Math 161 - Review for Final - KEY
1. Evaluate each of the following limits, or state that they do not exist.
3x 2 − 3
x→1 2x 2 − x − 1
3(x − 1)(x + 1)
lim
x→1 (2x + 1)(x − 1)
3(x + 1) 3(2)
lim
=
=2
x→1 2x + 1
3
a) lim
5x 4 + 5
c) lim
x →−∞ 5 − 5x 4
5x 4
3
x →∞ x − 2
b) lim
lim ( 5x ) = ∞
x→∞
tan(3x)
x
3sec 2 (3x)
lim
x→0
1
= 3(1) = 3
d) lim
x →0
5x 4
= −1
x→−∞ −5x 4
lim
e)
lim (sin2x − cos2x)
x →π / 2
sin(2x) − 2x
x →0
5x 2
f) lim
⎛ ⎛ π ⎞⎞
⎛ ⎛ π ⎞⎞
sin ⎜ 2 ⎜ ⎟ ⎟ − cos ⎜ 2 ⎜ ⎟ ⎟
⎝ ⎝ 2⎠⎠
⎝ ⎝ 2⎠⎠
2 cos(2x) − 2
x→0
10x
−4 sin(2x) 0
lim
=
=0
x→0
10
10
lim
sin (π ) − cos (π )
0 − (−1) = 1
2. Evaluate
dy
for each of the following functions.
dx
a) y = (5x3 – 2x)4
y′ = 4(5x 3 − 2x)3 (15x 2 − 2)
b) y = (x − 5) 3x + 2
y′ = (x − 5) ⋅ 12 (3x + 2)−1/2 (3) + (3x + 2)1/2 (1)
y′ = 12 (3x + 2)−1/2 [ 3(x − 5) + 2(3x + 2)]
y′ = 12 (3x + 2)−1/2 (9x − 11)
2
sin 2x
d) y = 5xe 4x
1− cos2x
(1 − cos(2x))(2 cos(2x)) − (sin(2x))(2 sin(2x))
y′ =
(1 − cos 2x )2
c) y =
y′ =
y′ =
y′ =
y′ =
y′ =
2 cos(2x) − 2 cos 2 (2x) − 2 sin 2 (2x)
(1 − cos 2x )2
2 cos(2x) − 2(cos 2 (2x) + sin 2 (2x))
(1 − cos 2x )2
2
2 cos(2x) − 2
2
y′ = 5e4 x (8x 2 + 1)
(1 − cos 2x )2
2(cos(2x) − 1)
(1 − cos 2x )2
−2
2
=
1 − cos(2x) cos(2x) − 1
e) y = ln(x4 + 4x)
4x 3 + 4
y′ = 4
x + 4x
f)
y = 23x − 5 + log 2 (x)
y′ = 2 3x − 5 (ln 2)(3) +
3. Evaluate each of the following integrals.
a)
∫
2
y′ = 5xe4 x (8x) + e4 x (5)
4x
dx
9−x
u = 9 − x2
du = −2xdx
du
dx =
−2x
4x du
∫ u1/2 ⋅ −2x
2
−2 ∫ u −1/2 du
u1/2
+C
1/ 2
−4u1/2 + C
2
−4 9 − x 2 + C
b)
∫ xe
4 x2
dx
u = 4x 2
du = 8xdx
du
dx =
8x
u du
∫ xe 8x
1 u
e du
8∫
1 u
e +C
8
2
e4 x
+C
8
1
x ln(2)
cos2x
c) ∫
dx
1 − sin2x
u = 1 − sin(2x)
du = −2 cos(2x)dx
du
dx =
−2 cos(2x)
cos(2x)
du
∫ u ⋅ −2 cos(2x)
−1 1
du
2 ∫u
−1
ln u + C
2
−1
ln 1 − sin(2x) + C
2
4
d)
∫ sin (4x) cos(4x)dx
3
π / 12
u = sin(4x)
du = 4 cos(4x)dx
du
dx =
4 cos(4x)
x = π /6
∫
u 3 cos(4x)
x = π /12
du
4 cos(4x)
x = π /6
1
u 3du
∫
4 x = π /12
1 u4
⋅
4 4
x = π /6
x = π /12
(sin(4x))4
16
π /6
π /12
sin(2π / 3) sin(π / 3)
−
=0
16
16
5
1
x
dx
4
∫ 5x
−3/2
dx
1
5x −1/2
−1 / 2
4
1
4
−10
x 1
−10 ⎛ −10 ⎞
−⎜
⎟ =5
4 ⎝ 1⎠
π/6
e)
∫x
f)
∫
ln(2x + 1)
dx
2x + 1
u = ln(2x + 1)
2dx
du =
2x + 1
(2x + 1)du
dx =
2
u (2x + 1)du
∫ 2x + 1 2
1
udu
2∫
1 u2
+C
2 2
( ln(2x + 1))2 + C
4
4. Solve each of the following problems.
a) Find the equation of the tangent line to the curve x 2 + xy + y 4 = 2 at the point (1, 1).
2x + x
dy
dy
+ y(1) + 4y 3
=0
dx
dx
dy
(x + 4y 3 ) = −2x − y
dx
dy −2x − y
=
dx x + 4y 3
−2(1) − 1 −3
m=
=
1 + 4(1)3
5
−3
y − 1 = ( x − 1)
5
−3
8
y=
x+
5
5
b) Consider the function f (x) = x 3 − 3x . Find the local maximum and minimum points for the
function and the x values where they occur. Find all inflection points for the graph and
determine where the graph is increasing, decreasing, concave up, or concave down.
f ′(x) = 3x 2 − 3 = 3(x 2 − 1)
f ′(x) = 0 ⇒ x = −1,1
Since f ′(x) < 0 on (–1, 1), f (x) is decreasing on (–1, 1).
Since f ′(x) > 0 on (−∞, −1) ∪ (1, ∞) , f (x) is increasing on (−∞, −1) ∪ (1, ∞) .
The function has a relative maximum of y = 2 at x = – 1 and
a relative minimum of y = –2 at x = 1.
f ′′(x) = 6x
f ′′(x) = 0 ⇒ x = 0
Since f ′′(x) < 0 on (−∞, 0) , f (x) is concave down on (−∞, 0) .
Since f ′′(x) > 0 on (0, ∞) , f (x) is concave up on (0, ∞) .
The function has an inflection point at (0, 0).
c) A spherical tootsie roll pop is giving up volume at a steady rate of 800 mm3 /min. How fast will
the radius be decreasing when the Tootsie Roll Pop is 20 mm across?
dV
= −800 mm3 /min
dt
V=
4 3
πr
3
dV
dr
= 4π r 2
dt
dt
dr
dt
dr −800 −2
=
=
≈ –0.64 mm/min
dt 400π π
−800 = 4π (10)2
d) A right triangle whose hypotenuse is 12 meters long is revolved about one of its legs to
generate a right circular cone. Find the radius, height, and volume of the cone of greatest
volume that can be made this way.
1
V = π r 2h
3
2
r + h 2 = ( 12 )2
r 2 + h 2 = 12
r 2 = 12 − h 2
1
V = π (12 − h 2 )h
3
π
V = (12h − h 3 )
3
dV π
= (12 − 3h 2 )
dh 3
dV
= 0 ⇒ h = 2, −2
dh
dV
changes sign from positive to negative at h = 2,
dh
there is an absolute maximum at h = 2 m.
r 2 = 12 − (2)2
Since
r2 = 8
r = 8 = 2 2 ≈ 2.82
1
V = π ( 8 )2 (2)
3
16π
V=
3
V ≈ 16.76 m3 .
e) Find the area of the region between the curve y = x 2 − 2x and the x-axis on [0, 4].
Note that the graph falls below the x-axis on [0, 2] and lies above the x-axis on [2, 4].
2
∫ (x
0
4
2
− 2x)dx + ∫ (x 2 − 2x)dx
2
2
4
⎛ x3
⎞
x3
− x2 + ⎜ − x2 ⎟
3
⎝ 3
⎠2
0
⎛ 03
⎞ 43
⎛ 23
⎞
23
− 22 − ⎜ − 02 ⎟ +
− 4 2 − ⎜ − 22 ⎟
3
3
⎝ 3
⎠
⎝ 3
⎠
8
64
8
−4 +
− 16 − + 4
3
3
3
=8
The area is 8 square units.
=
f) The acceleration of a particle is given by the formula a(t) = t + sin 2t . Find equations for
velocity (v) and position (s) if v(0) = 12 , and s(0) = 2 .
t 2 cos(2t)
v(t) = −
+C
2
2
1 0 cos(0)
= −
+C
2 2
2
C =1
t 3 sin(2t)
−
+t +C
6
4
0 3 sin(0)
2=
−
+0+C
6
4
C=2
t 2 cos(2t)
v(t) = −
+1
2
2
t 3 sin(2t)
s(t) = −
+t +2
6
4
s(t) =
5. a) A function f is continuous at a x = c in the interior of its domain if each of the following are
true.
1. lim f (x) exists.
x→c
2. f (c) is defined.
3. lim f (x) = f (c)
x→c
b) Sketch a graph of a function, which fails exactly 2 parts of the definition above. State which
parts of the definition fail.
The graph in #6 fails two parts of the definition (part 1 and part 3) at x = 1.
c) Sketch a graph of a function, which fails exactly 1 part of the definition above. State which
part of the definition fails.
The graph in #6 fails one part of the definition (part 3) at x = –2.
6. The graph of a function y = f(x) is shown. Answer each of the following.
4
a) lim f (x) = 3
b) lim− f (x) = 3
c) lim+ f (x) = 2
d) lim f (x) = ∞
x →−2
3
2
x →1
1
-3 -2
-1
x →1
1 2 3
x→−∞
e) lim f (x) = 1
x →∞
f) At what x-values is the function discontinuous? at x = –2, x = 1
g) Which, if any, of the discontinuities from part f are removable?
The discontinuity at x = –2 is removable since the limit exists
there.
7. How can a function be continuous at a point, but not differentiable there? Give two different
ways that this can happen.
There could be a sharp point or a vertical tangent line.
8. Calculus includes the study of limits, derivatives and integrals. Explain the connections between
limits, derivatives, indefinite integrals and definite integrals.
Derivatives are defined by a limit, the limit of a difference quotient. Indefinite integrals are
antiderivatives, so we just use the derivative process “backward” to determine indefinite
integrals. Definite integrals are the limit of a Riemann sum. In addition, the funcamental
theorem of calculus tells us that we can compute definite integrals by determining the
antiderivative, and then evaluating at the upper and lower limits and subtracting.
b
9. In the definition of the definite integral,
∫
a
n
f (x)dx = lim ∑ f (xk )Δx , explain what each of the
n→∞
k =1
following mean: Δx , f (xk ) , f (xk )Δx . Why do we take the limit as n → ∞ ?
Δx is the width of each subinterval or rectangle.
f (xk ) is the height of the kth rectangle.
f (xk )Δx is the area of the kth rectangle.
We take the limit so that we can get the exact value of the integral instead of an approximation.