Transcript f`(x) < 0

Chapter 3 – Spring 2014
Section 3.1 – Critical Numbers and
Absolute Extrema on an Interval
Critical Numbers
Critical numbers are x values on a continuous function at which f’(x) =
0 or f’(x) DNE.
What would these places look like on a graph?
1. Horizontal Tangents (Min/Max/half and half)
2. Vertical Tangents
3. Cusp
Find the critical numbers of the graph
graphically, then algebraically
f ( x)  2 x  3x
2
3
Critical Numbers: x=0 and x=.816
Uses of critical numbers
Critical numbers are locations where relative minimums and maximums
are possible.
They are also possible locations for absolute extrema of a function.
Locating Absolute Extrema on a closed
interval
The location of the absolute minimum and absolute maximum of a
function on a closed interval must be located at one of two places.
1. Critical Numbers
2. Endpoints
27
Find the absolute extrema of f ( x)  x  2 on
2x
the interval [2,6]
Critical Numbers and Absolute Extrema
Homework
P 169 (13,15,16,21,23,30)
Section 3.2 – Rolle’s Theorem and
Mean Value Theorem
Rolle’s Theorem
Rolle’s Theorem states that on a continuous and differentiable interval
between b and c, If f(b) = f(c), Then there is at least one number in between
(we’ll call it d) where f’(d) = 0.
Stated otherwise, if you place two points on a graph with the same y value,
no matter how you connect them there will be at least one point in between
with a horizontal tangent.
Why is Rolle’s Theorem Helpful
Rolle’s Theorem is typically used to prove that a function has to have a
relative minimum or a relative minimum. This is helpful when finding
a derivative and setting it equal to zero is very cumbersome.
Using Rolle’s Theorem
Consider the polynomial f(x)=(x-1)(x-3)(x-6)
Using Rolle’s Theorem, explain why f(x) must have at least two
locations where f’(x) = 0.
One what intervals are the horizontal tangents located? (___,___) and
(___,___)
Places Where Rolle’s Theorem Does Not Apply
Below is the graph of f(x) = cot(x) from –π to π . f(-π/2)=0 and f(π/2)=0.
Rolle’s Theorem states that if you place two points on a graph with the
same y value, no matter how you connect them there will be at least
one point in between with a horizontal tangent.
Why does Rolle’s Theorem not apply here?
The graph is not continuous.
Places Where Rolle’s Theorem Does Not Apply
The graph below is f(x) = lx-2l+1. Why does Rolle’s Theorem not apply
here?
The graph is not differentiable at
all point on the interval.
Determine whether Rolle’s Theorem applies
to the function. If it does, write any intervals
on which it applies. Then find all values of c
on your intervals at which f’(c)=0
f ( x)  x
3
2
1
Determine whether Rolle’s Theorem applies
to the function. If it does, write any intervals
on which it applies. Then find all values of c
on your intervals at which f’(c)=0
f ( x)  ( x  3)( x  1)
2
Rolle’s Theorem Homework
Rolle’s Theorem:
P 176 (1,2,8,11,18,19,29)
Mean Value Theorem
The mean value theorem is an extension of Rolle’s Theorem.
It states that on an interval (a,b) where the function is continuous and
differentiable, the exists some point c where
f (b)  f (a)
f '(c) 
ba
Stated otherwise, if you connect the to endpoints of a curve and find
the slope, there is at least one point on the curve who derivative
matches that slope.
Illustration of the Mean Value Theorem
According to the mean
value theorem there
must be at least one
other point between a
and b with a derivative
equal to that slope.
Our eyes can easily
confirm this is true by
locating the tangent
line with the same
slope.
Using the Mean Value Theorem
Given the function, f(x) = 3 – 8/x, does the Mean Value Theorem apply on
the interval (4,8). If it does, find the values of c where
f (b)  f (a)
f '(c) 
ba
The mean value theorem does apply on the interval (4,8). f’(c) = ¼ at c=5.66
Using the Mean Value Theorem
x2
Given the function, f(x) = x 2  1 , does the Mean Value Theorem apply on
the interval (0,4). If it does, find the values of c where
f (b)  f (a)
f '(c) 
ba
The mean value theorem does not apply on the interval (0,4) because
function is not continuous or differentiable at all points.
Places where the Mean Value Theorem does
not apply
The Mean Value Theorem has the same conditions as Rolle’s Theorem.
1. The function must be continuous.
2. The function must be differentiable.
Using the Mean Value Theorem - Speeding
Police recently installed two traffic cameras for recording license plate
numbers on a highway where the speed limit is 65 MPH. You are the
clerk for a county judge working on a case where a man received the
following ticket in the mail. The man is contesting the ticket on the
grounds that at no point did the police actually record him speeding
with a radar gun. Your job is to provide a legal basis for the ticket to
your boss.
Speeding Violation - $235
1/3/14
License Plate BQG-923
Mile Marker 13 @ 10:12:00 AM
Mile Marker 23 @ 10:20:30 AM
Mean Value Theorem HW
Mean Value Theorem:
P 177 (37,40,42,43,58,59)
Section 3.3 – Increasing and
Decreasing Functions and the First
Derivative Test
What is the Domain of the function
2(x  9)
f ( x)  2
x 4
2
Domain can be written in two ways: 1. Based on what is defined 2. Based on what isn’t defined. In Calculus
we usually write the domain based on what is defined.
Increasing and Decreasing Functions
If f(x) is a continuous and differentiable function on the interval (a,b),
Then:
1. f’(x) > 0 for all x in (a,b) iff f(x) is increasing on the interval (a,b)
2. f’(x) < 0 for all x in (a,b) iff f(x) is decreasing on the interval (a,b)
3. f’(x) = 0 for all x in (a,b) iff f(x) is constant on the interval (a,b)
Where is the function increasing or deceasing?
2(x  9)
f ( x)  2
x 4
2
We describe where the function is increasing and decreasing
the same way we write domain of a function
3 2
f ( x)  x  x
2
3
Determine where the function
is increasing and decreasing without a graph.
Functions change between increasing and decreasing at critical numbers, so let’s find the critical
numbers and then make a chart.
The critical numbers divide our intervals.
Interval
Test #
Sign of
f’(Test #)
Conclusion
First Derivative Test
The first derivative test requires an understanding of increasing and
decreasing functions to determine relative minimums and maximums.
A relative maximum is a point where the function is increasing on the left
and decreasing on the right.
A relative minimum is a point where the
function is decreasing on the left and
increasing on the right.
1
Determine where the function f ( x)  x 3  x is increasing and
decreasing. Then apply the first derivative test to determine
the location of any relative minimums or maximums.
Functions change between increasing and decreasing at critical
numbers, so let’s find the critical numbers and then make a chart.
The critical numbers divide the intervals.
Interval
Test #
Sign of
f’(Test #)
Conclusion
Curve Sketching Using the Derivative
x
-3
-2
-1
0
1
2
3
f’(x)
10
3
0
-1
0
3
10
Remember
f’(x) provides
the slopes,
not y values
Curve Sketching Using the Derivative
x
-3
-2
-1
0
1
2
3
f’(x)
-2
0
2
∞
2
0
-2
Increasing, Decreasing, and First Derivative
Test Homework.
Increasing and Decreasing Functions:
P 186 (3,6,7,11,13.80)
First Derivative Test:
P 186 (21,28,33,35,76)
Section 3.4 – Concavity and the
Second Derivative
Concavity: Concave Up vs. Concave Down
Relationship between concavity and the
second derivative
When a function is concave down the
first derivative is decreasing, so the
second derivative is negative.
When a function is concave up the first
derivative is increasing, so the second
derivative is positive.
Using Faces to Remember the Concavity and
nd
2 Derivative Connection
Describe the concavity of a function.
We express the concavity of a function the same way we express
domain, increasing, and decreasing by using interval notation.
Concave Up:
Concave Down:
Inflection Point
On a continuous function, a point is an inflection
point if the graph changes from concave up to
concave down (or vise versa) at that location.
An inflection point can also be described at a
location where to second derivative changes from
positive to negative (or vise versa).
At an inflection point, f’’(x)=0 or it does not exist.
Locate the Inflection Points on the Graph
Inflection Points:
Find the Inflection Points of f ( x)  x  4x
4
3
Let’s find the second derivative because either f’’(x)=0 or does not exist
at an inflection point.
Inflection points are x=0 and x=2
Determine the intervals on which the function
below is concave up and concave down.
f ( x)  2x3  3x2 12x  5
Concavity changes when f’’(x)=0 or f’’(x) DNE. Let’s find these places and then make a chart.
Interval
Test #
Sign of
f’’(Test #)
Conclusion
Determine the intervals on which the function
below is concave up and concave down.
x2  1
f ( x)  2
x 4
Concavity changes when f’’(x)=0 or f’’(x) DNE. Let’s find
these places and then we’ll make a chart on the next slide.
There are no points where f’’(x)=0, but f’’ DNE at x=-2 and x=2
x2  1
f ( x)  2
x 4
Interval
Test #
Sign of
f’’(Test #)
Conclusion
Second Derivative Test
The second derivative test applies the following understanding to
determine which critical numbers are relative minimums and which are
relative maximums.
A relative maximum is
always located on a
section of the graph
that is concave down,
f’’(x) < 0
A relative minimum is
always located on a
section of a graph that
is concave up, f’’(x) > 0
Find all Relative Extrema. Apply the Second
Derivative Test to determine Max’s from Min’s.
f ( x)  3x  5x
5
3
Relative extrema are located at critical numbers. Let’s locate them, then apply the
second derivative test to see which are relative maximums or minimums.
Applying the
nd
2
f ( x)  3x  5x
5
Point
Sign of f”(point)
Conclusion
3
Derivative Test
Sketch a Graph with the following Characteristics
• f(1) = f(5) = 0
• f’(x) < 0 when x < 3
• f’(3) = 0
• f’’(x) < 0 for all values of x.
Sketch a Graph with the following Characteristics
• f’(x) > 0 if x < 2
• f’(x) < 0 if x > 2
• f’’(x) < 0 while x ≠2
• f(2) is undefined & f’(2) DNE
Sketch a Graph with the following Characteristics
• f(1) = 2
• f’’(1) = 0
• f’(x) < 0 for all values of x
• f’’< 0 if x < 1
Find a, b, c, & d so the cubic
satisfies the given conditions.
Relative Maximum: (-1,1)
Relative Minimum: (1,-1)
Inflection Point: (0,0)
f ( x)  ax3  bx2  cx  d
Find a, b, c, & d so the cubic
satisfies the given conditions.
Relative Maximum: (2,2)
Relative Minimum: (0,-2)
Inflection Point: (1,0)
f ( x)  ax3  bx2  cx  d
Concavity, Inflection Point, and 2nd Derivative
Test Homework
Concavity, Inflection Point, and 2nd Derivative Test:
P 195 (2,5,10,15,19,26,29,31,37)
Curve Sketching:
P 196 (54,55,56)
Creating Equations with Initial Conditions:
P 196 (61,62)
Section 3.5 – Limits at Infinity
Limits at Infinity
Limits at Infinity look at what happens to a function as its x values increase
or decrease without bound.
Consider the function
3x 2
f ( x)  2
x 1
x
-∞←
-100
-10
-1
0
1
10
100
→∞
f(x)
3←
2.9997
2.97
1.5
0
1.5
2.97
2.9997
→3
Hopefully, you will notice that the limits as x approaches infinity and negative
infinity mirror what you learned in PreCalculus about horizontal asymptotes.
Rules for Finding Limits at Infinity
lim x  
x 
lim x  c  
x 
lim x  c  
x 
lim c  x  
x 
x
lim  
x  c
1
lim  0
x  x
lim c  c
x 
0
lim   FindLimit
x 
0

lim   FindLimit
x 

lim( f  g )  lim f  lim g
x 
x 
x 
Find the Limit at Infinity
4
lim(3  2 )
x 
x
Find the Limit at Infinity
lim(x  2)
3
x
Find the Limit at Infinity
3x  1
lim
x  x  1
The trick with rational functions is
to divide the top and bottom of
the fraction by x to the largest
power present.
Find the Limit at Infinity
3x  1
lim
x  x  1
2
The trick with rational functions is
to divide the top and bottom of
the fraction by x to the largest
power present.
Find the Limit at Infinity
3x  1
lim 2
x  x 1
The trick with rational functions is
to divide the top and bottom of
the fraction by x to the largest
power present.
Limits at Infinity of Sin and Cos
sin x
lim
x  x  1
To help us determine this limit,
let’s consider what we know
about the graph of sin x
Limits at Infinity of Sin and Cos
x
lim
x  x  cos x
To help us determine this limit,
let’s consider what we know
about the graph of cos x
Limits at Infinity Homework
Limits at Infinity: P 205 (3,5,17,21-25,31-34)
Section 3.6 – Curve Sketching
Curve Sketching Without Calculators
We will use the following concepts to aid in sketching curves.
• x and y intercepts
• Vertical asymptotes
• Increasing and decreasing
• Relative extrema
• Concavity
• Points of Inflection
• Horizontal Assymptotes
• Limits at infinity
Sketching a polynomial
f ( x)  x 12x  48x  64x  x( x  4)
4
x intercepts:
y intercepts:
Vertical asymptotes:
Horizontal asymptotes:
3
2
3
f ( x)  x 12x  48x  64x  x( x  4)
4
First Derivative:
Critical Numbers:
Second Derivative:
Points of Inflection:
3
2
3
f ( x)  x 12x  48x  64x  x( x  4)
4
3
2
3
• Make an ascending list of all vertical asymptotes, critical numbers,
and points of inflection.
• Make a chart compiling all the information
x value or interval
f(x)
f’(x)
f’’(x)
Graph features
Sketching a Radical Function
5
f ( x)  2 x 5 x
x intercepts:
y intercepts:
Vertical asymptotes:
Horizontal asymptotes:
3
4
3
5
f ( x)  2 x 5 x
First Derivative:
Critical Numbers:
Second Derivative:
Points of Inflection:
3
4
3
5
f ( x)  2 x 3 5 x
4
3
• Make an ascending list of all vertical asymptotes, critical numbers,
and points of inflection.
• Make a chart compiling all the information
x value or interval
f(x)
f’(x)
f’’(x)
Graph features
Sketching a Rational Function
x intercepts:
y intercepts:
Vertical asymptotes:
Horizontal asymptotes:
2( x 2  9)
f ( x)  2
x 4
2( x 2  9)
f ( x)  2
x 4
First Derivative:
Critical Numbers:
Second Derivative:
Points of Inflection:
2( x 2  9)
f ( x)  2
x 4
• Make an ascending list of all vertical asymptotes, critical numbers,
and points of inflection.
• Make a chart compiling all the information
x value or interval
f(x)
f’(x)
f’’(x)
Graph features
Curve Sketching Homework
Sketching a Curve: P 215 (7,11,19,31,9*)
Section 3.7 – Optimization
Optimization
Optimization is the process of applying our understanding of relative
maximums and minimums to word problems. In science, business, and
product design the best and worst scenarios are found by setting the
first derivative equal to zero.
Two numbers add up to 42. What is the most
their product could be?
Why did I
We are going to use the idea that relative extrema are found where
f’(x)=0 to solve this problem, but first we must write an equation.
First # = x
Second # = 42 – x
Therefore the
Product = x (42-x) = 42x-x2
We can graph this function to get a feel for what it looks like.
choose
Xmin=0
Xmax=42?
Let’s find the first derivative
Product = 42x-x2
Product’=42-2x
We find the relative extrema by setting Product’ = 0.
0=42-2x
2x=42
x=21
Remember the First # = x and the Second # = 42 – x
First # = 21, Second # = 21, with product of 441.
We do want to be sure this
is a maximum and not a
minimum. Let’s use the 2nd
derivative test.
Product’’= -2
So Product’’(21)=-2
Consider our smiley face
test. A negative 2nd
derivative is frowning,
concave down, and contains
a maximum.
You are given 4 feet of wire, with which you are to
make a square and circle. What size square and
circle will result in the most combined area?
Within the problem we find two equations.
4 ft = (circumference of circle) + (perimeter of square)
4 = 2πr + 4x
Area = (area of circle) + (area of square)
Area = πr2 + x2
Are there any restrictions on the value of x?
We need to substitute one equation into the other to eliminate a variable.
4 = 2πr + 4x
Area = πr2 + x2
4  2 r  4 x
4  4 x  2 r
4  4x
r
2
2  2x
r

Area   (
2  2x

)2  x 2
Area  2  2 x  x 2
Area  x  2 x  2
2
Area '  2 x  2
0  2x  2
2  2x
1 x
Let’s take a look at the 2nd derivative at
x=1 to verify x=1 is a max.
Area’’ = 2
Area’’(1) = 2
Since Area’’ is positive the face is
smiling, the graph is concave up, and
therefore x=1 is a minimum, instead of
a maximum!!
Let’s take a look at the graph.
Consider the restrictions…
On a closed interval where are the
possible locations of the extrema?
1. Critical Numbers
2. Endpoints.
The best solution here is x = 0. The
circle gets everything.
Optimization Homework
Optimizing: Worksheet