative maximum at that critical
point
If f(x) switches from decreasing to increasing at a critical
point, there is a relative minimum at that critical point.

Key Topics
 It might help to make a number line displaying
your findings
- - - | +++++++ | - - - - | +++++++ | - - - - - - - | +++
- to + means minimum
+ to - means maximum

First derivative test to check for
maximum and minimum points
 Let f be a twice differentiable function.
 Find f’(x), and put it equal to zero. Find the critical
points.
 Take test points, to check if f’(x) > 0 or < 0, for
increasing and decreasing function..
 If f ‘(x) changes sign from + to -, then x = c has a
maximum value, and if f changes sign from - to +,
then x = c has a minimum value.
 If f’(x) remains positive, then f is an increasing
function, and if it remains negative, it is a
deceasing function.

Exp: 5 (pg: 248)
Interval
Test values
Sign of f’(x)
Conclusion
Find the Relative Extrema of .
Hint: First find critical points using f’(x)=0.
Then find intervals where the function is inc/dec.
Let’s make a table:
35
x5x3)x(f 

4.1 Concavity- Up and Down
(pg:235)
CONCAVE UP CONCAVE DOWN
 A function f is
concave up on an
open interval if its
tangent lines have
increasing slopes on
that interval.
 In this case, the graph
of f lies above its
tangent lines.
 A function f is
concave down on an
open interval if its
tangent lines have
decreasing slopes on
that interval.
 In this case, the graph
of f lies below its
tangent lines.

concave up:
f’ is increasing.
tangent lines are
below the graph.
concave down:
f’ is decreasing.
tangent lines are
above the graph.

To determine the concavity of a
function using second derivative
Concave upward Concave downward
Think of
some
examples …

Finding Concavity
 Theorem 4.1.4 (pg:235)
 Let f be a function which is twice differentiable on
an open interval (a, b).
(a) If f”(x) > 0 for every x in (a, b), then f is concave
up on (a,b)
(b) If f”(x) < 0 for every x in (a, b), then f is concave
down on (a,b)

4.1 Inflection Points (pg:236)
If f is continuous on an open interval
containing a value x = d, and if f changes the
direction of its concavity at the point (d, f(d)),
then f has an inflection point at x = d.

Inflection Point
To find inflection points, find any point, c, in the domain
where ( ) 0 or ( )f x f x 
changes sign from the left to the right of c,
is undefined.
f If
Then (c, f(c)) is an inflection point of f.

Concavity
Test for concavity
Find the second derivative.
Put f ”(x) = 0 (to find points of inflection). Take test points.
a) If for every value of x in I, then f is concave up
on I.
b) If for every value of x in I, then f is concave
down on I.
0)(//
xf
( ) 0f x 

Second derivative test to check for
Concavity, and point of inflection
 Let f be a twice differentiable function.
 Find f”(x), and put it equal to zero. Find the points
of inflection x= a.
 Take test points, to check if f”(a) > 0 for concave
up or
f”(a) < 0, for concave down.
 Note: Find f”(x); is
undefined, gives points of inflection.
( ) 0 or ( )f x f x 

Example
Given , find the intervals on which f is
increasing/decreasing and concave up/down. Locate all
points of inflection.
13)( 23
 xxxf
Solution
13)( 23
 xxxf
2
( ) 3 6f x x x     3 2x x 
( ) 0f x  at x = 0 and at x = 2
0 2 
+ ++ - - - + ++
Increasing: (-∞, 0) U (2, ∞)
Decreasing: (0, 2)

Sign analysis of
 1
 , 1
 1, 
This will tell us where it is concave up and down:
¢¢f x( )= 6x -6 f “(x) = 0
at x = 1
- - - + ++
Concave Up:
Concave Down:

Questions: Find the interval where f is
increasing/decreasing/concave up/concave
down/ point of inflections, if any.
 Q:1
 Q: 2
 Q: 3
 Q:4
34)( 2
 xxxf
x
exxf 
 )1()(
3
)( xxf 
3/1
x3x2)x(f 

4.2 Second Derivative Test: (pg:247)

4.2 Second Derivative Test: (pg:247)
 Suppose f is twice differentiable at the point x = c.
 (i) If f ’(x) = o and f “(x) > 0, then f has a relative
minimum at x = c.
 (ii) If f ’(x) = o and f “(x) < 0, then f has a relative
maximum at x = c.
 (iii) If f ’(x) = o and f “(x) = 0, then the test is
inconclusive;
that is, f may have a relative maximum, a
relative minimum, or neither at x = c.

Exp: Find the relative extrema using second
derivative test: f”(x)< 0 ---Max and f”(x)> 0 ---
Min
Interval
Test values
Sign of f”(x)
Conclusion
3/22
)4()(  xxf

Test for Concavity
1. If f”(x) > 0 for all x in I, then f is concave up.
2. If f”(x) < 0 for all x in I, then f is concave down.
2nd Derivative Test (Use C.pts. of 1st Derivative)
1. If f”(c) > 0, then f(c) is a relative min.
2. If f”(c) < 0, then f(c) is a relative max.
3. If f”(c) = 0, then the test fails. No min. or max.

9
-9
-16
-27
Ex. 2 Determine where f(x) = x4 – 4x3 is
increasing, decreasing, has max’s or min’s, is
concave up or down, has inflection points.

Ques: 54 Ex:4.2 (pg:
253)
 Draw the graph of the
polynomial, and label
the coordinates of the
intercepts, stationary
points, and inflection
points.
 Check your work with a
graphing utility
(Desmos).
32
xx2x2)x(f 

For Graphing the function:
 Collect all the
information :
 1. Critical points: f '(x) =
0.
 2. Increasing /
Decreasing intervals;
 3. Points of inflection:
f"(x) = 0
 4. Concavity: up/down
 5. End behaviour: Limits
as x tends to plus /
minus infinity (
)
 6. x- intercepts (put y =
0) and y intercepts
 xorx
32
xx2x2)x(f 

Ex. 1 Determine where f(x) = 6(x2 + 3)-1 is
increasing, decreasing, has max’s or min’s, is
concave up or down, has inflection points.
f’(x) = -6(x2 + 3)-2(2x)
 22
3
12



x
x
C.point x= 0
0
+ 
 
.
0,
inc
  
.
,0
dec

1st der.
test
(0,2)
max.

f '(x) 
12x
x2
 3 
2
f "(x) 
x2
 3 
2
12  12x  2  x2
 3 2x 
x2
 3 
4
f "(x) 
12  x2
 3  x2
 3  4x2
 
x2
 3 
4


12  x2
 3 3x2
 3 
x2
 3 
4

36 x2
1 
x2
 3 
3
C.N.’s -1, 1
-1 1

concave up

down

up
 1,  1,1  ,1
Inflection points (-1, ) (1, )2
3
2
3

2nd Derivative Test Plug C.N.’s of 1st der. into 2nd
derivative.
C.N. from 1st der. was 0.  
 32
2
3
136
)("



x
x
xf
 0)0("f 0 is a maximum
Remember, a neg. in the 2nd der. means concave down.
Therefore,
the point is a maximum.

(0,2) max
(-1, 3/2) Inf. pt. (1, 3/2) Inf. pt.
There are no x-intercepts

4.3 Graphs of Rational functions:
(pg:254)
 Properties of Graphs:
 1. Symmetries (replace x by -x and y by -y)
 2. x- intercepts (put y = 0) and y- intercepts (put x
= 0)
 3. Relative extrema (maxima and minima)
 4. Concavity (up or down)
 5. Intervals of increase and decrease (
)
 6. Inflection points (Put and take test
points)
 7. Asymptotes (Horizontal: Take limit as )
 (Vertical: Put denominator = 0, after factoring
and cancellation)
 8. End behavior as
0)x(for0)x(f 
0)x(f 
 xorx

Example: pg:255
 Sketch a graph of


 (Then check and confirm on Desmos)
16x
8x2
y 2
2




x x
40 2x-
( )40 2A x x= -
2
40 2A x x= -
40 4A x¢ = -
0 40 4x= -
4 40x =
10x =
( )10 40 2 10A = - ×
( )10 20A =
2
200 ftA =
40 2l x= -
w x= 10 ftw =
20 ftl =
A Classic Problem
You have 40 feet of fence to enclose a rectangular garden
along the side of a barn. What is the maximum area that
you can enclose?
Example 1

What dimensions for a one liter cylindrical can will
use the least amount of material? (Least amount
of surface area)
We can minimize the material by minimizing the surface a
2
2 2A r rhp p= +
area of
ends
lateral
area
We need another
equation that
relates r and h:
2
V r hp=
( )3
1 L 1000 cm=
2
1000 r hp=
2
1000
h
rp
=
2
2
10 0
2
0
2A r r
rp
p p= + ×
2 2000
2A r
r
p= +
¢A = 4pr -
2000
r2
= 0
Example 2
Now plug in for “h”
2
2000
4 r
r
p=
3
500
r
p
=
5.42 cmr » ( )
2
1000
5.42
h
p
»
10.83 cmh »

Area and Perimeter – Example
Find the dimensions of a rectangular area of 225
square meters that has the least perimeter.
Drawing a figure might help.
l
w
A = l · w = 225
P = 2 · l + 2 · w
From the area equation solve for l and substitute
w
225
l 
P = 2 · l + 2 · w = w2
w
450

P'(w) =
2w2
- 450
w2
=
2(w -15)(w +15)
w2
= 0
There is a critical value at 15 and -15
So w = 15 meters,
and l = 15 meters
(square area)
Perimeter = 60 m

Maximizing Revenue and Profit
A company manufactures and sells x television sets per month.
The monthly cost and price-demand equations are:
R (x) = xp =
C (x) = 60,000 + 60x and p = 200 – x/50
a. Find the maximum revenue.
Maximum at R’(x) = 0
50
200
2
x
x 
R ‘ (x) =
25
200
x

x = 5,000
R(5000) = $500,000

Application #1 continued
P (x) =
b. Find the maximum profit, the production level that will realize
the maximum profit, and the price the company should charge
for each television set.
P ‘ (x) has a critical
value at x =
000,60140
50
2


x
x
P (3500) =
p (3500) =
C (x) = 60,000 + 60x and p = 200 – x/50
P ‘ (x) =
- x
25
+140 = 0
3500.
$185,000
$130 per TV.
50
200
2
x
x R (x) =
P(x)= R(x)-C(x)
P(x) = 200x -
x2
50
æ
è
ç
ö
ø
÷- 60000+60x( )

Derivatives in graphing-dfs

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