MATH 1325 Business Calculus Guided Notes LSC – North Harris By Isabella Fisher
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MATH 1325
Business Calculus
Guided Notes
LSC – North Harris
By Isabella Fisher
Section 2.1 – Functions and Theirs Graphs
A __________________ is a rule that assigns to each element in ____________ one and only one
element in _________________.
1 2 1 1
2 4 2 4
3 6 3 9
-1 -2 -1 1
The ___________________ of a function is the set of all ________________________________.
The ___________________ of a function is the set of all ________________________________.
We can find the value of an element/input in a function by substituting the input into the function and
simplifying (this is also known as finding the function value).
Ex1: Given 2( ) 3 6 3g x x x Find the following:
a: (0)g b: ( 1)g
c: ( )g a d: ( )g a
e: ( 1)g x
Set A Set B Set A Set B
Ex2: Given 2
2( )
1
ts t
t
Find the following:
a: (4)s b: (0)s
c: ( )s a d: (2 )s a
e: ( 1)s t
Sometimes we have functions that are defined by different rules over parts of the domain. They are
called ___________________________________________________________.
Ex3: Given
2 1 0( )
0
x if xf x
x if x
Find the following:
a: ( 2)f b. (2)f c: (0)f
Ex4: Find the domain, range and function values given a graph:
fD =_______________ fR =________________
Find (0)f =______
Find value for x for which: ( ) 5f x -- x=________
Find value for x for which: ( ) 0f x -- x=________
Recall to find domain
given the graph of a
function you:________
___________________
Recall to find range
given the graph of a
function you:________
___________________
If we are not given the graph a function we can determine the domain of a function as follows:
1. Begin by assuming that the domain is ALL REAL NUMBERS (_____________________________________)
2. Determine if there is any place where this function is not defined (gives you issues). These “issues” will
be excluded from your domain. The two types of functions that have issues are:
a. Rational functions because division by zero is undefined.
b. EVEN root functions because we cannot take the even root of a negative number.
A. To find the domain of rational functions:
1. Set the denominator equal to zero and solve for x.
2. The domain will be all real numbers EXCEPT where the denominator equals zero.
B. To find the domain of EVEN root functions:
i. If the radical is in the numerator:
Set the expression under the radical greater than or equal to zero and solve for x.
ii. If the radical is in the denominator:
Set the expression under the radical greater than zero and solve for x. (can’t be equal to
zero because it is in the denominator).
Find the domain of the following functions. .
EX5:
2 3( )
5
xf x
x
EX6:
2
2
3( )
1
xf x
x
EX7: ( ) 5f x x EX8: 1
( )5
f xx
EX9: 5
2( ) 2 1f x x EX10:
1( )
( 2)( 3)
xf x
x x
Graph the following functions and find the domain and range of each:
EX11: 2( ) 2 1f x x
EX12: ( ) 2f x x
EX13: ( ) 4f x x
EX14: 0
( )2 1 0
x if xf x
x if x
Section 2.2 – The Algebra of Functions
Composition of functions: Given 2 functions 𝑓(𝑥) and 𝑔(𝑥) we can determine:
i. (𝑓 ∘ 𝑔)(𝑥) = 𝑓(𝑔(𝑥)) which means to replace all the x’s in f(x) with the expression g(x) is equal to.
ii. (𝑔 ∘ 𝑓)(𝑥) = 𝑔(𝑓(𝑥)) which means to replace all the x’s in g(x) with the expression f(x) is equal to.
Use the given functions to find (𝑓 ∘ 𝑔)(𝑥) and (𝑔 ∘ 𝑓)(𝑥)
Ex1: 2( ) 1f x x x and
2( )g x x
Ex2: ( ) 2 3f x x and 2( ) 1g x x
Ex3: ( ) 1f x x and 1
( )1
g xx
Evaluate (2)h where h g f
Ex4: 3 2( ) 1f x x and
3( ) 3 1g x x
Ex5:
1( )
1f x
x
and
2( ) 1g x x
Find ( ) ( )f a h f a for the following functions
Ex6: 1
( ) 32
f x x
Ex7: 2( ) 2 1f x x x
Find ( ) ( )f a h f a
h
for the following functions ( 0where h )
Ex8: 2( ) 1f x x
Ex9: 2( ) 2 1f x x x
The _____________________ Cost Function is the variable cost plus the fixed cost.
The _____________________ Function is the revenue function minus the total cost function. (If this is
negative it represents a loss instead of a profit for a company.)
Ex10:
Ex11:
Section 2.3 – Functions and Mathematical Models
Objective: Look at the Cost, Revenue and Profit functions and evaluate function models
Analyze the demand and supply curve and determine equilibrium quantity and price
Section 2.4 – Limits
We say that a function, f , has a limit L as x approaches a lim ( )x a
f x L
if the value of ( )f x can be made as
close to the number L , taking x close to but not equal to a
So as ____ gets close to ____,from both sides, _______ is getting close to a value, _________.
NOTE: if you are NOT getting close to one y value, we say that the limit does not exist DNE
Use the graph to determine the limit:
EX1: 1
lim ( ) ______x
f x
2
lim ( ) ______x
f x
EX2: 3
lim ( ) ______x
f x
EX3:
2lim ( ) ______x
f x
Find a limit from a table of values
EX4:
2
1( )
1f x
x
;
1lim ( )x
f x
x
f(x)
EX4: 2 2
( )1
x xf x
x
;
1lim ( )x
f x
x
f(x)
To find the limit of a function as x approaches a finite value, ______________________ the given value into the function
and simplify.
Find the indicated limit
EX5: 2
lim 3x
EX6: 2
lim 3x
x
EX7: 2
1lim 1 2x
x
EX8: 2
2 1lim
2x
x
x
EX9: 2
1
8lim
2 4x
x
x
Sometimes the number you substitute into the function yields 0
0, this is called _____________________________ form
To find the limit of the function where the value causes you to be in indeterminate form you:
1. Factor and Reduce the function
2. Then take the limit
Find the indicated limit
EX10: 2
1
1lim
1x
x
x
EX11: 2
0limx
x x
x
Limit of a function at ± infinity
The function, f , has a limit L , as x increases without bound lim ( )x
f x L
if ( )f x is getting closer to L as x
approaches infinity.
Similarly: lim ( )x
f x M
if ( )f x is getting closer to M as x approaches negative infinity.
Find the indicated limit:
EX12: lim ( ) ______x
f x
lim ( ) ______x
f x
EX13: lim ( ) ______x
f x
lim ( ) ______x
f x
Theorem (property for a limit at infinity)
For all 0n 1
lim 0nx x
and
1lim 0
nx x
provided that
1nx
is defined.
To evaluate the limit for a rational function at ± infinity, divide each term in the numerator and denominator by the
highest power of x and use the theorem above.
Find the indicated limit
EX14: 2
1limx x
EX15: 3 2
3
3 1lim
1x
x x
x
EX16: 5 3
6 2
1lim
2 1x
x x x
x x
EX17: 3 2
3 2
2 5lim
7x
x x
x x x
EX18: 4 2
2
2lim
1x
x x
x x
EX19:
Section 2.5 – One Sided Limits and Continuity
One Sided Limits:
A function, f , has the right hand limit, L , as x approaches a from the right limx a
f x L
if the values of
f x can be made as close to L with x values sufficiently close to a (right of a )
Similarly:
A function, f , has the left hand limit, M , as x approaches a from the left limx a
f x M
if the values of
f x can be made as close to M with x values sufficiently close to a (left of a )
The limx a
f x L
if and only if limx a
f x L
AND limx a
f x L
Find the limit
EX1:
2
limx
f x
2
limx
f x
2
limx
f x
EX2:
1
limx
f x
1
limx
f x
1
limx
f x
EX3:
1
limx
f x
1
limx
f x
1
limx
f x
EX4:
3
limx
f x
3
limx
f x
3
limx
f x
Again we see that the limx a
f x L
if and only if limx a
f x L
AND limx a
f x L
To find the limits given a function we:
1. Substitute the given x into the function to find the value. If the value causes you to be in indeterminate form,
factor and reduce the function, then substitute.
2. Find the limit
Find the indicated limit
EX5: 1
lim 2 4x
x
EX6: 2
0
1lim
1x
x
x
EX7: 1
1lim
1x
x
x
EX8: 2
2
4lim
2x
x
x
EX9: Given 3
2 4 0( )
1 3 0
x if xf x
x if x
find the following
0
( )limx
f x
=_____________
0
( )limx
f x
=_____________
0
( )limx
f x
=_____________
A function, f , is continuous at a point, a , if it has no holes, jumps, breaks or gaps at a __________________________
__________________________________________________________________________________________________
You will usually encounter holes/jumps/breaks with _______________________________________________________
__________________________________________________________________________________________________
To show that a function is continuous at a point ALL the following points MUST be shown:
1. ( )f a is defined: _____________________________________________________________________________
2. The ( )limx a
f x
exists: ________________________________________________________________________
3. The ( ) ( )limx a
f x f a
Determine the values of x, if any, where the function is discontinuous. State the conditions that are violated if discontinuous.
EX10: 2 1
( )4 1
x if xf x
if x
EX11: 2
5 0
( ) 2 0
5 0
x if x
f x if x
x if x
EX12: 3 2( ) 2 1f x x x x
Find the values of x for which each function is continuous.
EX13: 2
( )2 1
xf x
x
EX14: 2
( )2 1
f xx
EX15: 1
( )2 1 1
x if xf x
x if x
Find the values of x for which each function is continuous.
EX16: 2
3( )
4
xf x
x
EX17: 2
2
2( )
3 2
x xf x
x x
Section 2.6 – The Derivative
A ___________________ line to a curve is a line that touches a curve, or a graph of a function, at only a single point.
What is the slope of the tangent line 1T ? What is the slope of the tangent line 2T ?
The derivative geometrically represents the slope of the tangent line at a point to a given function.
The derivative also represents the instantaneous rate of change of a function at a point.
The problem of finding the rate of change of 1 quantity with respect to another, is the same as finding the slope
of the tangent line to a curve
Given the following curve:
Find the slope of the line between the points P and Q
This the called the difference quotient.
We say that the slope of a tangent line to a graph at a point ,P x f x is given by
0limh
f x h f x
h
The difference quotient f x h f x
h
measures the the average rate of change of y with respect to x over an
interval ,x x h . Since we want to find the average rate of change at a single point, our h gets closer and closer to
zero (hence the limit) and this is referred to as the _____________________________________ rate of change of f at x.
The derivate of a function, f, with respect to x is called ____________________ (read _________________________)
The domain of f is the set of all x for which the limit exsists.
Find f x
Ex1: ( ) 2 7f x x Ex2: 2( ) 3 4f x x x
Find the slope of the tangent line to the graph of the function at the given point. Also determine the equation of the tangent line.
Ex3: 2( ) 3f x x x at 2, 10 Ex4:
1( )f x
x at
13,
3
Ex5: Let 2( ) 2 1f x x Find the following:
(a) Find the derivative, of f f
(b) Find the equation of the tangent line to the curve at the point (1,3)
(c) Sketch the graph
Ex6: Let 2( )f x x x Find the following
(a) Find the average rate of change of y with respect to x in the interval:
x=2 to x=3 x=2 to x=2.5 x=2 to x=2.1
(b) Find the instantaneous rate of change of y with respect to x at 2
Section 3.1 – Basic Rules of Differentiation
Recall that:
0
limh
f x h f xf x
h
We will learn “shortcuts” to finding the derivative of a function.
Note: said d, dx of f of xd
f xdx
represents the derivative of the function, f, with respect to x.
Rule 1 - The Derivative of a Constant: 0d
cdx
The derivative of a constant is ZERO
Find d
dx
EX1: 3f x EX2: 1.735f x
Rule 2 – The Power Rule: If n is any real number then, 1n ndx n x
dx
Find d
dx
EX3: 5f x x EX4: 2.1f x x
EX5: 7
4f x x
Rule 3 – The Derivative of a Constant multiple of a function: d d
c f x c f xdx dx
Find d
dx
EX6: 23f x x EX7: 127f x x
EX8: 1
39f x x EX9: 3f x x
EX10: 2f r r
Rule 4 – The Sum Rule: d d d
f x g x f x g xdx dx dx
Find d
dx
EX11: 25 3f x x x EX12: 25 3 7f x x x
EX13: 20.03 0.4 10f x x x EX14: 5
4 24 3 2f x x x
EX15: 3 24 3x x
f xx
EX16: 4 3
4 3 2f t
t t t
EX17: let 32 4f x x x find:
(a) 2f
(b) 0f (c) 2f
EX18: Find the slope and an equation of a tangent line to the graph of the function, f, at the specified point.
22 3 4 ; 2,6f x x x
Find the point on the graph of f where the tangent line is horizontal.
EX19: 3f x x EX20: 3 24f x x x
EX21: The demand function for walkie-talkies is given by: 20.1 0.4 35p f x x x where x is quantity
demanded in thousands and p is the unit price in dollars.
(a) Find f x
(b) What’s the rate of change of the unit price when quantity demanded is 10,000 units?
(c) What’s the unit price at that demand?
Section 3.2 – The Product and Quotient Rules
The Product Rule: given f x g x d
f x g x f x g x g x f xdx
(the first times the derivative of the second plus the second times the derivative of the first)
Find the derivative:
EX1: 22 1f x x x
EX2: 23 1 2f x x x
EX3: 3 2 21 2f w w w w w
EX4: 5 2 211 1 28
5f x x x x x
EX5: 25 1 2 1f x x x
The Quotient Rule: given
f x
g x
2
f x g x f x f x g xd
dx g x g x
(bottom times the derivative of the top minus top times the derivative of the bottom all over bottom squared)
Find the derivative:
EX6: 1
2f x
x
EX7: 2 4
1
sf s
s
Find the derivative and evaluate f x at the given point:
EX8: 4 22 1
xf x
x x
; 1x
Find the slope and the equation of the tangent line to the graph at the given point:
EX9: 2
1
xf x
x
;
42,
3
Find the points on the graph where the tangent line is horizontal:
EX10: 2 1
xf x
x
Find the points on the graph where the tangent line is equal to 12
:
EX11: 1
1
xf x
x
EX12: The demand function of exercise equipment is given by 2
500 20
0.01 1d x x
x
where x (measured in
units of a thousand) is the quantity demanded per week, and ( )d x is unit price in dollars. Find ( )d x
Section 3.3 – The Chain Rule
The Chain Rule: d
given g f x then g f x g f x f xdx
The General Power Rule: 1n nnd
f x then f x n f x f xdx
Find f
EX1: 4f x x EX2: 4
2 1f x x
EX3: 5
2 2f x x
EX4: 27f x x
EX5: 7 3 2f x x EX6: 1
2 3f x
x
EX7: 422 3 4f t t t EX8:
2 41 2 1f x x x
EX9: 3
3
2
xf x
x
EX10:
1
3 2
xf x
x
EX11: 2
2 1
1
xf x
x
Find the equation of the line tangent at the given point
EX12: 22 7 3,15f x x x at pt
Section 3.4 – Marginal Functions in Economics
The actual cost incurred in producing an additional unit of a certain commodity given that a plant is already at a certain level of operation is called the marginal cost. The marginal cost is approximated by the rate of change of the total cost function evaluated at the appropriate point. (The marginal cost function is the derivative of the total cost function).
Section 3.5 – Higher Order Derivatives
The 2nd derivative of a function, f , is the derivative of the first derivative. By definition, it is:
0
( ) ( )limh
f x h f x
h
(tells us acceleration)
In general the nth derivative is found by taking the derivative of the (n-1)st derivative.
To find f take the derivative of f
To find f take the derivative of f
To find ''''f take the derivative of f
Find the 1st and 2nd derivative of the following functions.
Ex1: 3 2( ) 2 3 1f x x x Ex2: 5 4 3 2( ) 1f x x x x x x
Ex3: 42( ) 3 1g t t t Ex4:
22( ) 1 1h x x x
Ex5: 5
2 2( ) 2 4h w w w Ex6: 2
( )1
tg t
t
Find the 3rd derivative of the following function
Ex7: ( ) 2 1f x x
Solve the following
Section 3.6 – Implicit Differentiation and Related Rates
Currently we only know how to differentiate if we have a function all in the same variable, i.e, 2( ) 2 16f x x
Implicit differentiation allows us to take the derivative of an equation that has more than one variable, i.e, 2 22 16x y . We call it implicit differentiation because the equation is not explicitly solved for y.
d
dx derivative of x
dx
dx derivative of x with respect to x
Steps for implicit differentiation: Find dy
dx by implicit differentiation
1. Differentiate both sides of the equation with respect to x. (make sure that the derivative of any
term involving y includes the factor dy
dx.
2. Solve the resulting equation for dy
dx in terms of x
and y.
Ex1: 2 22 16x y
Find dy
dx by implicit differentiation
Ex2: 3 3 4 0x y y Ex3: 2 2 6x xy
Ex4: 2 2 8x y xy Ex5:
2 2
1 11
x y
Ex6: 1
232 3x y x
Find the slope and an equation of a tangent line to the graph of the given equation at the specified point
Ex7: 2 24 9 36x y ; 0,2
Ex8: 2 2 16y x ; 2,2 5
Section 4.1 – Applications of the First Derivative
A function, f , is increasing on an interval ,a b if for every two numbers 1x and 2x in ,a b , 1 2f x f x
whenever 1 2x x (increasing if from left to right, the graph is going up)
A function, f , is decreasing on an interval ,a b if for every two numbers 1x and 2x in ,a b ,
1 2f x f x , whenever 1 2x x (decreasing if from left to right, the graph is going down)
A function, f , is constant on an interval ,a b if for every two numbers 1x and 2x in ,a b , 1 2f x f x ,
(constant if from left to right, the graph does NOT go up or down, it remains the same)
We can use f x to determine intervals of increasing/decreasing/constant:
If 0f x for every value of x in the interval ,a b then f is increasing on ,a b
If 0f x for every value of x in the interval ,a b then f is decreasing on ,a b
If 0f x for every value of x in the interval ,a b then f is constant on ,a b
To determine the intervals of increasing/decreasing/constant:
1. Find all the values of x for which 0f x and where f x is discontinuous (undefined) and identify the
open intervals determined by the numbers (these are called critical numbers)
2. Select a test value, c , in each interval found in step 1 and determine the sign (+/-) and determine the sign of
f c in that interval:
a. If 0f c the function f is increasing
b. If 0f c the function f is decreasing
Find the interval(s) where the function is increasing and the interval(s) where it is decreasing:
Ex1: 2( ) 3 4 9f x x x Ex2: 4 3( ) 4 10f x x x
Ex3: ( )1
xf x
x
Ex4:
35( ) 7f x x
Relative Extrema give us the highest points and lowest points in an interval.
Relative Maximum: A function, f , has a relative max at x c if there exists an open interval ,a b containing
c such that f x f c for all x in ,a b
Relative Minimum: A function, f , has a relative min at x c if there exists an open interval ,a b containing
c such that f x f c for all x in ,a b
To find the relative max/min of a function:
1. Determine the critical numbers (any number x in the domain of f , such that 0f x or f x DNE )
2. Determine the sign (+/-) of f x to the left and right of each critical number
a. If f x changes sign from positive to negative (+ to – ) then c is a relative max
b. If f x changes sign from negative to positive ( – to +) then c is a relative min
c. If f x DOESN’T change sign then f doesn’t have relative extrema at c
Find the x-value(s) of the relative maxima and relative minima, if any, of the function.
Ex5: 2( ) 6 6f x x x Ex6: 4 3( ) 4 8f x x x
Ex7: 9
( ) 2f x xx
Ex8: 2
3( ) 1f x x
Section 4.2 – Applications of the Second Derivative
The 2nd derivative determines concavity.
Concavity of a Function f : Let the function f be differentiable on an interval ,a b . Then,
1. The graph of f is concave upward on ,a b if f is increasing on ,a b .
2. The graph of f is concave downward on ,a b if f is decreasing on ,a b .
Theorem 2:
If 0f x for every value of x in ,a b then the graph of f is concave upward on ,a b
If 0f x for every value of x in ,a b then the graph of f is concave downward on ,a b
To determine intervals of concavity of the graph of f
1. Determine the values of x for which f is zero or is not defined, and identify open intervals determined by
these numbers.
2. Determine the sign (+/-) of f in each interval found in step 1 – compute f x
a. If 0f x then the graph of f is concave upward
b. If 0f x then the graph of f is concave downward
Determine where the function is concave upward and where it is concave downward.
Ex1: 2( ) 2 3 4f x x x Ex2: 4 3( ) 6 2 8f x x x x
Ex3: 3( )f x x
An inflection point is a point on the graph of a continuous function f , where the tangent line exists and the concavity
changes.
To determine inflection points:
1. Compute f x
2. Determine the numbers in the domain of f for which 0f x or where f x does not exist.
3. Determine the sign of f x to the left and right of each number, c (found in step 2). If there is a change in
the sign (+/-) as we move across x c , then c will be an inflection point ,c f c , which is found by
computing f c
Find the inflection point(s), if any, of the function
Ex4: 2
1( )f x x
x
Ex5: 4 3( ) 3 4 1g x x x
Instead of using the first derivative to determine relative maximums/minimums we can use the second derivative test.
Second Derivative Test:
1. Compute f x and f x
2. Find all the critical numbers of f at which 0f x )
3. Compute f c for each critical number c (the numbers we found in step 2)
a. If 0f c then f has a relative maximum at c
b. If 0f c then f has a relative minimum at c
c. If 0f c OR f c DNE, then the test has failed and is inconclusive
Use the Second Derivative Test to find the relative extrema, if any, of the function.
Ex6: 3 21( ) 2 5 10
3f x x x x Ex7: 2 16
( )f x xx
Section 4.3 – Curve Fitting
We will be learning how to sketch the graph of a function by using application of the 1st and 2nd derivatives. Vertical Asymptotes: A function, f , has a vertical asymptote at the line x a , if:
limx a
f x or
or limx a
f x or
What we will do to find vertical asymptotes is:
1. Factor and reduce the function if possible
2. Set the denominator equal to zero and solve for x. Any REAL NUMBERS you obtained is where the function has
a vertical asymptote.
1
f xx
Horizontal Asymptotes: A function, f , has a horizontal asymptote at the line y b , if:
limx
f x b
or limx
f x b
To find the horizontal asymptote:
1. Take the limit as the function approaches infinity.
1
f xx
Find the vertical and horizontal asymptotes of the following functions:
EX1: 3
1y
x EX2:
2 1
xy
x
EX3: 2
2 9
tg t
t
EX4:
2
52
2f x
x
Note: The graph of a polynomial function does NOT have asymptotes EX5: 23 5 9f x x x
To sketch the curve of a function:
1. Find the domain of the function
2. Find the x and y intercepts of the function
3. Determine the end behavior of x (behavior of f for large absolute values of x)
4. Find all the horizontal and vertical asymptotes of the function
5. Determine intervals of increasing and decreasing
6. Find the relative extrema
7. Determine concavity
8. Find the inflection points
9. Plot all the information above and any additional points necessary and sketch the graph.
Sketch the graph of the following functions:
EX6: 3 22 3 12 2f x x x x
a) Domain:__________________________
b) X-intercept(s):_____________________ y-intercept:_______________________
c) End Behavior:______________________
d) Horizontal Asymptote:________________ Vertical Asymptote:__________________
e) Increasing:________________________ decreasing:________________________
f) Relative Extrema:___________________
g) Concave up:________________________
Concave down:________________________
h) Inflection points:_______________________
EX7: 2
1g x
x
a) Domain:__________________________
b) X-intercept(s):_____________________
y-intercept:_______________________
c) End Behavior:______________________
d) Horizontal Asymptote:________________ Vertical Asymptote:__________________
e) Increasing:________________________ decreasing:________________________
f) Relative Extrema:___________________
g) Concave up:________________________
Concave down:________________________
h) Inflection points:_______________________
EX8: 1
1
xf x
x
a) Domain:__________________________
b) X-intercept(s):_____________________
y-intercept:_______________________
c) End Behavior:______________________
d) Horizontal Asymptote:________________ Vertical Asymptote:__________________
e) Increasing:________________________ decreasing:________________________
f) Relative Extrema:___________________
g) Concave up:________________________
Concave down:________________________
h) Inflection points:_______________________
EX9: 2
2 9
tf t
t
a) Domain:__________________________
b) X-intercept(s):_____________________
y-intercept:_______________________
c) End Behavior:______________________
d) Horizontal Asymptote:________________ Vertical Asymptote:__________________
e) Increasing:________________________ decreasing:________________________
f) Relative Extrema:___________________
g) Concave up:________________________
Concave down:________________________
h) Inflection points:_______________________
Section 4.4 – Optimization I
Absolute Extrema of a Function
o If f x f c for all x in the domain of f then f c is called the absolute maximum value of f
o If f x f c for all x in the domain of f then f c is called the absolute minimum value of f
The Extreme Value Theorem
o If a function f is continuous on a closed interval ,a b then f has both an absolute maximum value
and an absolute minimum value on ,a b (this is because on a closed interval there will be a highest
and lowest point on the graph)
We find absolute extrema on an open interval exactly as we did with relative extrema:
1. Find the first derivative
2. Find the critical numbers
3. Determine if there is a sign change (+/-) between the critical numbers
To find the absolute extrema of f on a closed interval ,a b then:
1. Find the first derivative
2. Find the critical numbers (______________________________________________________________________)
3. Compute f a and f b and #f critical s
4. The absolute maximum value and the absolute minimum value will correspond to the largest and smallest
numbers, respectively, found in step 3.
Find the absolute maximum value and the absolute minimum value, if any, of the function
Ex1: 23 4 9f x x x Ex2: 2 4 6g x x x on 0,5
Ex3: 133f x x Ex4: 2 1
xh x
x
Section 4.5 – Optimization II
Guidelines to solving optimization problems: 1. Assign a letter to each variable mentioned in the problem and draw a picture if necessary.
2. Find an expression for the quantity that will be optimized.
3. Use the conditions in the problem to find the quantity to be optimized as a function of 1 variable.
4. Optimize the function over its domain.
Section 5.2 – Logarithmic Functions
A logarithm of x to the Base b is defined by: logby x if and only if yx b 0, 1, 0b b and x
For Example: 32 8 thus 2log 8 3 12
19
3
thus
9
1 1log
3 2
Common logarithms:
log x is the same as 10log x - if no base is written the implied base is 10
ln x is the same as loge x - ln means log base e
Laws of Logarithms: Given m and n are positive real numbers and 0 1b and b then:
1. logb mn = log logb bm n
2. logb
m
n
= log logb bm n
3. logn
b m = logbn m
4. log 1 0b
5. log 1b b
Use the laws of logarithms to expand and simplify the expression.
EX1: 4log 1x x
EX2: 2
1log
1
x
x
EX3: 2
ln xxe EX4:
12
2 2ln
1
x
x x
Section 5.4 – Differentiation of Exponential Functions
The derivative of an Exponential Function is: x xde e
dx
The chain rule for Exponential Functions is: f x xde e f x
dx
Find the derivative of the following functions.
EX1: 3 xf x e EX2: 2xf x e
EX3: 22 xf x e x EX4: 2 uf u u e
EX5: x
xf x
e EX6:
1
x
x
ef x
e
EX7: 3
34 xf x e EX8: 22 1 sf s s e
Find the 2nd derivative of the following function
EX9: 23 5t tf t e e
Find the intervals where the following function is increasing and where it is decreasing
EX10: 2
2x
f x e
Determine the intervals where the function is concave down and/or concave up.
EX11: xf x xe
Section 5.5 – Differentiation of Logarithmic Functions
The derivative of ln x is: 1
ln 0d
x xdx x
The chain rule for ln x is:
ln 0
f xdf x f x
dx f x
The derivative of the function you’re taking the natural log of, divide by the function. Find the derivative of the following functions.
EX1: ln 1f x x EX2: 8lnf x x
EX3: 2
1lnf x
x
EX4: 2ln 4 6 3f x x x
EX5: 2
ln1
xf x
x
EX6: 2 lnf x x x
EX7: lnf x x EX8:
2ln xf x
x
EX9: 3
lnf x x EX10: lnxf x e x
Find the second derivative of the following function:
EX11: 2ln 2f x x
We can find the derivative of functions by taking the natural log of both sides of the equation and using the laws of logs to expand before taking the derivative.
Finding dy
dxby Logarithmic differentiation:
1. Take the ln of both sides of the equation and uses the laws of logarithms to write any complicated expressions as a sum/difference of simpler terms.
2. Differentiate both sides of the equation with respect to x
3. Solve for dy
dx
Use logarithmic differentiation to find the derivative of the following functions.
EX12: 2 3
1 2y x x EX13: 4
3 5 2 3y x x
EX14: 3xy
Section 5.6 – Exponential Functions as Mathematical Models
The exponential growth model is : 0
ktQ t Q e where 0 t
0Q is the amount of substance that is initially present when 0t and k is the constant of proportionality, called the
growth constant.
Section 6.1 – Antiderivaties and the Rules of Integration
-Recall that the derivative allowed us to find velocity given position. Now what we will do is go from knowing the
velocity to finding the position.
A function, F , is an antiderivative of f on an interval I if F x f x for all x in I
For example let’s say that 3 212 1
3F x x x x is an antiderivative of 2 4 1f x x x this is because
F x ____________________________ = f x
But note that if F x ________________________ it would also be the antiderivative of 2 4 1f x x x
Also if F x ________________________ we would get the same result.
The process of finding antiderivatives is called integration. f x dx F x C
Read: the indefinite integral of f of x with respect to x equals F of x plus c
The indefinite integral gives a family of functions because of the constant, thus when we take the antiderivative of a
function we will be adding a CONSTANT TERM, C to the antiderivative.
The function to be integrated is called the integrand and C is called the constant of integration.
Rules of integration:
1. The indefinite integral of a constant k dx kx C (where k is a constant term)
Ex1: 2 dx Ex2: dx Ex3: e dx
2. The power rule 11
1
n nx dx x Cn
(where n does not equal -1)
Ex4: 2x dx Ex5: 3x dx Ex6: 4x dx
3. The indefinite integral of a constant multiple of a function c f x dx c f x dx (where c is a constant)
Ex7: 2x dx Ex8: 2
35x dx Ex9: 2
2dx
x
4. The sum rule f x g x dx f x dx g x dx (we can integrate a term at a time)
Ex10: 2 3x x x dx
5. The indefinite integral of the exponential function x xe dx e C
Ex11: 1 xe dx
6. The indefinite integral of the function 1f x x 1 lnx dx x C (where x does not equal 0 )
Ex12: 5
dxx
Find the indefinite integral of the following functions
Ex13: 3
2 xx e dxx
Ex14: 3 3
2
t tdt
t
Ex15: 3 3 210u u u du
If we are given a point on the graph of f (an initial value), we can find the exact value of C for the antiderivative.
Find f by solving the initial value problem.
Ex16: 23 6 ; 2 4f x x x f
Ex17: 2 ; 0 2xf x e x f
Ex18: Find the function f given that the slope of the tangent line to the graph of f at any point ,x f x is
2 2 3x x and the graph of f passes through the point 1,2 .
Section 6.2 – Integration by Substitution
Integration by substitution allows us to integrate functions that we cannot separate into individual terms. It is related to
the chain rule for differentiating.
Steps for Integrating by Substitution:
1. Let u equal one of the functions, g x , in the integrand (usually the inside function, or the function that is
being raised to a power)
2. Find du g x dx
3. Use substitution to convert the ENTIRE into one only involving u
4. Evaluate the resulting integral using the rules for integration from section 6.1
5. Replace u with g x to obtain a final solution as a function in terms of x
Find the indefinite integral
Ex1: 7
24 2 1x x dx Ex2:
2
23
3 2
2
xdx
x x
Ex3: 2
3 1
xdx
x Ex4:
3
2
x x
x x
e edx
e e
Sometimes we will need to do a little work to get du to equal g x dx
Ex5: 2xe dx
Find the indefinite integral
Ex6:
3lnu
duu Ex7:
1
lndx
x x
Ex8: 43 1xx e du
Ex9: 4
tdt
t Hint:
41
4 4
t
t t
Fig A Fig B
Section 6.3 – Area and the Definite Integral
We are going to use Riemann Sums to approximate the area under the curve of a graph:
Let’s say we wanted to find the area under f(x) from a to b.
The area under the graph of a function can be found as follows:
Let f be a nonnegative continuous function on a closed interval ,a b . Then the area of the region under the graph of
f is 1 2 3 ... nA f x x f x x f x x f x x
where 1 2 3, , ,..., nx x x x are arbitrary points in the n subintervals of ,a b of equal width b a
xn
The sum on the right-hand side of this expression is called a Riemann sum, in honor of the German mathematician
Bernhard Riemann (1826-1866)
Ex1: Find an approximation of the area of the region R under the graph of f by computing the Riemann sum of f
corresponding to the partition of the interval into the subintervals shown in the figure. Use the midpoint of the
subintervals as the representative points.
Ex2:
Ex3:
Find an approximation of the area of the region R under the graph of the function f on the interval [a,b]. In each case,
use n subintervals and choose the representative points as indicated.
Ex4: 2( ) 4 ; [ 1,2] ; 6 ; left endpointsf x x n
Ex5: ( ) ; [0,3] ; 5; midpointsxf x e n
Section 6.4 – The Fundamental Theorem of Calculus
We will use the fundamental theorem of calculus to calculate the area of a region under a graph.
The Definite Integral:
Let function f be defined on ,a b . If 1 2 3lim ... nn
f x f x f x f x x
exists and is the same for all
representative points 1 2 3, , ,..., nx x x x in n subintervals of ,a b of equal width b a
xn
then this area is called the
Definite Integral denoted b
af x dx where the lower limit of integrationa & the upper limit of integrationb
The Fundamental Theorem of Calculus:
Let f be a continuous function on a closed interval ,a b , then b
af x dx F b F a
Where F is any antiderivative of f , that is F x f x
Find the area of the region under the graph of the function f on the interval ,a b
Ex1: 2 3 ; 1,2f x x
Ex2: 1
; 1,2f xx
Ex3: 31 ; 8, 1f x x
Evaluate the definite integral
Ex4: 4
1u du Ex5:
2
04 1x x dx
Ex6: 1 2
2
11x dx
Section 6.5 – Evaluating Definite Integrals
Properties of the Definite Integral: Let f and g be integrable functions. Then,
1. 0a
af x dx
2. b a
a bf x dx f x dx
3. b b
a acf x dx c f x dx where c is a constant
4. b b b
a a af x g x dx f x dx g x dx
5. b c b
a a cf x dx f x dx f x dx where a c b
Evaluate the definite integral
Ex1: 1 4
2 3
02 1x x dx Ex2:
32
13 2x x dx
Ex3: 1
0
1
2 1dx
x Ex4:
22
0
xxe dx
Find the area of the region under the graph of the function f on the interval ,a b
Ex5: 2
1; 1,2f x
x
The Average Value of a Function:
Suppose f is integrable on ,a b , then the average value of f on ,a b is: 1 b
af x dx
b a
Find the average value of the function f over interval ,a b
Ex6: 24 ; 2,3f x x Ex7: ; 0,4xf x e
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