Review Problems – Math115 Final Exam
(Final covers Sec 2.1-5.2, 5.4-6.5)
• Final Exam, Monday, Dec 12, 2016, 4:30PM - 7:00PM, Budig 120 unless Mallot 2001(
Jocobs), Mallot 2074( Pham), Snow 120( Beaty), Lindley 317 ( Serio). ( Your instructor
will announce the room in class.) Once you are in the auditorium, look for your instructor
who will direct you to your seat.
• Only simple graphing calculators (TI-84 plus and below) are allowed for the common exams. NO TI-NSPIRE ALLOWED.
• You can do all your work in exam booklets and circle one answer for each problem. Having
done with all the problems, you will need to complete the bubble sheet with a # 2 pencil.
• You will need to write your instructors name on margins of the scantron answer sheet.
• You are considered responsible to bring pencils and a calculator to the common exams. Pens
or pencils will not be provided for you, and interchanging calculators will be prohibited
during the exams.
• Bring your KU ID.
• The final exam is comprehensive and will consist of 12 True or False problems and 25
Multiple-Choice problems. Each True/False question will be worth 5 points and each MultipleChoice will be worth 10 points. The practice problems below are in addition to the practice
problems on the Midterm review:
Click for MIDTERM-REVIEW
• To study for this exam, start with this review and all of the previous reviews. Then review
your notes and homework assignments.
True or False Problems
1. T F If f 0 (c) = 0, then x = c is a critical number of f .
Solution: True By the definition, a CRITICAL number of a function f is any number x in
the domain of f such that f 0 (x) = 0 or f 0 (x) does not exists.
2. T F If f (c) is an absolute extrema of f (x), then x = c is a critical number of f (x).
Solution: False Some absolute extrema are achieved at end points.
3. T F If x = 1 is a critical number of f (x), then f (1) is either a relative maximum or a relative
minimum of f (x).
Solution: False Counter example: f (x) = (x − 1)3 . Graph it using your calculator and
observe that point (1, 0) is not a relative extremum. But f 0 (x) = 3(x − 1)2 so f 0 (1) = 0.
4. T F Any continuous function must have absolute extrema.
Solution: False Counter example: f (x) =
1
does not have any absolute extrema on (0, ∞)
x
5. T F Any continuous function on [a, b] has absolute extrema on [a, b].
Solution: True This is the extreme value theorem.
6. T F No continuous function on (a, b) can have absolute extrema on (a, b).
Solution:
3
− 6x2 + 11x − 6 on interval (0, 3) has absolute minimum:
False
√ Counter example: y = x √
2 3
2 3
−
and absolute maximum:
.
9
9
Graph using your calculator.
b
Z
7. T F If f is continuous on [a, b], then
f (x)dx exists.
a
Solution: True Any continuous function is integrable.
8. T F The area of the region bounded by
Z the graph of a nonnegative, continuous function f and
1
above the x-axis over the interval [0, 1] is
f (x)dx.
0
Page 2
Solution: True
9. T F If f 0 > 0 on [2, 4], then f is concave upward on interval [2, 4].
Solution: False Counterexample: f (x) = −x3 + 9x2 − 24x.
Note that f 0 (x) = −3x2 + 18x − 24 and f 0 > 0 on [2, 4].
Also, note that the theorem says that if f 00 > 0, then the graph of f is concave upward.
10. T F If f and g are both positive and increasing on (a, b), then f g is increasing on (a, b).
Solution: True If a < b then f (a) < f (b) ( because f is increasing) and g(a) < g(b)
(because g is increasing). So f (a)g(a) < f (b)g(b).
11. T F If f is concave up on (a, b), then the graph of −f is concave down on (a, b).
Solution: True
12. T F If c is a critical number of f where a < c < b and f 00 < 0 on (a, b), then f has a relative
maximum at x = c.
Solution: True f 00 (c) < 0 so by second derivative test, f has a relative maximum at x = c.
13. T F ex ey = exy
Solution: False ex ey = ex+y 6= exy
14. T F If x < y, then ex < ey .
Page 3
Solution: True f (x) = ex is strictly increasing.
15. T F ln(x3 ) = 3 ln(x) for x > 0.
Solution: True by Logarithmic rules.
16. T F (ln x)3 = 3 ln x for all x > 0.
Solution: False
17. T F ln a − ln b = ln(a − b) for all positive real numbers a and b.
Solution: False
ln a − ln b = ln(a/b) 6= ln(a − b).
2
18. T F ln(x2 ex ) = ln x2 + x2 for all x > π.
Solution: True
2
2
ln(x2 ex ) = ln(x2 ) + ln(ex ) = ln(x2 ) + x2 ln(e) = ln(x2 ) + x2 .
19. T F The function f (x) = ln |x| is continuous for all x < 0.
Solution: T
20. T F If f (x) = 3x , then f 0 (x) = x · 3x−1 .
Solution: False
This requires a logarithmic differentiation process:
Page 4
• ln(y) = ln(3x )
• ln(y) = x ln(3)
•
y0
= ln(3) ( Remember ln(3) is a number so a coefficient.)
y
• y 0 = ln(3)y
• y 0 = ln(3)3x .
21. T F If f (x) = eπ , then f 0 (x) = eπ .
Solution: False f (x) = eπ is a constant function. So f 0 (x) = 0
22. T F If f has a derivative for all x and h(x) = f (πx), then h0 (x) = πf 0 (πx).
Solution: True h(x) = f (g(x)) where g(x) = πx. By chain rule, h0 (x) = f 0 (g(x))g 0 (x) =
f 0 (pix)π
23. T F If f (x) = π 2 , then f 0 (x) = 2π.
Solution: False
f (x) = π 2 is a constant function so the derivative is 0.
24. T F If f (x) = ln 5, then f 0 (x) = 15 .
Solution: False f (x) = ln(5) is a constant function. Its derivative is zero.
25. T F If f (x) = ln ax , then f 0 (x) = ln a.
Solution: True
f (x) = x ln(a) so f 0 (x) = ln(a) .
Note that ln(a) is a coefficient.
Page 5
26. T F If f (x) =
1
−1
, then f 0 (x) =
.
ln 2x
2x ln 2x
Solution: False
f (x) = (ln(2) + ln(x))−1 so by chain rule:
1
−1
f 0 (x) = (−1)(ln(2x))−2 ( ) =
x
x(ln(x))2
27. T F If f (x) =
1
−1
, then f 0 (x) =
.
2x
e
2xe2x
Solution: False
f (x) = e−2x so by chain rule f 0 (x) = e−2x (−2) =
−2
e2x
1
−1
28. T F If f (x) = √ , then f 0 (x) = √ .
2x
2x 2x
Solution: True
f (x) = (2x)(−1/2) so f 0 (x) = (−1/2)(2x)(−3/2) (2) = −
−1
1
=
3/2
(2x)
(2x)(2x)1/2
29. T F If F and G are antiderivatives of f on an interval I, then F (x) = G(x) + C on I.
Solution: True
Any two antiderivatives of a function are only different by a constant.
Z 30. T F If f and g are both integrable, then
Z
Z
2f (x) − 3g(x) dx = 2 f (x)dx−3 g(x)dx.
Solution: True
Z Z Z Z
Z
2f (x) − 3g(x) dx =
2f (x) dx −
3g(x) dx = 2 f (x)dx − 3 g(x)dx.
Page 6
Z
31. T F If f and g are both integrable, then
Z
f (x)g(x)dx =
Z
f (x)dx
g(x)dx.
Solution: False
2
Z
(1 − x)dx, gives the area of the region under the graph of f (x) = 1 − x and above
32. T F
0
the x-axis on the interval [0, 2].
Solution: False
f (x) = 1 − x takes negative values on interval [1, 2].
33.
T F The total revenue realized in selling the first 5000 units of a product is given by
Z 5000
R0 (x) dx = R(5000) − R(0) with R(x) the total revenue.
0
Solution: True
Note that: R(0) = 0.
Z
2
√
34. T F
2
ex
dx = 0
1+x
Solution: True
Any definite integral with equal limits is zero.
Z
3
35. T F
1
dx
=−
x − 12
Z
1
3
dx
x − 12
Solution: True By properties of integral.
Z
1
√
x x + 1 dx =
36. T F
0
√
Z
x+1
0
1
√
2
x dx =
2
Page 7
Solution: False
Z
00
2
f 00 (x) dx = f 0 (2) − f 0 (0).
37. T F If f (x) is continuous on [0, 2], then
0
Solution: True f 0 is the antiderivative f 00 .
Z
38. T F If f is continuous on [a, b] and a < c < b, then
c
Z
f (x) dx =
b
c
Z
f (x) dx−
a
b
f (x) dx.
a
Solution: True
Z c
Z b
Z b
Z c
Z b
Z c
f (x) dx +
f (x) dx = −
f (x) dx.
f (x) dx −
f (x) dx =
f (x) dx −
a
a
a
a
Page 8
c
c
Multiple-Choice Problems
1. Let f (x) = ln(2 − x). The domain of f is
(A) (−∞, +∞)
(B) (−2, +∞) (C) (−∞, −2) (D) (−∞, 2) (E) (2, ∞)
Solution: D
ln is only defined on positive values. That is, inside ln should be positive. That is, 2 − x > 0.
That is, x < 2 or (−∞, 2)
2. Find the vertical asymptotes of function f (x) =
2+x
.
(1 − x)2
(A) x = −2 (B) x = 1 (C) y = 0 (D) y = −2 (E) y = 1
Solution: B The fraction is simplified so set the denominator equal to zero.
3. The equation of the tangent line to the graph of y = x ln x at the point (1,0) is
(A) y = x + 1
(B) y = x − 1 (C) y = (x + 1)e
(D) y = (x − 1)e (E) y − 1 = x + 1
Solution:
B
1
By product rule, f 0 (x) = ln(x) + x( ) = ln(x) + 1
x
0
Slope=f (1) = ln(1) + 1 = 1
Point-Slope: y − 0 = 1(x − 1)
4. The equation of the tangent line to the graph of y = ln(x2 ) at the point (2, ln 4) is
(A) y = x + 2 − ln 4 (B) y = 2(x − 2) − ln 4 (C) y = 2(x − 2) + ln 4
(D) y = x − 2 + ln 4
(E) y = x − 2 − ln 4
Solution:
D
Page 9
Simplify: y = 2 ln(x)
Take derivative: y 0
2
x
2
=1
2
Point-slope: y − ln(4) = x − 2 which gives y = x − 2 + ln(4).
m=
5. The equation of the tangent line to the graph of y = e2x−3 at the point ( 32 , 1) is
(A) y = 2e2x−3
(B) y = 2x − 4 (C) y = 2x − 2
(D) y = 2x − 3 (E) y = x −
Solution: C
Use chain rule, y 0 = e2x−3 (2)
m = e2(3/2)−3 (2) = e0 (2) = 2
Point-slope: y − 1 = 2(x − 3/2) which gives y = 2x − 2.
2
6. The equation of the tangent line to the graph of y = e−x at the point (1, 1/e) is
2
1
2
1
(A) y = − (x + 1) +
(B) y = − (x − 1) −
e
e
e
e
1
1
1
1
(D) y = (x − 1) +
(E) y = (x + 1) +
e
e
e
e
2
1
(C) y = − (x − 1) +
e
e
Solution: C
2
Use chain rule: y 0 = e−x (−2x) so m = e−1 (−2)(1) =
point-slope: y −
−2
.
e
1
2
−2
1
= (x − 1) which gives y =
(x − 1) +
e
e
e
e
√
7. The absolute minimum value of the function f (x) = 12 x2 − 2 x on [0, 3] is:
(A) −
3
2
(B) 0 (C)
√
9
− 2 3 (D) 1
2
(E) 2
Solution:
A
Page 10
1
2
√
1
x x−1
√
f (x) = x − √ =
x
x
0
Find all critical numbers:
f 0 does not exists at x = 0 which is in the domain.
f 0 = 0 gives x = 1.
list all critical numbers within the interval and the end points. {0, 1, 3}
Plug in the original:
f (0) = 0
√
1
f (1) = 12 − 2 1 =
2
√
1
f (3) = 32 − 2 3 =
2
−3
(absolute minimum.)
2
√
9
− 2 3 ' 1.036 ( absolute maximum.)
2
8. The absolute maximum value of the function f (x) =
(A) 0
(B) 1
(C) 2
(D) e
1
is:
1 + x2
(E) ln 2
Solution: B
Find the critical numbers:
−2x
f 0 (x) =
(1 + x2 )2
Domain of f is all real numbers. (Since 1 + x2 > 0 always.)
f 0 exists everywhere also.
The only critical number comes from f 0 (x) = 0 and that is x = 0.
No interval is given so see if the single critical number gives the maximum:
I use first derivative test here:
f0
+
0
0
−
So the absolute maximum is achieved at x = 0.
1
The value of absolute maximum is f (0) =
=1
1 + 02
9. The absolute maximum value of the function f (t) = te−t is:
Page 11
(A) 0
(B) 1
(C)
1
e
(D) e
(E) − 1
Solution: C
Use product rule: f 0 (x) = e−t − te−t = e−t (1 − t)
Set equal to zero to find the critical number: e−t (1 − t) = 0 gives t = 1.
See if the second derivative works and the only critical number gives the maximum.
f 00 (x) = −e−t (1 − t) − e−t = e−t (t − 2)
f 00 (1) = e−1 (−1) = −e−1 < 0 so x = 1 gives the relative maximum but since there is only
one critical number and no point of discontinuity, the relative max is absolute max.
Plug in the original to find the value of the max:
f (1) = (1)e−1 =
1
e
1
1
ln( ) on [1, 2] is:
t
t
1
(D) − 2 (E) −
2
10. The absolute minimum value of the function f (t) =
(A)
− ln 2
2
(B)
−1
2 ln 2
(C) − 2 ln 2
Solution:
A
Find the critical numbers:
1
( Note that f (t) = (− ln(t))
t
−1
1 −1
−1
By product rule: f 0 (t) = ( 2 (− ln(t)) + ( ))( ) = 2 (1 − ln(t))
t
t
t
t
0
Set f = 0 to get 1 − ln(t) = 0 and that is t = e.
Since the absolute value is being calculate on a closed interval and f is continuous on the
interval, use the following method. ( The absolute extremum theorem.)
List all critical numbers within the interval and the end points. ( Note that e > 2 and not in the
interval.) {1, 2}
1
f (1) = (− ln(1)) = 0 ( Absolute max)
1
1
− ln(2)
' −0.3466 ( absolute Min)
f (2) = (− ln(2)) =
2
2
Page 12
11. Let f (x) = 13 x3 − x2 + x − 6. Find the interval(s) where f is concave downward.
(A) (−∞, 1), (1, 2) (B) (−∞, 1), (2, ∞) (C) (−2, 1), (2, ∞) (D) (−∞, −2), (1, 2) (E) (−∞, 1)
Solution:
E
f 0 (x) = x2 − 2x + 1
Find the second derivative: f 00 (x) = 2x − 2.
Set 2x − 2 = 0 so x = 1.
f 00
−
0
1
+
12. Let f (x) = 13 x3 − x2 + x − 6. Find the inflection points.
(A) (1, f (1)) (B) ( 21 , f ( 12 )) (C) (0, f (0)) (D) (2, f (2)) (E) (3, f (3))
Solution: A
f 0 (x) = x2 − 2x + 1
Find the second derivative: f 00 (x) = 2x − 2.
Set 2x − 2 = 0 so x = 1.
f 00
−
0
1
+
At x = 1 a change of concavity happens and the graph has a tangent line at x = 1 so x = 1 is
the x-value of inflection point. Now the inflection point is (1, f (1)).
13. Let f (x) = x ln x. Determine the intervals where the function is decreasing.
1 1
1 1
1 1
1
1
(A) (−∞, ), ( , e) (B) (−∞, ), ( , ∞) (C) (0, ), ( , e) (D) (0, ), (e, ∞) (E) (0, )
e e
e e
e e
e
e
Solution: E
Use product rule: f 0 (x) = ln(x) + x(1/x) = ln(x) + 1.
Set equal to zero: ln(x) + 1 = 0 which solves to ln(x) = −1 and x = e−1 .
Page 13
Note that domain is where x > 0 ( because of ln). That is, (0, ∞).
f0
−
+
0
1
e
14. Find the derivative of function y = xln x .
(A) y 0 = (ln x)2
(B) y 0 =
(D) y 0 = ln x(xln x−1 )
2 ln x ln x
x
x
(C) y 0 = xln x
(E) y 0 = 2 ln x(xln x )
Solution: B
Use logarithmic differentiation:
• Take natural logarithm of both sides: ln(y) = ln(xln(x) )
• Simplify the right hand side:ln(y) = (ln(x))(ln(x)) = (ln(x))2
• Take derivative of both sides:
y0
1
2
= 2(ln(x))( ) = ( ) ln x ( We used a chain rule here
y
x
x
with ln as the inner function.)
• y 0 = ( 2(lnx x) )y
• y 0 = ( 2 lnx x )xln x
15. Find the derivative of function y = 10x .
(A) y 0 = 10x ln 10 (B) y 0 = 10x
(D) y 0 = x10x−1
(E) y 0 =
(C) y 0 = 1010 ln 10x
10x
10
Solution:
A Use logarithmic differentiation :
• Take natural log: ln(y) = ln(10x )
• Simplify the right hand side: ln(y) = x ln(10)
• Take the derivative:
y0
= ln(10) ( Note that ln(10) is a number hence counts as coeffiy
cient here.)
Page 14
• y 0 = y ln(10)
• y = 10x ln(10)
Alternative method you can use the formula:
If y = ax where a is a constant, y 0 = ln(a)ax .
16. An open box is to be made from a square sheet of tin measuring 12 in.×12 in. by cutting out
a square of side x inches from each corner of the sheet and folding up the four resulting flaps. To
maximize the volume of the box, x should be
(A) 1 in.
(B) 2 in.
(C) 3 in.
(D) 4 in.
(E) 5 in.
Solution: B
• The function of volume is V (x) = x(12 − 2x)2
• Derivative V 0 (x) = 2(12 − 2x)(−2)(x) + (12 − 2x)2 = (12 − 2x)(12 − 6x)
• V 0 (x) = 0 gives two answers: x = 6 ( Not acceptable, as then the volume of the box
will be zero) and x = 2.
• Now use the first derivative test to show that x = 2 gives a relative maximum in the
domain and since it is the only critical point in the domain:
f0
+
0
0
2
−
0
6
+
• observe that the function is highest within the domain of (0, 6) at x = 2.
17. A rectangular box is to have a square base and a volume of 20 ft3 . The material for the
base costs 30 cents/square foot, the material for the four sides costs 10 cents/square foot, and the
material for the top costs 20 cents/square foot. What are the dimensions of the box that can be
constructed at minimum cost ?
(A) 1 × 1 × 20 (B) 2 × 2 × 5 (C) 2.5 × 2.5 × 3.2
(D) 3 × 3 × 2.22 (E) 4 × 4 × 1.25
Page 15
Solution: B
y
x
x
•
• Constraint: x2 y = 20 so y =
20
x2
• Cost =0.3(x2 ) + 0.1(4xy) + 0.2(x2 )
C(x) = 0.5x2 +
• C 0 (x) = x −
8
x
8
x3 − 8
=
x2
x2
• The critical numbers: x = 0 and x = 2.
• x = 0 is not part of the domain of C.
• x = 2 is the only critical number in the domain and C 00 (x) = 1 −
(0, ∞). So x = 2 gives the absolute minimum.
• Find the other dimension: y =
16
> 0 on the domain
x3
20
=5
22
• So 2 × 2 × 5 is the answer.
18. A particle starts at the point (5,0) at t=0 and moves along the x-axis in such a way that at time
t > 0 its velocity v(t) is given by v(t) = t/(1 + t2 ). Determine the maximum velocity attained by
the particle.
(A)
2
5
(B)
2
3
(C)
3
10
(D)
1
2
(E)
4
17
Solution: D
v 0 (t) =
(1 + t2 ) − 2t(t))
1 − t2
=
(1 + t2 )2
(1 + t2 )2
Page 16
Two critical numbers: t = −1( Not acceptable time value.) and t = 1.
v 00 (t) =
2t3 − 6t
.
(1 + t2 )3
v 00 (1) = −0.5 < 0 therefore t = 1 gives the absolute maximum in the domain.
1
The value of the absolute min is f (1) = .
2
19. A particle starts at the point (5,0) at t=0 and moves along the x-axis in such a way that at time
t > 0 its velocity v(t) is given by v(t) = t/(1 + t2 ). Determine the position of the particle at t=6.
√
√
√
√
√
(A) ln 5 + 5 (B) ln 2 + 5 (C) ln 10 + 5 (D) ln 17 + 5 (E) ln 37 + 5
Solution: E
To find the position, integrate:
Z
t
dt
s(t) =
(1 + t2 )
Use u-substitute u = 1 + t2 and du = 2tdt.
Z
Z 1
( 2 )du
t
dt
=
= (1/2) ln |u| + C = (1/2) ln(t2 + 1) + C = ln(t2 + 1)(1/2) + C
2
(1 + t )
u
Initial value (5, 0). Plug in : 5 = ln(02 + 1)1/2 + C so C = 5
So s(t) = ln(t2 + 1) + 5
√
That is s(6) = ln(62 + 1)1/2 + 5 = ln( 37) + 5
20. A particle starts at the point (5,0) at t=0 and moves along the x-axis in such a way that at time
t > 0 its velocity v(t) is given by v(t) = t/(1 + t2 ). Find the limiting value of the velocity as t
increases without bound.
(A) 0
(B) 2
(C) 5
(D) 10 (E) 17
Solution:
A lim
t→∞ t2
t
=0
+1
21. A farmer wishes to enclose a rectangular piece of land along the straight part of a river. No
fencing is needed along the river bank. The farmer has 2000 meters of fencing. What is the largest
area that can be enclosed?
Page 17
(A) 50, 000 m2
(B) 1, 0002 m2
(D) 2(1, 000)2 m2
(C) 2(5002 ) m2
river
(E) 6(5002 ) m2
x
y
Solution: C
• Constraint: 2x + y = 2000. Solve for y = 2000 − 2x.
• A(x) = x(2000 − 2x) = 2000x − 2x2
• A0 (x) = 2000 − 4x
• A0 (x) = 0 gives x = 500
• A00 (x) = −4 < 0 (concavity downward) so the only critical number gives the maximum.
• The maximum is A(500) = 2000(500) − 2(500)2 = 2(5002 ) m2 .
22. The number of lamps a company sells is a function of the price charged, and it can be approximated by the function N (x) = 200 + 50x + 36.5x2 − x3 , where x is the price (in dollars) charged
for each lamp and 0 ≤ x ≤ 37. The price to maximize the number of lamps sold should be
(A) $13 (B) $20 (C) $25 (D) $37 (E) $30
Solution: C
N 0 (x) = 50 + 73x − 3x2
Set N 0 (x) = 0 and solve the quadratic equation to get:
x = −2/3 and x = 25. Negative value is not defined for price.
To see what the absolute max is check the critical number and the end points:
N (0) = 200
N (25) = 8637.5
N (37) = 1365.5
So maximum was achieved at x = 25 dollars.
23. Use differentials to estimate the change in
√
x2 + 5 when x increases from 2 to 2.123.
(A) 0.083 (B) 0.082 (C) 0.081 (D) 0.080
Page 18
(E) 0.084(D) None of the above
Solution: B
∆y ' dy = f 0 (x)∆x
x
f 0 (x) = (1/2)(x2 + 5)−1/2 (2x) = √
2
x +1
2
f 0 (2) =
3
2
dy = (2.123 − 2) = 0.082
3
24. The velocity
of a car (in feet/second) t seconds after starting from rest is given by the function
√
v(t) = 2 t (0 ≤ t ≤ 30). Find the car’s position at any time t.
2
(A) t3/2 + C
3
4
(B) t3/2
3
4
2
(C) t1/2 +
3
3
4
(D) t1/2
3
2
2
(E) t3/2 +
3
3
Solution: B
To find the position, take antiderivative.
Z
Z
Z √
4t3/2
1/2
3/2
+C
s(t) = v(t)dt = 2 t dt = 2t dt = 2t /(3/2) + C =
3
Use the initial value (0, 0), to find C:
0=
4(0)3/2
+ C gives C = 0
3
4t3/2
So s(t) =
3
Z
25. Evaluate
√
x − 2ex dx.
2
(A) x3/2 − 2ex
3
2
(B) x3/2 − 2ex + C
3
3
(D) x2/3 − 2ex + C
2
3
(C) x2/3 − 2ex
2
2
2ex+1
(E) x3/2 −
+C
3
x+1
Page 19
Solution: B
Z
Z
√
x3/2
2x3/2
x
− 2ex + C =
− 2ex + C
( x − e ) dx = (x1/2 − 2ex ) dx =
3/2
3
Z
2
xe−x dx.
26. Evaluate
2
(A) − e−x + C
1
(B) − e−x + C
2
2
(C) (1 − 2x2 )e−x + C
1
2
(D) − e−x + C
2
2
(E) − 2e−x + C
Solution:
D
Use integration by u-substitution: u = −x2 and du = −2x dx
Z
2
eu (−1/2)du = (−1/2)eu + C = (−1/2)e−x + C
27. If F 0 (x) =
√
x and F (1) = 1, then F (4) =
(A) 2/5 (B) 2 (C) 17/3 (D) 1/5 (E) 23/2
Solution: C
Z
Find antiderivative : F (x) =
√
x dx =
Z
x
1/2
2x3/2
dx =
+C
3
Plug in the initial value (1, 1) ( F (1) = 1):
1=
2(1)3/2
2
+ C = + C Gives C =
3
3
1
3
and
2x3/2
1
F (x) =
=
3
3
So F (4) =
2(43/2 ) 1
17
+ =
3
3
3
Page 20
8
Z
28. Calculate
1
8
1/3
4x + 2 dx.
x
(A) 49 (B) 50 (C) 51 (D) 52 (E) 54
Solution:
D
8 8
4x4/3 8x−1 8 4/3
dx =
4x + 8x
=
+
= 3x −
4/3
−1 1
x 1
1
4/3
4/3
= (3(8 ) − 8/8) − 3(1 ) − 8/1 = 47 − (−5) = 52
Z
8
−2
1/3
3
Z
|1 − x| dx.
29. Evaluate
0
(A) 3/2 (B) 5/2 (C) 7/2 (D) 9/2 (E) 11/2
Solution: B
1 − x if x ≤ 1
Notice that |1 − x| =
x − 1 if x > 1
So break the integral into appropriate integrals:
Z 3
Z 1
(x − 1) dx =
(1 − x) dx +
1
0
1 3
2
2
x − x /2 + x /2 − x = (1 − 12 /2) − (0 − 02 /2) + (32 /2 − 3) − (12 /2 − 1) = 5/2
0
Z
30. Calculate
1
4
√
(x + 1) x dx.
0
(A)
272
15
√
40 √
16
(B) 80 (C) 5 5 − 5 (D)
5−
3
15
Page 21
√
(E) 5 5
Solution: A
5/2
4
Z 4
x
x3/2 3/2
1/2
(x + x )dx =
+
=
=
5/2
3/2 0
0
4
2x5/2
2x3/2 16
64
+
5
3
= 5 + 3 − (0 + 0) = 272/15
0
Z 2
3
x(x − 1) + 1 dx =
31.
(A)
x2 (x2 − 1)4
x2 (x2 − 1)4
+x (B)
+C
8
4
(C)
(x2 − 1)4
+x+C
8
(D)
(x2 − 1)4
+x
8
(E)
(x2 − 1)4
+C
4
Solution:
C
Separate into two integrals.
Z Z Z
2
3
2
x(x − 1) + 1 dx =
x(x − 1) dx + 1 dx
Use u-substitute u = x2 − 1 and du = 2x dx to do the first part.
Z Z
(x2 − 1)4
u4
2
+C
x(x − 1) dx = u3 (1/2)du = (1/2) + C =
4
8
The entire integral is =
(x2 − 1)4
+x+C
8
1
32. Let R be the region under the graph of f (x) = on [1, 5]. Find an approximation of the area
x
using Riemann sums for 4 equal subintervals and left hand endpoints.
(A) ln(5) (B)
77
60
(C)
25
12
Solution: C
∆x =
5−1
=1
4
Page 22
(D)
496
315
(E)
58
45
Now A ' f (1)∆x + f (2)∆x + f (3)∆x + f (4)∆x = (1/1)(1) + (1/2)(1) + (1/3)(1) +
(1/4)(1) = 25/12
33. Find horizontal asymptote of the graph of the function f (x) = e−x − 1
(A) y = 1 (B) y = −1 (C) x = 1 (D) x = −1 (E)
Solution: B
limx→∞ f (x) = limx→∞ (e−x ) − 1 = −1
34. If you want to calculate the Riemann sum of x5 over the interval [2,5] using six subintervals of
equal length and choosing the right endpoints, which of the following should you calculate?
(A) 25 + (2.5)5 + 35 + (3.5)5 + 45 + (4.5)5
(B) (2.5)5 + 35 + (3.5)5 + 45 + (4.5)5 + 55
1
(C) (25 + (2.5)5 + 35 + (3.5)5 + 45 + (4.5)5 )
2
1
(D) (25 + (2.5)5 + 35 + (3.5)5 + 45 + (4.5)5 + 55 )
2
1
(E) ((2.5)5 + 35 + (3.5)5 + 45 + (4.5)5 + 55 )
2
Solution: E
5−2
1
=
6
2
A ' ∆x f (2.5) + f (3) + f (3.5) + f (4) + f (4.5) + f (5) =
∆x =
5
5
2
5
5
5
5
(1/2) (2.5) + (3) + (3.5) + (4) + (4) + (4.5) + (5)
35. The area under the curve y = πx3 and above the x-axis on the interval [1,2] is
(A) 2π
(B)
7π
4
(C)
15π
4
(D) 4π
(E) 7π
Page 23
Solution:
C
Z 2
1
2
x4 15π
πx dx = π = π(24 /4 − 14 /4) =
4 1
4
3
36. The area under the curve y = 16 − x2 and above the x-axis is
(A)
64
3
(B) 64
(C)
128
3
(D)
256
3
(E)
16
3
Solution: D
Graph the function or find the x-intercepts and where the function is positive.
x-intercepts: 16 − x2 gives x = ±4. and f is positive in between.
4
Z 4
43
(−4)3
x3 2
A=
(16 − x ) dx = 16x − = (16(4) − ) − 16(−4) −
=
3 −4
3
3
−4
128/3 − (−128/3) =
256
3
16. f
14.
12.
10.
8.
6.
4.
2.
−6.−4.−2.
2. 4. 6.
37. Find the area of the region under the graph of y = x(x2 − 1)3 + 1 and above the x-axis on the
interval [0, 2] is
(A) 2 (B) 4 (C) 8 (D) 10 (E) 12
Page 24
Solution: E
Check
Find the indefinite integral. ( We did in Problem 31 .) And plug in the integral limits.
2
2
2
Z 2
4
(x − 1)4
(2
−
1)
2
3
x(x − 1) + 1 dx =
A=
+ x =
+ 2 − 1/8 + 0 =
8
8
0
0
80
+ 2 = 12
8
Page 25
38.Which of the following statements is true for the function represented in the graph below in
Figure 2?
3.5
3.
f
2.5
2.
1.5
1.
0.5
0.5
1.
1.5
2.
Figure 2
00
(A) f 0 > 0 on (0, 0.5), f
00
(B) f 0 > 0 on (0, 0.5), f
00
(C) f 0 < 0 on (0, 0.5), f
00
(D) f 0 < 0 on (0, 0.5), f
(E) None of the above
00
> 0 on (0, 1) and f
00
< 0 on (0, 1) and f
00
> 0 on (0, 1) and f
00
< 0 on (0, 1) and f
< 0 on (1, 2).
> 0 on (1, 2).
< 0 on (1, 2).
> 0 on (1, 2).
Solution:
B
The graph is increasing on interval (0, 0.5) and concave downward on interval (0, 1) and upward on interval (1, 2).
39. Find the vertical asymptotes of function f (x) =
(A) x = −1
2−x
.
(1 + x)2
(B) x = 2 (C) y = −1 (D) y = 2 (E) x = 0
Solution: A
The vertical asymptotes for rational functions come from zeros of the denominator. 1 + x = 0
gives x = −1.
Page 26
40. The horizontal asymptote of graph of the function g(x) =
(A) y = −e4
(B) y = −1/e2
(C) y = 0
(D) y = e
e 2 x3 + 1
is the line
ex3 − e4
(E) y = e + 1
Solution: D
Horizontal asymptote of a rational function is obtained by finding limit value as x → ∞:
e2 x3 + 1
e2
=e
=
x→∞ ex3 − e4
e
lim g(x) = lim g(x) = lim
x→∞
a→−∞
41. How many vertical asymptotes does the graph of y =
(A) 0
(B) 1
(C) 2
(D) 3
x17
have?
(x − 1)(x − 2)(x − 3)
(E) 4
Solution: D
The denominator has three factors and therefore three zeros.
42. If lim f (x) = 17 then what sort of asymptote(s) must the graph of y = f (x) have?
x→−∞
(A) a horizontal asymptote of y = 17
(B) a vertical asymptote at x = 17
(C) no vertical asymptote
(D) no horizontal asymptote
Solution:
A
43. Find
dy
in terms of x and y when x and y are related by the equation x + y + ey = 3.
dx
(A) −
2
ey
(B) −
1
ey
(C) − (1 + ey ) (D) −
Page 27
1
1 + ey
(E) 2 − ey
Solution: D
Take the implicit derivative of the equation:
dy
dy
+ ey
=0
dx
dx
dy
Solve for
dx
−1
dy
=
dx
1 + ey
1+
44. Find an equation of the tangent line to the graph of y = e2x − ln(x2 + 1) at the point (0, 1).
(A) y = 2x + 1
(B) y = 2x − 1 (C) y − 1 = 2e2x −
(D) y − 1 = 2e2x −
x2
1
x
+1
2x
x (E) y = x + 1
+1
x2
Solution: A
Find the derivative: y 0 = e2x (2) −
Find the slope: m = 2e0 −
(2x)
x2 + 1
2(0)
=2
02 + 1
Point-slope: y − 1 = e(x − 0) which gives y = 2x + 1.
45. Let f (x) = 10x9 − 9x10 . Then f has inflection point(s) at x =:
(A) 0, 1 (B) 0,
8
9
(C) 1,
9
8
8
9
(D) 0, , 1 (E) 0, 1,
9
8
Solution: B
f 0 (x) = 90x8 − 90x9
f 00 (x) = 720x7 − 810x8
Set f 00 equal to zero and make a number line.
72x7 − 810x8 = 90x7 (8 − 9x) = 0 gives x = 0 and x = 8/9
Page 28
f 00
+
0
0
−
0
8/9
+
So there is a change in concavity at both values x = 0 and x = 8/9.
46. Which of the following statements is true for the function represented in the graph below in
Figure 3?
2.
1.
−2.
−1.
f
1.
2.
−1.
−2.
Figure 3
(A)f has one inflection point, one relative minimum, and no relative maximum in the interval
(−2, 2).
(B) f has one inflection point, one relative maximum, and one relative minimum in the interval
(−2, 2).
(C) f has two inflection points and one relative minimum in the interval (−2, 2).
(D) f has two inflection points and one relative maximum in the interval (−2, 2).
Page 29
(E) None of the above
Solution:
B The inflection point is (0, 0).
Relative maximum at a value less than 0 and a relative minimum for a value more than 0.
47. The velocity (in feet/second) of a hawk flying straight down towards its prey is
v(t) = −3t, where t is the number of seconds since it started flying straight down. If the hawk
starts flying from 200 feet, i.e. h(0) = 200, what is its height h at t seconds?
(A) 1.5t + C
(B) 200 + 1.5t (C) − 1.5t + 200
(D) 200 − 1.5t
(E) − 1.5t2 + 200
Solution: E
Find the antiderivative:
Z
−3t2
s(t) = (−3t) dt =
+C
2
Plug in the initial value: (0,200):
−3(02 )
+ C gives C = 200
2
−3t2
So s(t) =
+ 200
2
200 =
48. A company estimates that the marginal cost of manufacturing its product is estimated by the
following in dollars/month when the level of production is x items/month. C 0 (x) = 0.002x + 130
The fixed costs incurred are $6500/month. Find the total monthly cost C(x) incurred in manufacturing x items/month.
(A) C(x) = 0.001x2 + 130x + 6500
(B) C(x) = 0.002x2 + 130x + 6500
(C) C(x) = 0.001x2 + 130x
(D) C(x) = 0.002x2 + 130x
(E) None of the above.
Solution:
A
Page 30
Find the antiderivative:
Z
(0.002x + 130) dx = 0.001x2 + 130x + C
Plug in the initial value: (0, 6500) to find C:
6500 = 0.001(0)2 + 130(0) + C gives C = 6500.
49. If F 0 (x) = ex + x and F (0) = 3, then F (2) equals
(A) e2 + 2
(B) e2 + 4
(C) e2 + 5
(D) e2 + 6
(E) e2 + 8
Solution: B
Z
Find the antiderivative: F (x) =
Use the initial value: 3 = e0 +
02
2
(ex + x) dx = ex +
x2
+C
2
+ C. This gives 3 = 1 + C so C = 2
Plug c = 2 back in the F :
x2
2
F (x) = ex +
+2
Find F (2) = e2 + 2 + 2 = e2 + 4
Z
50. Calculate
1
4
1
1 + 2 dx.
x
(A) 0 (B)
1
4
(C)
15
4
(D)
17
4
(E)
21
4
Solution: C
4
Z 4
1
1
15
15
−2
−1
=
1+x
dx = x + x /(−1) = (4 − ) − 1 −
=
−0=
4
1
4
4
1
1
51. At 8:00 am a wading pool is filled with water. The temperature (in Farenheit) of the water in
the pool t hours after it is filled is given by g(t) = t3 − 3t2 + 58 for 0 ≤ t ≤ 5. What is the average
temperature of the water between 9:00 am and noon?
(A) 43.5265 (B) 58 (C) 58.25 (D) 174.75 (E) 232
Page 31
Solution: C
9am matches t = 1 and noon is t = 4.
4
Z 4
1
1 1 4
1
3
2
3
(t − 3t + 58)dt =
t − t + 58t = (232 − 57.25) = 58.25
gave =
4−1 1
3 4
3
1
x+1
52. The area of the region bounded by the graph of y =
and the x-axis from x = 1 to x = e
x
is
(A) 1 (B) e (C) e + 1 (D) e − 1 (E) e + 2
Solution: B
e
Z e
Z e
x+1
1
A=
dx =
1+
dx = (x + ln |x|) = (e + ln(e)) − 1 + ln(1) =
x
x
1
1
1
e+1−1=e
53. Snow is falling so that t hours after noon the rate at which it is falling is
much snow (in cm) fell between 1 p.m. and 9 p.m.?
(A) 8 (B)
26
3
(C)
52
3
√
t cm per hour. How
(D) 26 (E) 39
Solution: C
Z
Accumulation=
1
9
9
√
t3/2 t dt =
= (2/3)93/2 − (2/3)13/2 = (2/3)(27) − 2/3 = 52/3
3/2 1
54. Bacteria in refrigerated milk grows exponential by doubling every 24 hours. The bacteria count
in milk starts at 500 and milks spoils when it reaches a count of 4,000,000. How many days will it
take for milk to spoil?
(A) ln(4000) (B)
ln(8000)
ln(2)
(C)
ln(4, 000, 000)
ln(500)
Solution: B
Page 32
(D) 12 (E) 9.5
• Model: Q(t) = Q0 ekt
• Find k: Use the information that the numbers doubles everyday.
2Q0 = Q0 ek(1) and solve k = ln(2).
• Replace k in the model and also replace Q0 = 500:
Q(t) = 500eln(2)t
• Use the information in the question.
4, 000, 000 = 500eln(2)t and solve: 800 = eln(2)t and ln(800) = ln(2)t
ln(8000)
t=
days
ln(2)
55. The half life of polonium-210 is 140 days. How long will it take 300 micrograms of polonium210 to decay to 60 micrograms?
(A) 140
ln 2
ln 5
(B) 80
ln 2
ln 5
(C)
− ln 2
140
(D) 80
ln 5
ln 2
(E) 140
ln 5
ln 2
Solution: E
• Model: Q(t) = Q0 e−kt
• Half life life is 140 days:
(0.5)Q0 = Q0 e−k(140)
ln(1) − ln(2)
ln(2)
ln(1/2)
=
=
k=
−140
−140
140
ln(2)
• replace k back; Q(t) = Q0 e− 140 t
• Read the information of the question; Q0 = 300 and Q(t) = 60.
• 60 = 300e−0.005t so t =
ln(1/5)
− ln(2)
140
= 140
ln(1) − ln(5)
ln(5)
= 140
− ln(2)
ln(2)
56. A riverbank is eroding exponentially so that every year it loses 10% of its soil. How much of
its soil will it have in 20 years?
(A) about 20% (B) about 12% (C) about 10% (D) about 6.5% (E) about 5.2%
Page 33
Solution: B
• Model: Q(t) = Q0 e−kt
• Find k: Use the information that it loses 10% of its soil. That is 90% of soil is remaining
after a year is passed.
0.9Q0 = Q0 e−k(1) and solve k = − ln(0.9).
• Replace k in the Q(t) = Q0 eln(0.9)t
• Plug in 20 years Q(t) = Q0 eln(0.9)(20) ' 0.12Q0
57. A riverbank is eroding exponentially so that every year it loses 10% of its soil. How long will
it take for it to lose half its soil?
(A) about 20 years (B) about 12 years (C) about 10 years (D) about 6.5 years (E) about 5.2
years
Solution: D
• Model: Q(t) = Q0 e−kt
• Find k: Use the information that it loses 10% of its soil. That is 90% of soil is remaining
after a year is passed.
0.9Q0 = Q0 e−k(1) and solve k = − ln(0.9).
• Replace k in the Q(t) = Q0 eln(0.9)t
• Use the information in the question:
0.5Q0 = Q0 E ln(0.9)t solve to get t =
ln(0.5)
' 2.6 years
ln(0.9)
58. A population of bacteria is growing exponentially so that it triples every 20 hours. How long
does it take to double? (Pick the closest approximation.)
(A) 5.7 hours (B) 6.3 hours (C) 8.9 hours (D) 10 hours (E) 12.6 hours
Solution: E
• Model: Q(t) = Q0 ekt .
Page 34
• Use the information, it triples ever 20 hours, to find k:
ln(3)
' 0.055
3Q0 = Q0 ek(20) and solve for k. k =
20
• Replace k back in formula:
Q(t) = Q0 e0.005t
• Use the information in the question:
2Q0 = Q0 e0.055t gives t = ln(2)/0.055 ' 12.6 hours
Page 35
Answers
True or False
1. T
2. F
3. F
4. F
5. T
6. F
7. T
16. F
8. T
17. F
9. F
11. T
12. T
13. F
14. T
15. T
18. T
21. F
22. T
23. F
24. F
25. T
26. F
27. F
28. T
31. F
32. F
33. T
34. T
35. T
36. F
37. T
38 T
10. T
19. T
29. T
20. F
30. T
Multiple Choice
1. D
2. B
3. B
4. D
5. C
6. C
7. A
8. B
9. C
10. A
11. E
12. A
13. E
14. B
15. A
16. B
17. B
18. D
19. E 20. A
21. C
22. C
23. B
24. B
25. B
26. D
27. C
28. D
29. B
30. A
31. C
32. C
33. B
34. E
35. C
36. D
37. E
38. B
39. A
40. D
41. D
42. A
43. D
44. A
45. B
46. B
47. E
48. A
49. B
50. C
51. C
52. B
53. C
54. B
55. E
56. B
57. D
58. E
Page 36
Some Road maps to problem solving: (Caution: very informal)
Domain 1
• Exclude zeros of the denominator.
p
• Exclude all that makes under the square root negative. ( (g(x)) solve g(x) ≥ 0.)
• Exclude all that makes inside the logarithm negative or zero. ( ln(g(x)) solve g(x) > 0)
Asymptotes
2, 33, 39, 40, 41, 42
• Finding vertical asymptotes:
P (x)
If f (x) =
is a rational function where P (x) and Q(x) are polynomials.
Q(x)
Then, the line x = a is a vertical asymptote if Q(a) = 0 and P (a) 6= 0. ( Zeros of the denominator.)
If f (x) = ln(g(x)) then the solutions to g(x) = 0 are the vertical asymptotes.
• 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→−∞
Note that if you are finding a horizontal asymptote of a rational function, then the asymptotes is the
same for both ends.
If you are finding horizontal asymptote of a exponential function, the asymptotes may only exist on
one side.
• 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 ±∞.
Extremums
True/false, 8, 9, 10, 17,18
• For relative extremum, use number line with f 0 and arrows. ( Top of the hill and bottom of the
valley.)
Or use second derivative test if f 0 (c) = 0.
• Finding absolute extrema :
Step 1: Take the derivative and find the critical points.
– If over a closed interval:
Step 2: Plug in the critical numbers in the interval and the end points in the original f . Largest
is the absolute max and smallest is the absolute min. The x-value associated with those are the
x-values for max and min.
– If not over a closed interval
Step 2: First check to see if only one relative max and min exists by using first derivative f 0
and arrows on the number line. If one extremum of either kind exists, then that is the absolute
extremum.
Step 3: If step 2 is not true, graph and find the max and min.
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• Optimization word problems. 16, 17, 21,22
Concavity and inflection points
11, 12, 45, 46
• Use second derivative and corresponding number line.
• For inflection point, make sure the concavity changes.
• concavity upward (smiley face) when f 00 > 0 (positive) and downward (Frowny face) when f 00 < 0
(negative).
Increasing/ decreasing intervals. 13
• Use first derivative f 0 and the number line.
• f 0 > 0 means the graph of f increasing. f 0 < 0 means the graph is decreasing.
Tangent lines (good practice for derivatives as well).
Exponential and logarithmic functions
3, 4, 5, 6, 44
• Know your exponential and logarithmic rules.
• Know exponential growth and decay word problems. (Q(t) = Q0 ekt / Q(t) = Q0 e−kt ) 54, 55, 56,
57
• Know your exponential differentiation:
d
f (x) )
dx (e
= ef (x) f 0 (x)
• Know differentiation of logarithms (simplify first) :
d
dx [ln f (x)]
=
f 0 (x)
f (x)
(f (x) > 0)
3, 4, 5, 6, 44
• Know the logarithmic differentiation: True/false, 14, 15
1. Take the natural log of both sides
2. Simplify RHS.
3. Differentiate both side of the equation.
4. RHS= y 0 /y
5. Multiply the right side by y.
• Domain, asymptotes, intercepts and all.
• Implicit differentiation with such functions.43
Integration
• Know antiderivative. True/false, 25, 26, 31
• Know definite integrals and areas. True/False, 28, 29, 30, 35, 36 37, 50, 52, 53
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• Know Riemann sums.32, 34
• Know average value formula
1
b−a
Rb
a
f (x)dx. 51
• Know how integrate to find the area under the curve. (Refer to definite integral.)
• Initial value problems ( both regular and word problems.) 19, 24, 27, 47, 49
• U-substitution. 19, 26, 31
The relationship between position, velocity and acceleration:18, 19, 20, 24, 47
d
d
dt
dt
s(t) =
⇒
v(t) =
⇒
a(t)
R
... dt
R
... dt
a(t) ====⇒ v(t) ====⇒ s(t)
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