Calculus Chapter 3 3-1 Name ________________________________ Date ______________ Class ____________ Goal: To find limits of functions Section 3-1 Intro to Limits Definition: Limit We write lim () x c fx L → = or () fx L → as x c → if the functional value () fx is close to the single real number L whenever x is close, but not equal, to c (on either side of c). Definition: One sided limits lim () x c f x L − → = is the limit of the function as x approaches the value c from the left. lim () x c f x L + → = is the limit of the function as x approaches the value c from the right. Properties of Limits: 1. lim x c k k → = for any constant k 2. lim x c x c → = 3. lim[ () ( )] lim () lim () x c x c x c f x gx fx gx → → → + = + 4. lim[ () ( )] lim () lim () x c x c x c f x gx fx gx → → → − = − 5. lim () lim () x c x c kf x k fx → → = for any constant k 6. lim[ () ( )] lim ( ) lim () x c x c x c fx gx fx gx → → → ⋅ = ⋅ 7. lim () () lim () lim () x c x c x c fx fx gx gx → → → ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ provided lim () 0 x c gx → ≠ 8. lim () lim () n n x c x c fx fx → → = (the limit value must be positive for n even.)
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Calculus Chapter 3
3-1
Name ________________________________ Date ______________ Class ____________
Goal: To find limits of functions
Section 3-1 Intro to Limits
Definition: Limit We write lim ( )
x cf x L
→= or ( )f x L→ as x c→
if the functional value ( )f x is close to the single real number L whenever x is close, but not equal, to c (on either side of c). Definition: One sided limits lim ( )
x cf x L
−→= is the limit of the function as x approaches the value c from the left.
lim ( )
x cf x L
+→= is the limit of the function as x approaches the value c from the right.
Properties of Limits: 1. lim
x ck k
→= for any constant k
2. limx c
x c→
=
3. lim[ ( ) ( )] lim ( ) lim ( )x c x c x c
f x g x f x g x→ → →
+ = +
4. lim[ ( ) ( )] lim ( ) lim ( )x c x c x c
f x g x f x g x→ → →
− = −
5. lim ( ) lim ( )x c x c
kf x k f x→ →
= for any constant k
6. lim[ ( ) ( )] lim ( ) lim ( )x c x c x c
f x g x f x g x→ → →
⋅ = ⋅
7. lim ( )( )
lim( ) lim ( )
x c
x cx c
f xf x
g x g x→
→→
⎛ ⎞=⎜ ⎟⎝ ⎠
provided lim ( ) 0x c
g x→
≠
8. lim ( ) lim ( )n nx c x c
f x f x→ →
= (the limit value must be positive for n even.)
Calculus Chapter 3
3-2
1 - 5 Find each limit if it exists 1.
6lim (4 5)x
x→
+ (4(6) 5) 24 5 29= + = + =
2.
3lim 5
xx
→−5( 3) 15= − = −
3.
3lim (2 7)x
x x→
+3 3
lim lim (2 7) 3[2(3) 7)] 3(13) 39x x
x x→ →
= ⋅ + = + = =
4. 1
8lim
2x
x
x→−
+⎛ ⎞⎜ ⎟⎝ ⎠+
1
1
lim 8 1 8 77
lim 2 1 2 1x
x
x
x→−
→−
+ − += = = =+ − +
5. 5
lim 5 11x
x→−
− + 5( 5) 11 25 11 36 6= − − + = + = =
6 - 8 Find the value of the following limits given that
3lim ( ) 6x
f x→
= and 3
lim ( ) 2.x
g x→
= −
6.
3lim 7 ( )x
f x→
7 6 42= ⋅ =
7.
3lim[3 ( ) 2 ( )]x
f x g x→
− 3(6) 2( 2) 18 4 22= − − = + =
8. 3
2 ( )lim
3 ( )x
f x
g x→
⎛ ⎞⎜ ⎟⎝ ⎠
2(6) 122
3( 2) 6= = = −
− −
Calculus Chapter 3
3-3
9. Let 2 2 if 1
( ) .4 1 if 1
x xf x
x x
⎧⎪ + <= ⎨− ≥⎪⎩
Find:
a)
1lim ( )
xf x
−→
Since we are looking for the left–hand limit, use the top function.
2 2
1 1lim ( ) lim ( 2) ((1) 2) 3
x xf x x
− −→ →= + = + =
b)
1lim ( )
xf x
+→
Since we are looking for the right–hand limit, use the bottom function.
1 1lim ( ) lim (4 1) (4(1) 1) 3
x xf x x
+ +→ →= − = − =
c)
1lim ( )x
f x→
Since the left– and right–hand limits are the same value,
1lim ( ) 3.x
f x→
=
d) (1)f The value of 1 is defined in the bottom function, therefore (1) 4(1) 1 3.f = − =
10. Let 5 6 if 2
( ) .2 4 if 2
x xf x
x x
− ≤⎧= ⎨ + >⎩
Find:
a)
2lim ( )
xf x
−→
Since we are looking for the left–hand limit, use the top function.
2 2lim ( ) lim (5 6) 5(2) 6 4
x xf x x
− −→ →= − = − =
b)
2lim ( )
xf x
+→
Since we are looking for the right–hand limit, use the bottom function.
2 2lim ( ) lim (2 4) (2(2) 4) 8
x xf x x
+ +→ →= + = + =
Calculus Chapter 3
3-4
c) 2
lim ( )x
f x→
Since the left– and right–hand limits are not the same value,
2lim ( )x
f x→
does
not exist. d) (2)f The value of 2 is defined in the top function, therefore (2) 5(2) 6 4.f = − =
11. Let 2 4 5
( ) .1
x xf x
x
⎛ ⎞+ −= ⎜ ⎟−⎝ ⎠ Find
a)
1lim ( )x
f x→
Substituting in a value of 1 will result in a zero in the denominator, therefore we must try to remove the problem by first simplifying the function:
2 4 5 ( 5)( 1)
( ) 5,1 1
x x x xf x x
x x
+ − + −= = = +− −
1x ≠
Therefore,
1 1lim ( ) lim( 5) 1 5 6x x
f x x→ →
= + = + =
b) 1
lim ( )x
f x→−
2 2
1
4 5 ( 1) 4( 1) 5 8lim 4
1 1 1 2x
x x
x→−
+ − − + − − −= = = =− − − −
c) 5
lim ( )x
f x→−
2 2
5
4 5 ( 5) 4( 5) 5 0lim 0
1 5 1 6x
x x
x→−
+ − − + − −= = = =− − − −
Calculus Chapter 3
3-5
12. Let 2
( ) .2
xf x
x
⎛ ⎞−= ⎜ ⎟−⎝ ⎠
Find
a)
2lim ( )
xf x
−→
As x approaches 2 from the left side, the value of the numerator will be positive and the value of the denominator will be negative (because the x values are smaller than 2). The limit will therefore, be a value of –1 (because they are opposite signs). b)
2lim ( )
xf x
+→
As x approaches 2 from the right side, the value of the numerator will be positive and the value of the denominator will be positive (because the x values are larger than 2). The limit will therefore, be a value of 1 (because they are the same sign). c)
2lim ( )x
f x→
does not exist because the left–hand and right–hand limits are
different values.
Calculus Chapter 3
3-6
Calculus Chapter 3
3-7
Name ________________________________ Date ______________ Class ____________
Goal: To find limits of functions as they approach infinity
Section 3-2 Infinite Limits and Limits at Infinity
Limits of Power Functions at Infinity: If p is a positive real number and k is any real number except 0, then
1. lim 0px
k
x→−∞= 2. lim 0
px
k
x→∞=
3. lim p
xkx
→−∞= ±∞ 4. lim p
xkx
→∞= ±∞
provided that px is a real number for negative values of x. The limits in 3 and 4 will be either positive or negative infinity, depending on k and p. Limits of Rational Functions at Infinity:
If 1
1 1 01
1 1 0
( ) , 0, 0,m m
m mm nn n
n n
a x a x a x af x a b
b x b x b x b
−−
−−
+ + + += ≠ ≠
+ + + +
then lim ( ) limm
mnx x n
a xf x
b x→∞ →∞= and lim ( ) lim
mm
nx x n
a xf x
b x→−∞ →−∞=
There are three cases to consider: 1. If ,m n< then lim ( ) lim ( ) 0.
x xf x f x
→∞ →−∞= =
2. If ,m n= lim ( ) lim ( ) .m
x x n
af x f x
b→∞ →−∞= =
3. If ,m n> then the limit will be or ,∞ −∞ depending on the values of m, n,
,ma and .nb
Calculus Chapter 3
3-8
1 - 3 Find each limit. Use or ∞ −∞when appropriate.
1. ( )2
xf x
x=
+
a)
2lim ( )
xf x
−→−
If the value is substituted into the function, we would have a zero in the denominator. Since there is no way to remove the problem, we will look at what happens around the value of –2 when approached from the left. These values are smaller than –2, therefore the denominator will approach 0 from the negative side (that is, the denominator will always be negative). Since the numerator will always be negative,
2lim ( ) .
xf x
−→−= ∞
b)
2lim ( )
xf x
+→−
If the value is substituted into the function, we would have a zero in the denominator. Since there is no way to remove the problem, we will look at what happens around the value of –2 when approached from the right. These values are larger than –2, therefore the denominator will approach 0 from the positive side (that is, the denominator will always be positive). Since the numerator will always be negative,
2lim ( ) .
xf x
+→−= −∞
c)
2lim ( )
xf x
→− does not exist because the left–hand and right–hand limits are
different infinite limits.
2. 2
3 4( )
( 3)
xf x
x
−=−
a)
3lim ( )
xf x
−→
If the value is substituted into the function, we would have a zero in the denominator. Since there is no way to remove the problem, we will look at what happens around the value of 3 when approached from the left. These values are smaller than 3, but the denominator value is being squared and will always be positive. Since the value of the numerator is 5 as the x value approaches 3,
3lim ( ) .
xf x
−→= ∞
Calculus Chapter 3
3-9
b) 3
lim ( )x
f x+→
If the value is substituted into the function, we would have a zero in the denominator. Since there is no way to remove the problem, we will look at what happens around the value of 3 when approached from the right. These values are larger than 3, but the denominator value is being squared and will always be positive. Since the value of the numerator is 5 as the x value approaches 3,
3lim ( ) .
xf x
+→= ∞
c)
3lim ( )x
f x→
= ∞ because the left– and right–hand limits both approach infinity.
3. 2 3 5
( )5
x xf x
x
+ −=+
a)
5lim ( )
xf x
−→−
If the value is substituted into the function, we would have a zero in the denominator. Since there is no way to remove the problem, we will look at what happens around the value of –5 when approached from the left. These values are smaller than –5, therefore the denominator will approach 0 from the negative side (that is, the denominator will always be negative). Since the numerator is positive around –5,
5lim ( ) .
xf x
−→−= −∞
b)
5lim ( )
xf x
+→−
If the value is substituted into the function, we would have a zero in the denominator. Since there is no way to remove the problem, we will look at what happens around the value of –5 when approached from the right. These values are larger than –5, therefore the denominator will approach 0 from the positive side (that is, the denominator will always be positive). Since the numerator is positive around –5,
5lim ( ) .
xf x
+→−= ∞
c)
5lim ( )
xf x
→− does not exist because the left–hand and right–hand limits are
different infinite limits.
Calculus Chapter 3
3-10
4 - 6 Find each function value and limit. Use or ∞ −∞where appropriate.
4. 2 3
( )3 8
xf x
x
+=−
a) (20)f
2(20) 3 43
(20) 0.82693(20) 8 52
f+= = ≈−
b) (200)f
2(200) 3 403
(200) 0.68073(200) 8 592
f+= = ≈−
c) lim ( )x
f x→∞
2
3= because the function is a rational expression and ,m n= the
limit is the ratio of the coefficients.
5. 2
5( )
3 2 2
xf x
x x
−=+ +
a) (10)f
2
10 5 5(10) 0.0155
3223(10) 2(10) 2f
−= = ≈+ +
b) (100)f
2
100 5 95(100) 0.0031
30, 2223(100) 2(100) 2f
−= = ≈+ +
c) lim ( )
xf x
→∞0= because the function is a rational expression and .m n<
Calculus Chapter 3
3-11
6. 3 2 1
( )3 8
x xf x
x
+ −=−
a) ( 10)f −
3( 10) 2( 10) 1 1021
( 10) 12.30123 8( 10) 83
f− + − − −− = = ≈ −
− −
b) ( 100)f −
3( 100) 2( 100) 1 1,000,201
( 100) 1245.58033 8( 100) 803
f− + − − −− = = ≈ −
− −
c) lim ( )
xf x
→−∞= −∞ because the function is a rational expression, ,m n> and
when x becomes increasingly negative, the value of the function will be increasingly negative. 7 - 9 Find the vertical and horizontal asymptotes for the following functions.
7. 3
( )3
xf x
x=
+
Vertical asymptotes are found by setting the denominator equal to zero. Therefore, there would be a vertical asymptote at 3 0 3.x x+ = ⇒ = − Horizontal asymptotes are found by finding the limit of the function as it goes to infinity. The limit of this function is 3 because m n= and the limit is the ratio of the coefficients. Therefore, the horizontal asymptote is 3.y =
8. 2
2
3( )
4
xf x
x
−=+
Vertical asymptotes are found by setting the denominator equal to zero. Since the denominator cannot be zero, there are no vertical asymptotes. Horizontal asymptotes are found by finding the limit of the function as it goes to infinity. The limit of this function is 1 because m n= and the limit is the ratio of the coefficients. Therefore, the horizontal asymptote is 1.y =
Calculus Chapter 3
3-12
9. 2
2
4 5( )
8 15
x xf x
x x
+ −=+ +
The function can be reduced as follows:
2
2
4 5 ( 5)( 1) 1( ) ,
( 5)( 3) 38 15
x x x x xf x
x x xx x
+ − + − −= = =+ + ++ +
5x ≠ − and 3x ≠ −
Vertical asymptotes are found by setting the denominator equal to zero. Therefore,
there would be a vertical asymptote at 3 0 3.x x+ = ⇒ = −
Horizontal asymptotes are found by finding the limit of the function as it goes to infinity. The limit of this function is 1 because m n= and the limit is the ratio of the coefficients. Therefore, the horizontal asymptote is 1.y =
Calculus Chapter 3
3-13
Name ________________________________ Date ______________ Class ____________
Goal: To determine if functions are continuous at specific points and intervals
Section 3-3 Continuity
Definition: Continuity A function f is continuous at the point x c= if 1. lim ( )
x cf x
→exists 2. ( )f c exists 3. lim ( ) ( )
x cf x f c
→=
Continuity Properties: 1. A constant function ( ) ,f x k= where k is a constant, is continuous for all x.
2. For n a positive integer, ( ) nf x x= is continuous for all x. 3. A polynomial function is continuous for all x.
4. A rational function is continuous for all x except those values that make a denominator 0.
5. For n an odd positive integer greater than 1, ( )n f x is continuous wherever
( )f x is continuous.
6. For n an even positive integer, ( )n f x is continuous wherever ( )f x is continuous and nonnegative.
Constructing Sign Charts:
1. Find all partition numbers. These are all the values that make the function discontinuous or 0.
2. Plot the numbers found in step 1 on a real–number line, dividing the number line into intervals.
3. Select a test value in each open interval and evaluate ( )f x at each test value to determine whether ( )f x is positive or negative. 4. Construct a sign chart, using the real–number line in step 2.
Calculus Chapter 3
3-14
1 - 5 Using the continuity properties, determine where each of the functions are continuous.
1. 3 2( ) 3 4 7f x x x x= − + + Since the function is a polynomial function, it is continuous for all x.
2. 3 2
2
5 3( )
8 15
x x xf x
x x
+ + −=− +
Since the function is a rational function, it is continuous for all x except when the
denominator is 0. Therefore, 2 8 15 0,x x− + = or ( 5)( 3) 0,x x− − = when 3x = or 5.x = So the function is continuous for all x except 3x = or 5.x =
3. 2
6 1( )
6
xf x
x
−=+
Since the function is a rational function, it is continuous for all x except when the denominator is 0. The denominator cannot have a value of 0, therefore the function is continuous for all x.
4. 3( ) 10f x x= − Since the function is an odd positive root, the function is continuous for all values of x where the radicand is continuous. The radicand is a polynomial function, therefore the function is continuous for all x.
5. 2( ) 25f x x= − Since the function is an even root, the function is continuous for all values of x where
the radicand is continuous and nonnegative. Therefore, 2 25 0x − ≥ and the points in question are 5.± For all values between –5 and 5, the radicand would be negative, therefore, the function is continuous on the interval ( , 5] [5, ).−∞ − ∪ ∞
Calculus Chapter 3
3-15
6 - 7 Use a sign chart to solve each inequality. Express answers in inequality and interval notation.
6. 24 29 7 0x x− + < Find the partition numbers:
24 29 7 0
(4 1)( 7) 0
x x
x x
− + =− − =
Therefore, the partition numbers are 14 and 7.
The number line would be broken into three parts involving the intervals
1 14 4( , ), ( ,7), and (7, ).−∞ ∞ Pick test values. We choose 0, 1, and 8 for the test values. Find
the function values using these three test values.
2(0) 4(0) 29(0) 7
(0) 7
f
f
= − +=
2(1) 4(1) 29(1) 7
(1) 4 29 7
(1) 18
f
f
f
= − += − += −
2(8) 4(8) 29(8) 7
(8) 256 232 7
(8) 31
f
f
f
= − += − +=
Only the test value 1 makes the inequality true, so the solution is 14 7,x< < or 1
4( ,7).
Calculus Chapter 3
3-16
7. 3 26
03
x x
x
+ >+
Find the partition numbers:
3 2
2
6 0
( 6) 0
x x
x x
+ =
+ =
3 0x + =
Therefore, the partition numbers are 0, –6, and –3. The number line would be broken into four parts involving the intervals ( , 6), ( 6, 3), ( 3,0), and (0, ).−∞ − − − − ∞ Pick test values. We choose –7, –4, –1, and 1 for the test values. Find the function values using these four test values.
3 2( 7) 6( 7) 49( 7) 12.25
7 3 4f
− + − −− = = =− + −
3 2( 4) 6( 4) 32
( 4) 324 3 1
f− + −− = = = −
− + −
3 2( 1) 6( 1) 5
( 1) 2.51 3 2
f− + −− = = =
− +
3 2(1) 6(1) 7(1) 1.75
1 3 4f
+= = =+
The test values of –7, –1, and 1 make the original inequality true, so the solution is
6 or 3,x x< − > − or ( , 6) ( 3, ).−∞ − ∪ − ∞
Calculus Chapter 3
3-17
Name ________________________________ Date ______________ Class ____________
Goal: To find the first derivative of a function using the four step process.
Section 3-4 The Derivative
Definition: Average Rate of Change For ( ),y f x= the average rate of change from x a= to x a h= + is
( ) ( ) ( ) ( )
, 0( )
f a h f a f a h f ah
a h a h
+ − + −= ≠+ −
where h is the distance from the initial value of x to the final value of x. Definition: Instantaneous Rate of Change For ( ),y f x= the instantaneous rate of change at x a= is
0
( ) ( )lim ,h
f a h f a
h→
+ − if the limit exists.
This formula is also used to find the slope of a graph at the point ( , ( ))a f a and to find the first derivative of a function, ( ).f x Procedure: Finding the first derivative: 1. Find ( ).f x h+ 2. Find ( ) ( ).f x h f x+ −
3. Find ( ) ( )
.f x h f x
h
+ −
4. Find 0
( ) ( )lim .h
f x h f x
h→
+ −
Calculus Chapter 3
3-18
1 - 3 Use the four step procedure to find ( )f x′ and then find (1),f ′ (2),f ′ and (3).f ′ 1. ( ) 6 9f x x= − Step 1: ( ) 6( ) 9 6 6 9f x h x h x h+ = + − = + − Step 2: ( ) ( ) 6 6 9 (6 9) 6f x h f x x h x h+ − = + − − − =
Step 3: ( ) ( ) 6
6f x h f x h
h h
+ − = =
Step 4: 0 0
( ) ( )lim lim 6 6h h
f x h f x
h→ →
+ − = =
( ) 6,f x =′ (1) 6,f =′ (2) 6,f =′ and (3) 6.f =′
2. 2( ) 2 4 7f x x x= − + −
Step 1: 2
2 2
2 2
( ) 2( ) 4( ) 7
2( 2 ) 4( ) 7
( ) 2 4 2 4 4 7
f x h x h x h
x xh h x h
f x h x xh h x h
+ = − + + + −
= − + + + + −
+ = − − − + + −
Step 2: 2 2 2
2
( ) ( ) 2 4 2 4 4 7 ( 2 4 7)
( ) ( ) 4 2 4
f x h f x x xh h x h x x
f x h f x xh h h
+ − = − − − + + − − − + −
+ − = − − +
Step 3: 2( ) ( ) 4 2 4
( ) ( )4 2 4
f x h f x xh h h
h hf x h f x
x hh
+ − − − +=
+ − = − − +
Step 4: 0 0
0
( ) ( )lim lim ( 4 2 4)
4 2(0) 4
( ) ( )lim 4 4
h h
h
f x h f xx h
hx
f x h f xx
h
→ →
→
+ − = − − +
= − − ++ − = − +
Therefore, ( ) 4 4f x x= − +′ (1) 4(1) 4
(1) 0
f
f
= − +′=′
(2) 4(2) 4
(2) 4
f
f
= − +′= −′
(3) 4(3) 4
(3) 8
f
f
= − +′= −′
Calculus Chapter 3
3-19
3. 2
( )5
xf x
x=
−
Step 1: 2( ) 2 2
( )( ) 5 5
x h x hf x h
x h x h
+ ++ = =+ − + −
Step 2: 2 2 2
( ) ( )5 5
(2 2 )( 5) 2 ( 5)
( 5)( 5) ( 5)( 5)
10( ) ( )
( 5)( 5)
x h xf x h f x
x h xx h x x x h
x h x x h x
hf x h f x
x h x
++ − = −+ − −+ − + −= −
+ − − + − −−+ − =
+ − −
Step 3: ( ) ( ) 10
( 5)( 5)
( ) ( ) 10
( 5)( 5)
f x h f x h
h x h x h
f x h f x
h x h x
+ − −=+ − −
+ − −=+ − −
Step 4: 0 0
20
( ) ( ) 10lim lim
( 5)( 5)
10
( 0 5)( 5)
( ) ( ) 10lim
( 5)
h h
h
f x h f x
h x h x
x x
f x h f x
h x
→ →
→
+ − −=+ − −−=
+ − −+ − −=
−
Therefore, 2
10( )
( 5)f x
x
−=′−
2 2
10 10 10 5(1)
16 8(1 5) ( 4)f
− − −= = = = −′− −
2 2
10 10 10 10(2)
9 9(2 5) ( 3)f
− − −= = = = −′− −
2 2
10 10 10 5(3)
4 2(3 5) ( 2)f
− − −= = = = −′− −
Calculus Chapter 3
3-20
4. The profit, in hundreds of dollars, from the sale of x items is given by
2( ) 2 5 6P x x x= − + a. Find the average rate of change of profit from x = 2 to x = 4. b. Find the instantaneous rate of change equation using the four–step procedure.
c. Using the equation found in part b, find the instantaneous rate of change when x = 2 and interpret the results.
Solution: a. First find the function values at 2 and 4. Note that 4 2 2.h = − =
2(2) 2(2) 5(2) 6
(2) 8 10 6
(2) 4
P
P
P
= − += − +=
2(4) 2(4) 5(4) 6
(4) 32 20 6
(4) 18
P
P
P
= − += − +=
Now use the average rate of change formula:
( ) ( ) (2 2) (2)
2(4) (2)
218 4
72
P a h P a P P
hP P
+ − + −=
−=
−= =
b. Step 1: 2
2 2
2 2
( ) 2( ) 5( ) 6
2( 2 ) 5( ) 6
( ) 2 4 2 5 5 6
P x h x h x h
x xh h x h
P x h x xh h x h
+ = + − + +
= + + − + +
+ = + + − − +
Step 2: 2 2 2
2
( ) ( ) 2 4 2 5 5 6 (2 5 6)
( ) ( ) 4 2 5
P x h P x x xh h x h x x
P x h P x xh h h
+ − = + + − − + − − +
+ − = + −
Step 3: 2( ) ( ) 4 2 5
( ) ( )4 2 5
P x h P x xh h h
h hP x h P x
x hh
+ − + −=
+ − = + −
Calculus Chapter 3
3-21
Step 4: 0 0
0
( ) ( )lim lim (4 2 5)
4 2(0) 5
( ) ( )lim 4 5
h h
h
P x h P xx h
hx
P x h P xx
h
→ →
→
+ − = + −
= + −+ − = −
Therefore, ( ) 4 5P x x= −′ c. (2) 4(2) 5 3P = − =′ means that when the second item was sold, your profit increased by $300. 5. The distance of a particle from some fixed point is given by
2( ) 5 2s t t t= + + where t is time measured in seconds. a. Find the average velocity from t = 4 to t = 6. b. Find the instantaneous rate of change equation using the four–step procedure.
c. Using the equation found in part b, find the instantaneous rate of change when t = 4 and interpret the results.
Solution: a. First find the function values at 4 and 6:
2(4) (4) 5(4) 2
(4) 16 20 2
(4) 38
s
s
s
= + += + +=
2(6) (6) 5(6) 2
(6) 36 30 2
(6) 68
s
s
s
= + += + +=
Now use the average rate of change formula. Note that 6 4 2.h = − =
( ) ( ) (4 2) (4)
2(6) (4)
268 38
152
s t h s t s s
hs s
+ − + −=
−=
−= =
Calculus Chapter 3
3-22
b. Step 1: 2
2 2
( ) ( ) 5( ) 2
( ) 2 5 5 2
s t h t h t h
s t h t th h t h
+ = + + + +
+ = + + + + +
Step 2: 2 2 2
2
( ) ( ) 2 5 5 2 ( 5 2)
( ) ( ) 2 5
s t h s t t th h t h t t
s t h s t th h h
+ − = + + + + + − + +
+ − = + +
Step 3: 2( ) ( ) 2 5
( ) ( )2 5
s t h s t th h h
h hs t h s t
t hh
+ − + +=
+ − = + +
Step 4: 0 0
0
( ) ( )lim lim (2 5)
2 0 5
( ) ( )lim 2 5
h h
h
s t h s tt h
ht
s t h s tt
h
→ →
→
+ − = + +
= + ++ − = +
Therefore, ( ) 2 5s t t= +′ c. (4) 2(4) 5 9s = + =′ means that after 2 seconds the particle is traveling at 9 units per second.
Calculus Chapter 3
3-23
Name ________________________________ Date ______________ Class ____________
Goal: To find the first derivatives using the basic properties
1 - 6 Find the indicated derivatives.
1. y′ for 8y x=
Use theorem 2 to find the derivative: 78y x=′
2. dydx
for 91x
y =
Convert the problem to the form nx and then use theorem 2.
991
xy x−= = 10
10 99dydx x
x−= − = −
Section 3-5 Basic Differentiation Properties
Notation: If ( ),y f x= then ( ), , dydx
f x y′ ′ all represent the derivative of f at x.
Theorems: 1. If ( ) ,y f x C= = then ( ) 0f x =′ (Constant Function Rule)
2. If ( ) ,ny f x x= = where n is a real number, then 1( ) nf x nx −=′ (Power Rule) 3. If ( ) ( ),y f x ku x= = then ( ) ( )f x ku x=′ ′ (Constant Multiple Property)
4. If ( ) ( ) ( ),y f x u x v x= = ± then ( ) ( ) ( )f x u x v x= ±′ ′ ′ (Sum and Difference Property)
Calculus Chapter 3
3-24
3. 2(5 6 3)ddu u u− +
Use theorem 4 to break the original function into pieces, then use a combination of theorems 1–3.
2 2(5 6 3) 5 6 3
10 6 0
10 6
d d d ddu du du du
u u u u
u
u
− + = − +
= − += −
4. ( )f x′ if 2.3 1.2( ) 7 5f x x x x= + − + Use theorem 4 to break the original function into pieces, then use a combination of theorems 1–3.
2.3 1.2
1.3 0.2
1.3 0.2
( ) ( ) 7 ( ) ( ) (5)
2.3 7(1.2) 1 0
2.3 8.4 1
f x f x f x f x f
x x
x x
= + − +′ ′ ′ ′ ′
= + − +
= + −
5. 41( 6 5 8)d
dx xx x+ − +
Use theorem 4 to break the original function into pieces, then use a combination of theorems 1–3.
4 4
12
12
5
1 1
4
5 12
34
( 6 5 8) 6 5 8
6 5 8
4 6( ) 5 0
5
d d d d ddx dx dx dx dxx x
d d d ddx dx dx dx
xx
x x x x
x x x
x x
−
−−
+ − + = + − +
= + − +
= − + − +
= − + −
6. ( )f x′ if 34 1
5 5 5( ) 4 2 5f x x x x−= + − + Use theorem 4 to break the original function into pieces, then use a combination of theorems 1 - 3.
34 1
5 5 5
61 25 5 54 12 2
5 5 5
( ) ( ) 4 ( ) 2 ( ) (5)f x f x f x f x f
x x x
−
− − −
= + − +′ ′ ′ ′ ′
= + +
Calculus Chapter 3
3-25
7. Given the function 2( ) 2 8 3f x x x= − + a. Find ( ).f x′ b. Find the slope of the graph of f at 3.x = c. Find the equation of the tangent line at 3.x = d. Find the value of x where the tangent is horizontal. Solution:
a. 2
2
( ) 2 8 3
( ) 2 ( ) 8 ( ) (3)
( ) 4 8
f x x x
f x f x f x f
f x x
= − +
= − +′ ′ ′ ′= −′
b. Substitute the value into the equation in part a to find the slope. ( ) 4 8
(3) 4(3) 8
4
f x x
f
m
= −′= −′=
c. Substitute the given value into the original function to find the y value of the point and then use that point with the slope found in part b. The point is (3, –3). 1 1( )
( 3) 4( 3)
3 4 12
4 15
y y m x x
y x
y x
y x
− = −− − = −
+ = −= −
d. The tangent is horizontal when the first derivative has a value of 0. ( ) 4 8
0 4 8
8 4
2
f x x
x
x
x
= −′= −==
Calculus Chapter 3
3-26
8 - 9 Use the following information for both problems: If an object moves along the y axis (marked in feet) so that its position at time x (in seconds) is given by the indicated function, find: a. The instantaneous velocity function '( )v f x= b. The velocity when 0x = and 4x = c. The time(s) when 0v =
8. 2( ) 3 12 8f x x x= − − Solution:
a. 2
2
( ) 3 12 8
( ) 3 ( ) 12 ( ) (8)
( ) 6 12
6 12
f x x x
f x f x f x f
f x x
v x
= − −
= − −′ ′ ′ ′= −′= −
b. ( ) 6 12
( ) 6 12
(0) 6(0) 12
(0) 12
f x x
v x x
v
v
= −′= −= −= −
( ) 6 12
( ) 6 12
(4) 6(4) 12
(4) 12
f x x
v x x
v
v
= −′= −= −=
c. ( ) 6 12
0 6 12
12 6
2
v x x
x
x
x
= −= −==
Calculus Chapter 3
3-27
9. 3 2212( ) 30f x x x x= − +
Solution:
a. 3 2212
3 2212
2
2
( ) 30
( ) ( ) ( ) 30 ( )
( ) 3 21 30
3 21 30
f x x x x
f x f x f x f x
f x x x
v x x
= − +
= − +′ ′ ′ ′
= − +′
= − +
b. 2
2
2
( ) 3 21 30
( ) 3 21 30
(0) 3(0) 21(0) 30
(0) 0 0 30
(0) 30
f x x x
v x x x
v
v
v
= − +′
= − +
= − += − +=
2
2
2
( ) 3 21 30
( ) 3 21 30
(4) 3(4) 21(4) 30
(4) 48 84 30
(4) 6
f x x x
v x x x
v
v
v
= − +′
= − +
= − += − += −
c. 2
2
2
( ) 3 21 30
0 3 21 30
0 3( 7 10)
0 3( 5)( 2)
2,5
v x x x
x x
x x
x x
x
= − +
= − +
= − += − −=
10. Find ( )f x′ if 2( ) (3 5)f x x= − First simplify the function by using the FOIL to expand it.
2 2( ) (3 5) (3 5)(3 5) 9 30 25f x x x x x x= − = − − = − + Now use theorem 4 to break the original function into pieces, then use a combination of theorems 1–3.
2
2
( ) 9 30 25
( ) 9 ( ) 30 ( ) (25)
( ) 18 30
f x x x
f x f x f x f
f x x
= − +
= − +′ ′ ′ ′= −′
Calculus Chapter 3
3-28
11. Find ( )f x′ if 9 5( ) .xx
f x −=
First simplify the function.
19 5 9 5( ) 9 5x xx x x
f x x−−= = − = −
Now use theorem 4 to break the original function into pieces, then use a combination of theorems 1 - 3.
2
1
1
2
5
( ) 9 5
( ) (9) 5 ( )
( ) 0 5
( )x
f x x
f x f f x
f x x
f x
−
−
−
= −
= −′ ′ ′
= +′
=′
Calculus Chapter 3
3-29
Name ________________________________ Date ______________ Class ____________
Goal: To use differentials to solve problems
1. Given the function 32 ,y x= find , , and yx
x y ∆∆∆ ∆ given 1 3x = and 2 6.x =
When 1 3,x = 31 2(3) 54.y = = When 2 6,x = 3
2 2(6) 432.y = =
6 3
3
x
x
∆ = −∆ =
432 54
378
y
y
∆ = −∆ =
3783 126y
x∆∆ = =
2. Given the function 32 ,y x= find , , and yx
x y ∆∆∆ ∆ given 1 2x = and 2 5.x =
When 1 2,x = 31 2(2) 16.y = = When 2 5,x = 3
2 2(5) 250.y = =
5 2
3
x
x
∆ = −∆ =
250 16
234
y
y
∆ = −∆ =
2343 78y
x∆∆ = =
Section 3-6 Differentials
Definition: Differentials If ( )y f x= defines a differentiable function, then the differential dy or df is defined as the product of ( )f x′ and dx, where .dx x= ∆ Symbolically, ( )dy f x dx= ′ or ( )df f x dx= ′ where .dx x= ∆ Recall that ( ) ( )y f x x f x∆ = + ∆ −
Calculus Chapter 3
3-30
3. Given the function 3 25 6 8 11,y x x x= − + + find .dy
3 2
2
2
5 6 8 11
15 12 8
(15 12 8)
dydx
y x x x
x x
dy x x dx
= − + +
= − +
= − +
4. Given the function 312(6 ),xy x= − find .dy
312
3 4112
2 313
2 313
(6 )
6
18
(18 )
x
dydx
y x
y x x
x x
dy x x dx
= −
= −
= −
= −
5. Given the function 2 325 7 ,y x x= − − find and dy y∆ given 4x = and 0.1.dx x= ∆ =
2 3
2
2
25 7
14 3
( 14 3 )
dydx
y x x
x x
dy x x dx
= − −
= − −
= − −
2
2
( 14 3 )
[ 14(4) 3(4) ](0.1)
( 56 48)(0.1)
( 104)(0.1)
10.4
dy x x dx
dy
dy
dy
dy
= − −
= − −= − −= −= −
2 3 2 3
( ) ( )
(4 0.1) (4)
(4.1) (4)
(25 7(4.1) (4.1) ) (25 7(4) (4) )
161.591 ( 151)
10.591
y f x x f x
y f f
y f f
y
y
y
∆ = + ∆ −∆ = + −∆ = −
∆ = − − − − −∆ = − − −∆ = −
Calculus Chapter 3
3-31
6. A company will sell N units of a product after spend x thousand dollars in advertising, as given by
2120N x x= − 10 60x≤ ≤ Approximately what increase in sales will result by increasing the advertising budget from $15,000 to $17,000? From $25,000 to $27,000?
2120
120 2
(120 2 )
dNdx
N x x
x
dN x dx
= −
= −
= −
(120 2 )
(120 2(15))(2)
180
dN x dx
dN
dN
= −= −=
(120 2 )
(120 2(25))(2)
140
dN x dx
dN
dN
= −= −=
Therefore, increasing the advertising budget from $15,000 to $17, 000 will result in an increase of 180 units and an increase from $25,000 to $27,000 will only result in a 140 unit increase. 7. The average pulse rate y (in beats per minute) of a healthy person x inches tall is given approximately by
590
yx
= 30 75x≤ ≤
Approximately how will the pulse rate change for a change in height from 49 inches to 52 inches?
1
2590
590y xx
−= =
1
2
32
32
590
295
( 295 )
dydx
y x
x
dy x dx
−
−
−
=
= −
= −
3
2
32
295343
885343
( 295 )
( 295(49) )(3)
( )(3)
2.58
dy x dx
dy
dy
dy
−
−
= −
= −
= −
= − ≈ −
Therefore, if a person grows from 49 to 52 inches, their pulse rate would decrease by approximately 2.5 beats per minute.
Calculus Chapter 3
3-32
Calculus Chapter 3
3-33
Name ________________________________ Date ______________ Class ____________
Goal: To solve problems involving marginal functions in business and economics
1 - 10 Find the indicated function if cost and revenue are given by
2( ) 3000 20 0.003C x x x= − + and 2( ) 5000 100R x x x= − 1. Marginal cost function
2( ) 3000 20 0.003
( ) 20 0.006
C x x x
C x x
= − += − +′
2. Average cost function
2( ) 3000 20 0.003 3000( ) 20 0.003C x x x
x x xC x x− += = = − +
Section 3-7 Marginal Analysis in Bus. And Econ.
Definition: Marginal Cost, Revenue, and Profit If x is the number of units of a product produced in some time interval, then total cost = ( )C x and marginal cost = ( )C x′ total revenue = ( )R x and marginal revenue = ( )R x′ total profit = ( ) ( )R x C x− and marginal profit = ( ) ( )R x C x−′ ′ Definition: Marginal Average Cost, Revenue, and Profit If x is the number of units of a product produced in some time interval, then
Cost per unit: average cost = ( )C xx
C = and marginal average cost = ( ) ( )ddx
C x C x′ =
Rev. per unit: average revenue = ( )R xx
R = and marginal average revenue = ( ) ( )ddx
R x R x′ =
Profit per unit: average profit = ( )P xx
P = and marginal average profit = ( ) ( )ddx
P x P x′ =
Calculus Chapter 3
3-34
3. Marginal average cost function
2
3000
1
2
3000
( ) 20 0.003
( ) 3000 20 0.003
( ) 3000 0.003
( ) 0.003
x
x
C x x
C x x x
C x x
C x
−
−
= − +
= − +
′ = − +
′ = − +
4. Marginal revenue function
2( ) 5000 100
( ) 5000 200
R x x x
R x x
= −= −′
5. Average revenue function
2( ) 5000 100( ) 5000 100R x x x
x xR x x−= = = −
6. Marginal average revenue function
( ) 5000 100
( ) 100
R x x
R x
= −
′ = −
7. Profit function
2 2
2
( ) ( ) ( )
(5000 100 ) (3000 20 0.003 )
( ) 100.003 5020 3000
P x R x C x
x x x x
P x x x
= −
= − − − +
= − + −
8. Marginal profit function
2( ) 100.003 5020 3000
( ) 200.006 5020
P x x x
P x x
= − + −= − +′
9. Average profit function
2( ) 100.003 5020 3000 3000( ) 100.003 5020P x x x
x x xP x x− + −= = = − + −
Calculus Chapter 3
3-35
10. Marginal average profit function
2
2
3000
3000
( ) ( ) ( )
( ) ( 100) ( 0.003)
( ) 100.003
x
x
P x R x C x
P x
P x
′ ′ ′= −
′ = − − − +
′ = −
11. Consider the revenue (in dollars) of a stereo system given by
1000( ) 1000x
R x x= +
a. Find the exact revenue from the sale of the 101st stereo.
b. Use marginal revenue to approximate the revenue from the sale of the 101st stereo.
Solution:
a. To find the exact revenue, find the revenue from the 101st and 100th and subtract their values:
1000
1000101
( ) 1000
(101) 1000(101)
(101) 101,009.90
xR x x
R
R
= +
= +
=
1000
1000100
( ) 1000
(100) 1000(100)
(100) 100,010
xR x x
R
R
= +
= +
=
(101) (100) 101009.90 100010 999.90R R− = − = Therefore, the actual revenue from the 101st stereo was $999.90. b. Find the marginal revenue formula and then substitute in 101.
2
1000
1
2
1000
( ) 1000
( ) 1000 1000
( ) 1000 1000
( ) 1000
x
x
R x x
R x x x
R x x
R x
−
−
= +
= +
= − +′
= − +′
2
2
1000
1000(101)
( ) 1000
(101) 1000
( ) 0.098 1000
( ) 999.90
xR x
R
R x
R x
= − +′
= − +′
= − +′=′
Therefore, the approximate revenue from the 101st stereo was $999.90.
Calculus Chapter 3
3-36
12. The total cost (in dollars) of manufacturing x units of a product is: ( ) 10,000 15C x x= + a. Find the average cost per unit if 300 units are produced.
b. Find the marginal average cost at a production level of 300 units and interpret the results.
c. Use the results in parts a and b to estimate the average cost per unit if 301 units are produced.
Solution: a. The cost of producing 300 units is (300) 10,000 15(300) 14,500.C = + =
Therefore the average cost is 14500300(300) $48.33.C = =
b. The average cost function is ( ) 10,000 15 1( ) 10000 15.C x xx x
C x x+ −= = = +
Therefore, the marginal average cost function for producing the 300th unit
would be:
2
1
2
10000
( ) 10000 15
( ) 10000
( )x
C x x
C x x
C x
−
−
= +
′ = −
= −
2
2
10000
10000(300)
( )
(300)
(300) 0.11
xC x
C
C
= −
= −
= −
This means that the average cost is decreasing by $0.11 per unit produced. c. The average cost per unit for the 301st unit would be $48.33 – $0.11 = $48.22.
Calculus Chapter 3
3-37
13. The total profit (in dollars) from the sale of x units of a product is:
2( ) 30 0.03 200P x x x= − + a. Find the exact profit from the 201st unit sold. b. Find the marginal profit from selling the 201st unit. Solution:
a. To find the exact profit, find the profit from the 201st and 200th and subtract their values:
2
2
( ) 30 0.03 200
(201) 30(201) 0.03(201) 200
(201) 5017.97
P x x x
P
P
= − +
= − +=
2
2
( ) 30 0.03 200
(200) 30(200) 0.03(200) 200
(200) 5000
P x x x
P
P
= − +
= − +=
(201) (200) 5017.97 5000 17.97P P− = − =
Therefore, the actual profit from the 201st unit was $17.97. b. Find the marginal profit formula and then substitute in 201.
2( ) 30 0.03 200
( ) 30 0.06
P x x x
P x x
= − += −′
( ) 30 0.06
(201) 30 0.06(201)
(201) 17.94
P x x
P
P
= −′= −′=′
Therefore, the approximate profit from the 201st stereo was $17.94.
Calculus Chapter 3
3-38
14. The total cost and revenue (in dollars) for the production and sale of x units are given, respectively, by:
( ) 32 36,000C x x= + and 2( ) 300 0.03R x x x= − a. Find the profit function P(x).
b. Determine the actual cost, revenue, and profit from making and selling 101 units.
c. Determine the marginal cost, revenue, and profit from making and selling the 101st unit.
Solution: a.
2
2
( ) ( ) ( )
( ) (300 0.03 ) (32 36,000)
( ) 0.03 268 36,000
P x R x C x
P x x x x
P x x x
= −
= − − +
= − + −
b. ( ) 32 36,000
(101) 32(101) 36,000
(101) 39,232
C x x
C
C
= += +=
( ) 32 36,000
(100) 32(100) 36,000
(100) 39,200
C x x
C
C
= += +=
2
2
( ) 300 0.03
(101) 300(101) 0.03(101)
(101) 29,993.97
R x x x
R
R
= −
= −=
2
2
( ) 300 0.03
(100) 300(100) 0.03(100)
(100) 29,700
R x x x
R
R
= −
= −=
( ) ( ) ( )
(101) (101) (101)
(101) 29,993.97 39,232
(101) 9238.03
P x R x C x
P R C
P
P
= −= −= −= −
( ) ( ) ( )
(100) (100) (100)
(100) 29,700 39, 200
(100) 9500
P x R x C x
P R C
P
P
= −= −= −= −
Actual Cost Actual Revenue Actual Profit (101) (100)
39, 232 39, 200
32
C C−−
(101) (100)
29,993.97 29,700
293.97
R R−−
(101) (100)
9238.03 ( 9500)
261.97
P P−− − −
Calculus Chapter 3
3-39
c. Marginal functions would be:
( ) 32 36,000
( ) 32
C x x
C x
= +=′
2( ) 300 0.03
( ) 300 0.06
R x x x
R x x
= −= −′
( ) ( ) ( )
( ) (300 0.06 ) (32)
( ) 0.06 268
P x R x C x
P x x
P x x
= −′ ′ ′= − −′= − +′
Marginal values for the 101st unit are: ( ) 32
(101) 32
C x
C
=′=′
( ) 300 0.06
(101) 300 0.06(101)
(101) 293.94
R x x
R
R
= −′= −′=′
( ) 0.06 268
(101) 0.06(101) 268
(101) 261.94
P x x
P
P
= − +′= − +′=′
Calculus Chapter 3
3-40
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