Download class exam II review

Second in-class exam, 3rd exam review
This exam is over the material in Chapter 4 and Sections 5.1, 5.2, 5.4 and 5.5. Study
this review, the material for quiz 4,quiz 5, quiz 6 and the appropriate homework assignments.
My recommendation: Solve each problem on paper. Then check your answers against the solutions.
•
Properties of graph:
Intervals of concavity, first and second derivative tests. Relative/absolute extrema.
Asymptotes.
1. Let f (x) = x200 − 200x
(a) Where is f concave up? concave down?
(b) Where does f have a relative maximum? a relative minimum?
Solution:
Take the first derivative. f 0 (x) = 200x199 − 200. Find the critical numbers: 200x199 − 200 = 0 to find
x = 1. ( The only real root.)
Find the second derivative. f 00 (x) = (200)(199)x198 .
(a) Find the sign for f 00
f 00
+
+
0
0
The sign of f 00 is positive everywhere. So the function is concave upward on all values of x.
(b) Put the critical number on the number line above to use the second derivative test.
f 00
+
0
0
+
1
The value of f 00 at the critical number is positive ( smiley face) and f has a relative minimum at x = 1.
Plug in to find the value is f (1) = (1)200 − 200(1) = −199. Relative minimum point is (1,-199) .
No Relative maximum because f 00 is concave upward everywhere.
2. (a) Find the relative maximum and relative minimum f (x) = 2x3 − 3ax2 where a > 0 is a constant.
(b) what is the inflection point of the graph?
Solution: (a) f 0 (x) = 6x2 − 6ax . Set equal to zero. 6x2 − 6ax = 0 gives 6x(x − a) = 0 so the critical
numbers are 0 and a
f 00 (x) = 12x − 6a
f 00 (0) = 12(0) − 6a = −6a < 0 so f has a relative maximum at x = 0
Find the value of function at x = 0, f (0) = 2(0)3 − 3a(0)2 = 0.
That is, (0, 0) is the relative maximum point.
f 00 (a) = 12(a) − 6a = 6a > 0 so f has a relative minimum at x = a
Find the value of function at x = a, f (a) = 2(a)3 − 3a(a)2 = 2a3 − 3a3 = −a3
That is (a, −a3 ). is the relative minimum point.
(b) Set the second derivative equal to zero. f 00 (x) = 12x − 6a = 0 implies x = a/2. Use test points 0 and
a. f 00 (0) = −6a < 0 and f 00 (a) = 6a > 0.
f 00
−−
0
a/2
+
So concavity changes as we move across a/2. Find the y-value for a/2: f (a/2) = 2(a/2)3 − 3(a(a/2)2 =
a3 /4 − 3a3 /4 = −a3 /2. That is,
(a/2, −a3 /2) is the inflection point.
x2 + b2
, where b > 0.
b2 − x2
(a) Find all vertical asymptotes of the graph of y = f (x).
(b) Find all horizontal asymptotes of the graph of y = f (x).
3. Let f (x) =
Solution:
(a) To find the vertical asymptotes, set the denominator equal to zero. b2 − x2 = 0 has two solutions
x = ±b. That is, the vertical asymptotes are x=b and x=-b .
(b) Horizontal asymptotes are obtained from the behaviour as x → ±∞:
x2 + b2
lim
= −1
x→∞ b2 − x2
x2 + b2
lim
= −1
x→−∞ b2 − x2
The horizontal asymptote is y=-1
x+b
where b > 0.
x−b
(a) Find all vertical asymptotes of the graph of y = f (x).
(b) Find all horizontal asymptotes of the graph of y = f (x).
(c) Find all relative maxima and all relative minima of the graph of y = f (x).
(d) Where is f increasing? Decreasing?
(e) Where is the graph of y = f (x) concave up? Where is it concave down?
(f) Sketch the graph of y = f (x).
4. Let f (x) =
Solution:
(a) To find the vertical asymptotes set the denominator equal to zero. x − b = 0 gives x=b .
(b) To find the horizontal asymptotes, find the limits as x → ±∞:
x+b
x+b
lim
= 1 and the same about lim
=1
x→∞ x − b
x→−∞ x − b
so the horizontal asymptote is y=1 .
(x − b)(1) − (x + b)(1)
−2b
=
.
(x − b)2
(x − b)2
f 0 (x) = 0 has no answer because the numerator is a constant.
f 0 (x) doesn’t exist at x = b. But x = b is not in the domain of f so it is not a critical number. So no
relative minima or maxima.
(d) Finding the interval of increasing/ decreasing:
Two test points are 0 and 2b.
(c) Find all critical numbers. f 0 (x) =
Page 2
−2b
−2
=
< 0.
(0 − b)2
b
f 0 (0) =
−2b
−2
=
< 0.
(2b − b)2
b
f 0 (2b) =
f0
−−
0
−−
b
2b
Decreasing on intervals (∞, b) and (b, ∞) .
No interval of increasing.
(e) f 00 (x) =
4b
(x − b)3
Two test points are 0 and 2b.
f 00 (0) =
4b
−4
= 2 < 0.
(0 − b)3
b
f 00 (2b) =
f 00
4b
4
= 2 > 0.
(2b − b)3
b
−−
0
+
b
2b
f is concave downward on interval (−∞, b) and concave upward on interval (b, ∞) .
(f) Find a couple of points by plugging in 0 and 2b in the original.
f (0) =
0+b
= −1 so first point is (0, −1) .
0−b
f (2b) =
2b + b
= 3 so second point is (2b, 3) .
2b − b
Find x-intercepts/ y-intercepts:
Y-intercept comes from setting x = 0. We already have done that and we know
point (0, −1) or y = −1 is the y-intercept .
x+b
= 0. That is the numerator is equal to zero. x + b = 0
x−b
point (−b, 0) or x = −b is the x-intercept.
x-intercept comes from setting y = 0. That is,
gives x = −b and
Use the three points and information in part (a)-(e) to sketch the graph
Page 3
(2b, 3)
y=1
f (x) =
x+b
x−b
x=b
(−b, 0)
(0, −1)
5. Let f (x) = ex − xe100 .
(a) Does the graph of y = f (x) have any relative maxima or relative minima? If so, what are they?
(b) Where is the graph of y = f (x) concave up? Where is it concave down?
Solution:
(a) f 0 (x) = ex − e100 Find the critical points: ex − e100 = 0 gives x = 100.
Use the first derivative test to see if a max or min. The two test points are
x = 0 and x = 200.
f 0 (0) = 1 − e100 < 0 and f 0 (200) = e200 − e100 > 0
−−
+
100
So there is a relative min at x = 100. The value of min is f (100) = e100 − 100e100 = −99e100
(b) f 00 (x) = ex
Page 4
f 00 is never zero and exists everywhere. So the sign doesn’t change. It is always positive. Therefore, the
function is concave upward everywhere. (−∞, ∞) .
6. Sketch the graph of y = ln(ax) where a > 1.
Solution:
• Domain is x > 0.
1
x
No critical numbers in the domain (0, ∞). So the sign of f 0 does not change. The sign of f 0 is always
positive on the domain.
So f is always increasing.
−1
• f 00 (x) = 2 is always negative. So f is concave downward everywhere.
x
• y = ln x + ln a so y 0 =
• No horizontal asymptotes
since lim f (x) does not exists.
x→∞
The vertical asymptote is t (as for any ln function) the solution to inside equal to zero. That is,
ax = 0. Vertical asymptote x = 0
• Find a few points on the graph.
What is the x-intercept: where y = 0, ln x + ln a = 0 which is ln x = − ln a.
1
1
. So ( , 0) is the x-intercept.
a
a
There is no y-intercept because x = 0 is not in the domain.
Find one more point at least. Plug in x = 1 to get point (1, ln a). Note that ln a > 0
By the log of difference rule, ln x = ln( a1 ). That is, x =
(1, a)
( a1 , 0)
1
a
1
a
y = ln(ax)
7. Sketch the graph of y = ln(ax) where 0 < a < 1.
Solution:
The process is identical to the process above. Except the point 1/a > 1 and ln a < 0 so
Page 5
( a1 , 0)
a
1
(1, a)
1
a
y = ln(ax)
8. Sketch the graph of the function y = f (x) where we know the following:
(i) limx→−∞ f (x) = 1
(ii) limx→∞ f (x) = −1
(iii) f 0 (−1) = f 0 (3) = 0
(iv) f 00 (−2) = f 00 (1) = f 00 (4) = 0
(v) f 00 (x) > 0 on intervals (−∞, −2) and (1, 4)
(vi) f 00 (x) < 0 on intervals (−2, 1) and (4, ∞)
(vii) f (−2) = 2, f (−1) = 3, f (1) = 0,f (3) = −3 and f (4) = −2
Solution:
• (i) and (ii) tell us two horizontal asymptotes. y = 1 and y = −1.
• Use (v) and (vi) to get the following for f 00
f 00
−
+
−2
−
+
1
4
• (iii) gives us the critical numbers −1, 3. Using the above number line, x = −1 gives a relative
maximum. So the relative max is point (−1, 3) . Using the above number line, x = 3 gives a
relative minimum. So the relative min is point (3, −3) .
• By (iv), (v) and (vi), f has inflection points at x = −2, 1, 4.
By (vii), they are (−2, 2), (1, 0) and (4, −2).
Page 6
(−1, 3)
(−2, 2)
x=1
(1, 0)
x = −1
(4, −2)
(3, −3)
Page 7
9. Sketch a graph of a function f with the following properties:
(i) f 0 (−1) = 0, f 0 (1) = 0 and f 0 (3) = 0.
(ii) f 00 (x) > 0 on intervals (−∞, 0) and (2, ∞).
(iii) f 00 (x) < 0 on interval (0, 2).
(iv) f 00 (0) = f 00 (2) = 0.
(v) f (−1) = −1, f (0) = 1, f (1) = 4, f (2) = 3 and f (3) = 3.
(vi) lim f (x) = ∞ and lim f (x) = ∞.
x→∞
x→−∞
Solution:
Parts (ii) and (iii) give the following
f 00
−
+
+
0
2
Part (i) gives three critical numbers: {−1, 1, 3}
Using the sign number line for f 00 and second derivative test: f has relative max at x = 1 and relative
min at x = −1, 3. By part (v), relative min at (−1, 1) and (3, 3) an relative max at (1, 4).
Again using the above sign number line, f has inflection points at x = 0, 2. By (v), they are (0, 1) and
(2, 3).
f
(1, 4)
4
(2, 3)
3
(3, 3)
2
1 (0, 1)
−2.
0
−1.
−1
(−1, −1)
−2
−3
−4
Page 8
1.
2.
3.
10. Consider the following graph of the function y = f (x):
y = f (x)
6.
4.
2.
−1.
0
1.
2
−2.
−4.
Mark all of the following statements which are correct.
(i) f 00 (−0.5) < 0
(ii) f 00 (−0.5) = 0
(iii) f 00 (−0.5) > 0
(iv) There is an inflection point at x = 0.
(v) There is an inflection point between x = 0 and x = 1.
(vi) f 00 (1) < 0
(vii) f 00 (1) = 0
(viii) f 00 (1) > 0
(ix) There is an inflection point between x = 1 and x = 2.
Solution: (iii) The graph is concave upward at x = −0.5.
(viii) The graph is concave upward at x = 1.
There are no inflection points because the concavity does not change.
Page 9
•
Absolute Maximum and minimum.
11. Find the absolute maximum and absolute minimum of the function f (x) = 12 e100x − e50x in the interval [−1, 2].
Solution:
Find all critical numbers:
f 0 (x) = 50e100x − 50e50x Set equal to zero to get critical numbers. 50e100x − 50e50x = 0. Now solve
e100x = e50x which gives 100x = 50x and so the only critical number x = 0 .
Now find all y-values for the set of end points and critical number. {−1, 0, 2}.
f (−1) = 21 e−100 − e−50 , f (0) 12 − 1 = − 12 and f (2) = 12 e200 − e100
Use your calculator or math logic to see that f (0) < f (−1) < f (2).
So Absolute minimum is (0, − 12 ) and Absolute maximum (2, 12 e200 − e100 ) .
12. Find the absolute maximum and absolute minimum of the function f (x) = x4 − 200x2 in the interval [−20, 20].
Solution: Find all critical numbers:
f 0 (x) = 4x3 − 400x . Set equal zero. 4x3 − 400x = 0 gives solutions x = −10, 0, 10 .
Now the set of all critical numbers and end points is {−20, −10, 0, 10, 20}
Find all y-values:
f (−20) = (−20)4 − 200(−20)2 = 80, 000
f (−10) = (−10)4 − 200(−10)2 = −10, 000
f (0) = (0)4 − 200(0) = 0
f (10) = (10)4 − 200(10) = −10, 000
f (20) = (20)4 − 200(20) = 8, 000
f (−10) = f (10) < f (0) < f (20) = f (−20)
So Absolute minimum at x = −10, 10 is −10, 000
and absolute maximum at x = −20, 20 is 80, 000 .
•
Optimization word problems.
13. A rectangular box with a square base is to be constructed so its surface area is 400 square inches.
y
x
x
(a) What dimensions will maximize its volume? And what is the maximum volume?
(b) How do you know that your answer in (a) is a maximum?
Solution:
(a)
• Find the constraint. The surface area equal to 400 square inches. That is, 2x2 + 4xy = 400 simplify
to x2 + 2xy = 200.
Page 10
• Solve the constraint for y: 2xy = 200 − x2 . and y =
200 − x2
.
2x
• Find the function that is being optimization. V (x) = x2 y.
• Replace y in the function by the value solved from the constraint. V (x) = x2
V (x) = 200x −
200 − x2
.
2x
x3
.
2
2
• Take the derivative to find the critical numbers. V 0 (x) = 100 − 3x2 and set equal to zero and solve.
r
200
x=±
.
3
Negative
values are dimension values so the answer only critical number within the domain: x =
r
200
.
3
• This gives you a relative maximum because the second derivative is negative at the critical number:
V 00 (x) = −3x
r
√
200
00
) = − 600 < 0
V (
3
r
r
r
200
200
200 3
• Find the value of y-value: V (
) = 100(
)−(
) /2 ' 544 .
3
3
3
r
200
' 8.2.
(b) We used the second derivative test to show that the relative maximum when x =
3
q
200 − ( 200
)2
3
q
' 8.2
and y =
2 200
3
and V 00 (x) < 0 for all x > 0. That is, f is concave downward on the domain. Which means the
relative max is the absolute max.
14. Postal regulations specify that a parcel sent by priority mail may have a combined length and girth of no more
than 144 in. Find the dimensions of a rectangular package that has a square cross section and largest volume
that may be sent by priority mail. Hint: The length plus the girth is 4x + l.
x
l
x
Solution:
• Constraint: Girth 4x + l = 144 solve for l to get l = 144 − 4x
• The function: V (x) = x2 .l = x2 (144 − 4x) = 144x2 − 4x3
• The derivative: V 0 (x) = 288x − 12x2 set equal to zero to get critical numbers: x = 0 which renders
volume 0 ( not acceptable) and x = 288
= 24
12
So back to l = 144 − 4x = 144 − 96 = 48
length 48in
Page 11
width 24in
height 24in
• What is the volume of such a package?
24 × 24 × 48 = 27, 648 in3
• Is this the maximum?
Check the critical numbers and end points. Domain is [0, 144/4].
V (0) = 0, V (36) = 0 and V (24) = 27, 648 in2 .
So 27, 648 is the maximum volume.
15. By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the
resulting flaps, an open box may be made. If the cardboard is 16 in. long and 12 in. wide, find the dimensions
of the box that will yield the maximum volume. (Round your answers to two decimal places.)
x
16
x
w = 12 − 2x
12
l = 16 − 2x
Solution:
• Constraints: 2x + l = 16 and 2x + w = 12
So the base of the box has dimensions l = 16 − 2x, and w = 12 − 2x.
• The height of the box is x. Recall that the formula for the volume of a rectangular box is:
V (x) = (16 − 2x)(12 − 2x)x = 4x3 + −56x2 + 192x
(where we have expanded the product to make finding the derivative simpler.)
Note that the domain of V (x) is [0, 6] ( Because all sides are supposed to be non-negative.)
• Taking the derivative, we find: V 0 (x) = 12x2 + −112x + 192
To find critical points, set equal to zero:
12x2 + −112x + 192 = 0
√
112± 1122 −4(12)(192)
Use quadratic formula, x =
2(12)
x = 2.26 , or x = 7.07.
We see that the value x = 7.07 lies outside the domain for V (x).
• Does the critical number give the absolute max?
We now compute the values of V (x) for the remaining critical number and at the endpoints of the
domain {0, 2.06, 6}, list the function values:
V (2.26) = 194.07
V (0) = 0
Page 12
V (6) = 0
So absolute maximum volume, 194.07 occurs at the critical number in the domain for V (x),
that is, at x = 2.26 .
• What are the dimensions of the box?
x=2.26 , w = 12 − 2(2.26) = 7.48
and l = 16 − 2(2.26 = 11.48 units.
16. Linda’s Lemonade Stand sells glasses of lemonade at baseball games for $1 a glass. At that price she sells 500
glasses of lemonade a game. She’s noticed that for every 5 cents she raises the price, she sells 10 fewer glasses.
(a) What price gets her the maximum revenue? (Round off to the nearest penny). And what is the maximum
revenue (round off to the nearest penny)?
(b) How do you know that your answer in (a) is a maximum? An absolute maximum?
Solution:
• Find the independent variables. x: Number of increase in price by 5 cents.
• Find a relationship between the variables.
Number of items sold = 500 − 10x and price= 1 + 0.05x.
• Function to be optimized: R(x) = (500 − 10x)(1 + 0.05x)
• Take the derivative of it: R0 (x) = (−10)(1 + 0.05x) + (500 − 10x)(0.05) = −10 − 0.5x + 25 − 0.5x =
15 − x.
• Set equal to zero to find the critical numbers: 15 − x = 0 gives x = 15.
• Use the second derivative to see if a maximum: R00 (x) = −1 < 0 everywhere so critical number gives
absolute maximum.
• Use the original to find the maximum value: R(15) = (500 − (10)(15))(1 + 0.05(15)) = 612.5
17. You are constructing a garden which will be separated into 4 plots as shown:
x
y
You will surround the garden by a rectangular fence, and separate the plots with fencing material. You have
100 yards of fencing material to use.
(a) What are the dimensions that will maximize the area of the garden? And what is the maximum area?
(b) How do you know that your answer in (a) is a maximum? An absolute maximum?
Solution: (a)
• Find the constraint. The limited fences. 5x + 2y = 100.
• Solve for y =
100 − 5x
5
= 50 − x .
2
2
• Find the function that is optimized. The area. A(x) = xy
• Replace the y by the value found. A(x) = x(50 − 5x/2).
So A(x) = 50x − 5x2 /2 .
Page 13
• Take the derivative A0 (x) = 50 − 5x and set equal zero.
50 − 5x = 0 gives critical number: x = 10
• The second derivative is A00 (x) = −10 < 0 so the ”only” critical value gives the absolute maximum.
• The value of maximum is A(10) = 50(10) − 5(10)2 /2 = 250 .
(b) Because the second derivative is negative everywhere and there is only one critical number.
18. A rectangular box is to have a volume of 1.6 f t3 . For some reason, it has been decided that the height of
the box is to be 2 times its width. The material for the sides and bottom costs 40 cents per square foot and
the material for the top costs 20 cents a square foot. What are the dimensions of the enclosure that can be
constructed at a minimum cost?
Solution:
• Constraints z = 2x and xyz = 1.6 which gives y = .8/x2
• The function to be optimized: C(x) = 40 × (2xz + 2yz + xy) + 20xy =
!
40 ×
(2x)(2x) + (2)(0.8/x2 )(2x) + (x)(0.8/x2 )
+ 20(x(0.8/x2 ) = 160x2 + 128/x + 32/x + 16/x =
160x2 + 176/x
320x3 − 176
set it equal to zero to find the critical number x ' .819.
x2
• Find the second derivative:
352
C 00 (x) = 320 + 3 is always positive for positive numbers. So the minimum is rendered at critical
x
number x = 0.819.
• C 0 (x) = 320x − 176/x2 =
• Dimensions:
x = 0.819
z = 2 × x ' 1.638
y ' 1.19
•
Exponential and logarithmic rules.
19. Simplify the following expressions:
2y+1 e3x
ex 2y
(b) (22x 33x 22x−1 )3
(a)
(c) ln((e3x ex+1 )1/3 )
√ √
ex 2x
(d) ln(
)
2e
Solution: (a)
2y+1 e3x
= 2y+1 e3x e−x 2−y = 2y+1 2−y e3x e−x = 2y+1−y e3x−x = 2e2x .
ex 2y
(b) (22x 33x 22x−1 )3 = 26x 39x 26x−3 = 26x 26x−3 36x = 26x+6x−3 39x = 212x−3 39x .
!
!
(c) (1/3) ln(e3x ex+1 ) = (1/3) ln(e3x ) + ln(ex+1 )
= (1/3) 3x + (x + 1)
= (1/3)(4x + 1) .
√ √
√
√
ex 2x
(d) ln(
) = ln( ex ) + ln( 2x ) − ln(2e) = (1/2) ln(ex ) + (1/2) ln(2x ) −
2e
!
(x/2) + (x ln(2)/2) − ln(2) − 1 = or (ln(2) + 1) (1/2)x − 1
Page 14
!
ln(2) + ln(e)
=
•
Solving exponential and logarithmic equation.
20. Solve 2x
2
+1
= 25
Solution: x2 + 1 = 5 so x2 = 4 which gives x = ±2 .
x
21. Solve log10 ( 10
)=2
x
= 102
10
Solution:
x = 1000 .
22. Solve
50
= 20.
1 + 4e2t
Solution:
• Multiply both side by the denominator:
• 50 = 20(1 + 4e2t ) That is 50 = 20 + 80e2t
• Solve for the exponential term: 30 = 80e2t
3
• Then e2t =
8
3
• Take ln of both side: 2t = ln( )
8
3
• t = (1/2) ln( ) ' −0.49 .
8
•
Derivative of Exponential and logarithm.
23. Find the derivatives of the following functions:
2
3
(a) f (x) = ex −x
(b) g(x) = ln(x2 − x3 )
(c) h(x) = ln(x2 ex )
(d) j(x) = e1/(x
2
+1)
Solution: (a) f 0 (x) = ex
(b) g 0 (x) =
2
−x3
(2x − 3x2 ) .
2x − 3x2
.
x2 − x3
(c) If you use
derivative of inside
2xex + x2 ex
2+x
which simplifies to h0 (x) =
then h0 (x) =
inside
x2 ex
x
Or you can use the simplified form h(x) = 2 ln(x) + x and get h0 (x) =
2
(d) j 0 (x) = e1/(x
+1)
(−2x/(x2 + 1)2 )
Page 15
2
x
+ 1 . ( The same answer.)
24. The number of widgets an employee can produce t months after starting to work at the widget factory is
Q(t) = 100 − 25e−0.3t per day. (a) When an employee is very experienced, how many widgets per day can
she be expected to produce? (b) What is the rate of change in number of widget produced at t = 2.
Solution:
(a) Very experience means t → ∞ so lim 100 − 25e−0.3t = 100
t→∞
(b) Q0 (t) = 7.5e−0.3t and f 0 (2) = 7.5e−0.6 .
•
Using the logarithmic differentiation to find the derivative.
25. Using logarithmic differentiation, find the derivative of f (x) = (x − 1)(x + 1)2 (x − 2)3 (x + 2)4
Solution: Product rule is messy for this function so we try to switch to another method.
Take the logarithm of both sides.
!
ln(f (x)) = ln (x − 1)(x + 1)2 (x − 2)3 (x + 2)4
= ln(x − 1) + 2 ln(x + 1) + 3 ln(x − 2) + 4 ln(x + 2)
Take the derivative of both sides:
f 0 (x)
1
2
3
4
=
+
+
+
f (x)
x−1
x+1
x−2
x+2
Solve for f 0 (x):
!
f 0 (x) =
1
x−1
+
2
x+1
+
3
x−2
+
4
x+2
f (x) =
Replace f (x) by its value:
!
0
f (x) =
1
x−1
+
2
x+1
+
3
x−2
+
4
x+2
!
2
3
(x − 1)(x + 1) (x − 2) (x + 2)
26. Using logarithmic differentiation, find the derivative of f (x) = xx
Solution: Take the logarithm of both sides:
ln(f (x)) = x ln(x) and take the derivative:
f 0 (x)
= ln(x) + x(1/x) = ln(x) + 1
f (x)
Solve for f 0 (x):
!
!
f 0 (x) =
ln(x) + 1
f (x)
Replace f (x) for what it is:
!
!
f 0 (x) =
ln(x) + 1
xx
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4
SHORT NOTES:
• Finding vertical asymptote of a rational function
P (x)
is a rational function where P (x) and Q(x) are polynomials.
Suppose f (x) =
Q(x)
Then, the line x = a is a vertical asymptote if Q(a) = 0 and P (a) 6= 0. ( Zeros of the denominator.)
• Horizontal asymptote
The line y = b is a horizontal asymptote of a function f if either lim f (x) = b or lim f (x) = b
x→∞
x→−∞
• In general the behavior at infinities is the limit as x → ±∞ infinities.
It could be a number ( including 0) which gives the horizontal asymptote or it could be ±∞.
• The Extreme Value Theorem
If a function is continuous on a closed interval [a, b], then f has both an absolute maximum and absolute
minimum value on [a, b].
• Finding the absolute Extrema of f on a closed interval [a, b]
1. Make sure that the function is continuous on the interval.
2. Find the critical numbers of f that lie in (a, b).
3. Compute the values of f at each Critical number and find f (a) and f (b).
4. Absolute max is largest number computed in (c) and absolute min is the smallest one.
• An easy method of finding whether absolute min or max is achieved at a critical number c.
Take the second derivative,
• If the second derivative is always positive on the domain, then absolute min is achieved at c.
• If the second derivative is always negative then the absolute max is achieved at c.
• Laws of Exponents
Let a and b be positive numbers and let x and y be real numbers. Then,
1. bx .by = bx+y
bx
2. y = bx−y
b
3. (bx )y = bxy
4. (ab)x = ax bx
a x
ax
5. ( ) = x
b
b
• Laws of Logarithms
If m and n are positive numbers and b > 0, b 6= 1, then
1. logb mn = logb m + logb n
2. logb
m
n
= logb m − logb n
n
3. logb m = n logb m
4. logb 1 = 0
5. logb b = 1
• Laws of exponents in English. (These only serve to help you to memorize the laws.)
1. Product of two exponentials with like base is the base to sum of the exponents.
2. Quotient of two exponentials with like base is the base to difference of the exponents.
3. To power to power is to product of the powers.
4. Product to a power is first to the power times the second to the power.
5. Quotient to a power is top to the power divided by the bottom to the power.
• Laws of logarithm in English (These only serve to help you to memorize the laws.)
1. log of product is equal to sum of logs.
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2. log of quotients is equal to difference of logs.
3. log of a number to a power is the power times the log of the number.
4. log of 1 is zero.
5. log of base is 1.
• The Chain Rule for Exponential and logarithmic Functions
If f (x) is a differentiable function, then
d
• dx
(ef (x) ) = ef (x) f 0 (x)
•
d
[ln f (x)]
dx
=
f 0 (x)
f (x)
(f (x) > 0)
• Logarithmic Differentiation:
1. Take the natural log of both side of the equation and simplify the complicated log into logs with simpler
inside functions.
d
y0
2. Differentiate both side of the equation. Remember Right hand side:
(ln(y) = .
dx
y
3. Multiply the right side bu y and the left side by the rule for the function y.
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