Download Mr. Benson BC Calculus Midterm Practice Problems Answers Solve

Mr. Benson
BC Calculus
Midterm Practice Problems Answers
Solve the problem.
1) Assume that a watermelon dropped from a tall building falls y = 16t2 ft in t sec. Find the watermelon's average
speed during the first 6 sec of fall.
Avg speed =
y(6) - y(0) 576 - 0
=
= 96 ft/sec
6-0
6-0
Determine the limit by substitution.
2) lim (x3 + 5x2 - 7x + 1)
x 2
8 + 20 - 14 + 1 = 15
x3 - 6x + 8
x-2
3) lim
x 0
0-0+8
= -4
0-2
Determine the limit algebraically, if it exists.
x+6
4) lim
(x
- 6)2
x 6
12
0
vertical asymptote
limit = DNE
5)
lim
x -7
6) lim
x 0
x2 + 11x + 28 (x + 7)(x + 4)
=
= x + 4 = -7 + 4 = -3
x+7
(x + 7)
sin 6x 6 2
= =
9x
9 3
Find the limit.
7) Let lim f(x) = 8 and lim g(x) = -6. Find lim [f(x) + g(x)]2 .
x 8
x 8
x 8
(8 + (-6))2 = (2)2 = 4
1
Determine the limit graphically, if it exists.
lim f(x).
8) Find lim f(x) and
x 2x 2+
x
lim f(x) = 5
2-
x
lim f(x) = -3
2+
Evaluate or determine that the limit does not exist for each of the limits (a) lim f(x), (b) lim f(x), and (c) lim f(x)
x 1
x 1x 1+
for the given function f.
9)
-2x - 2,
f(x) = 1,
-4x + 8,
for x < 1,
for x = 1,
for x > 1
a) -4
b) 4
c) DNE
Find the vertical asymptotes of the graph of f(x).
1
10) f(x) =
2
x - 16
x = 4, -4
Find the asymptote(s) of the given function.
x+8
horizontal asymptotes(s)
11) g(x) =
x2 - 2
x
1
lim
= =0
2
x
x
x
y=0
2
12) h(x) =
32x2
horizontal asymptotes(s)
8x2 - 5
3x2
lim
=4
8x2
x
y=4
Find the intervals on which the function is continuous.
2
13) y =
(x + 4)2 + 8
(
, ) or all reals
14) y =
5x + 9
9
[- , ) or x
5
-
9
5
Find the points of discontinuity. Identify each type of discontinuity.
x-6
x-6
=
15) y =
2
(x
6)(x - 8)
x - 14x + 48
infinite jump at x = 8 because y(8) =
hole at x = 6 because y(6) =
2
0
0
0
Find the equation for the tangent to the curve at the given point.
16) f(x) = -1- x2 at x = 5
f'(x) = -2x
m = f'(5) = -10
y1 = f(5) = -1 - (5)2 = -26
y - y1 = m(x - x1 )
17) f(x) =
y + 26 = -10(x - 5)
y = -10x + 24
48
at x = 1
x
f'(x) = -
48
x2
m = f'(1) = -
48
= -48
(1)2
48
y1 = f(1) =
= 48
1
y - y1 = m(x - x1 )
y - 48 = -48(x - 1)
y = -48x + 96
3
Find the equation of the normal line to the indicated curve at the given point.
18) y = 3x2 at x = 2
f'(x) = 6x
f'(2) = 12
m=-
1
12
y1 = f(2) = 3(2)2 = 12
y - y1 = m(x - x1 )
y - 12 = -
1
(x - 2)
12
y=-
1
73
x+
12
6
Solve the problem.
d
(7x5 - 1x 3 - 3x + 2).
19) Find
dx
35x4 - 3x2 - 3
The graph of a function is given. Choose the answer that represents the graph of its derivative.
20) (C)
Solve the problem.
21) Find an equation of the tangent line to the graph of y = 6 x - x + 9 at the point
(36, 9).
f'(x) =
3
-1
x
m = f'(36) =
3
3
1
-1= -1=6
2
36
y - y1 = m(x - x1 )
1
y - 9 = - (x - 36)
2
1
y = - x + 27
2
22) Find the equation of the normal line to the curve y = 5x - 4x2 at the point (-1, -9).
f'(x) = 5 - 8x
f'(-1) = 13
m=-
y - y1 = m(x - x1 )
1
13
y+9=-
1
(x + 1)
13
y=-
1
105
x13
13
4
Find dy/dx.
23) y =
1 6 1 3
x - x
2
3
dy/dx = 3x5 - x2
Find dy/dx using the product rule.
24) y = (2x - 8)(6 - 4x3 )
dy/dx = (2x - 8)(-12x2 ) + (2)(6 - 4x3 ) = -24x3 + 96x2 + 12 - 8x3 = -32x3 + 96x2 + 12
25) y = 8x cos (7x - 2)
dy/dx = 8x (-7sin(7x - 2)) + 8 cos (7x - 2) = -56 xsin (7x - 2) + 8 cos (7x - 2)
Find dy/dx using the quotient rule.
4x + 4
26) y =
9x - 1
dy/dx =
27) y =
(9x - 1)(4) - (4x + 4)(9) 36x - 4 - 36x - 36
-40
=
=
(9x - 1)2
(9x - 1)2
(9x - 1)2
x2
3 - 9x
dy/dx =
(3 - 9x)(2x) - (x2)(-9) 6x - 18x2 + 9x2 -9x2 + 6x
=
=
(3 - 9x)2
(3 - 9x) 2
(3 - 9x)2
Find the fourth derivative of the function.
28) y = 5x5 + 5x4 - 3x
y' = 25x4 + 20x3 - 3
y'' = 100x3 + 60x2
y''' = 300x2 + 120x
y(4) = 600x + 120
Find dy/dx.
29) y = 5 tan6 x = 5 (tan x)6
dy/dx = 30 tan 5 x sec2x
30) y =
sin x
10x
dy/dx =
10x cos x - 10 sin x x cos x - sinx
=
100x2
10x2
5
31) y =
2 + sin 2x
dy/dx =
1
(2 cos 2x) =
2 2 + sin 2x
cos 2x
2 + sin 2x
1
= (7 - 8x)-1/2
7 - 8x
32) y =
dy/dx = -
1
(-8) =
2(7 - 8x)3/2
4
(7 - 8x) 3/2
Find y .
33) y = 5 sin(4x + 7)
y' = 20 cos (4x + 7)
y'' = -80 sin (4x + 7)
Find an equation for the line tangent to the curve at the point defined by the given value of t.
34) x = sin t, y = 7 sin t, t =
3
3
x( ) = sin ( ) =
3
3
2
y( ) = 7 sin ( ) =
3
3
y' =
7 3
2
dy/dt 7 cos t
=
=7
dx/dt
cos t
y - y1 = m(x - x1 )
y-
7 3
3
= 7(x )
2
2
y = 7x
Find dy/dx by implicit differentiation. If applicable, express the result in terms of x and y.
35) 8y -3xy - 9 = 0
8y' - 3xy' - 3y = 0
8y' - 3xy' = 3y
y'(8 - 3x) = 3y
3y
y' =
8 - 3x
6
At the given point, find the slope of the curve, the line that is tangent to the curve, or the line that is normal to the curve,
as requested.
36) x2 + y2 - 2x + 4y = 8, tangent at (4, 0)
2x + 2yy' - 2 + 4y' = 0
2yy' + 4y' = 2 - 2x
y'(2y + 4) = 2 - 2x
2 - 2x
y' =
2y + 4
m = y'(4, 0) =
2 - 8 -6
3
=
=0+4
4
2
y - y1 = m(x - x1 )
3
y - 0 = - (x - 4)
2
3
y=- x+6
2
Use implicit differentiation to find dy/dx and d2 y/dx 2.
37) xy - x + y = 6
xy' + y - 1 + y' = 0
xy' + y' = 1 - y
y'(x + 1) = 1 - y
1-y
y' =
x+1
y'' =
(x + 1)(-y') - (1 - y)(1)
=
(x + 1)2
(x + 1)(-
1- y
)-1+y
x+1
(x + 1)2
=
Find the derivative of the given function.
38) y = tan-1 11x
y' =
1
( 11x)2 + 1
( 11) =
11
2
11x + 1
Find dy/dx.
39) f(x) = 7e-8x
f'(x) = -56e-8x
40) y = ln (x - 8)
dy/dx =
1
x-8
Use logarithmic differentiation to find dy/dx.
41) y = x7x
ln y = ln x7x = 7x ln x
1
1
y' = (7x)( ) + (7)(ln x) = 7 + 7 ln x
y
x
y' = (7 + 7 ln x)y = (7 + 7 ln x) x7x
7
-1 + y - 1 + y
2y - 2
=
(x + 1)2
(x + 1)2
Solve the problem.
42) A particle moves along a line according the the postion function s(t) = t2 + 4t. How far has the particle travelled
in the first 4 seconds?
s(0) = (0)2 + 4(0) = 0
s(4) = (4)2 + 4(4) = 32
v(t) = s'(t) = 2t + 4 = 0
s(4) - s(0) = 32
t = -2 but t
0
43) For what value of x does the function f(x) = x3 - 9x2 - 120x + 6 have a local minimum?
f '(x) = 3x2 - 18x - 120 = 3(x2 - 6x - 40) = 3(x - 10)(x + 4) = 0
-4 10
+ | - | +
min at x = 10
x = -4, 10
44) Find the following for the graph of y = x3 - 4x2 + 4x + 2:
a) Critical points
b) Identity types of Extrema
c) Increasing and decreasing intervals
d) Inflection points
e) Concave up and concave down intervals
y = x3 - 4x2 + 4x + 2
y' = 3x2 - 8x + 4 = (3x - 2)(x - 2) = 0
x = 2/3, 2
2/3
2
max at 2/3 and min at 2, inc: ( , 2/3) (2, ) and dec: (2/3, 2)
+ | - | +
y'' = 6x - 8 = 0
x = 4/3
4/3
- | +
inflection pt at 4/3, up: (4/3, ) and down: ( , 4/3)
45) Find two non-negative numbers x and y whose sum is 100 and for which x2 y is a maximum.
x + y = 100
y = 100 - x
P = x2 y = x2(100 - x) = 100x2 - x3
P' = 200x - 3x2 = 0
0 200/3 100
X | + | - | X
x = 0,
200
3
max at x = 200/3
Give an appropriate answer.
46) Find the value or values of c that satisfy
y = 100/3
f(b) - f(a)
48
= f (c) for the function f(x) = x +
on the interval [6, 16].
b- a
x
f(b) - f(a) f(16) - f(6) 19 - 14
5
1
=
=
=
=
b- a
16 - 6
10
10 2
f'(x) = 1 -
48
x2
1
48
=12
c2
48 1
=
c2 2
c2 = 96
c=4 6
9.798
8
Find the extreme values of the function on the interval and where they occur.
47) g(x) = -x2 + 11x - 30 on 5 x 6
g'(x) = -2x + 11 = 0
11
2
5
X | +
| -
x=
11
2
6
| X
max at
11
, min at x = 5, 6
2
Find the extreme values of the function and where they occur.
48) y = x3 - 3x2 + 7x - 10
y' = 3x2 - 6x + 7 = 0 Never
49) y =
y' =
No extrema
8x
2
x +1
(x2 + 1)(8) - (8x)(2x) 8x2 + 8 - 16x2 -8x 2 + 8
=
=
=0
(x2 + 1)2
(x2 + 1)2
(x2 + 1)2
-1
- |
1
+ | -
x = 1, -1
min at x = -1, max at x = 1
Use the First Derivative Test to determine the local extrema of the function, and identify any absolute extrema.
50) y = 4x3 + 7
y' = 12x2 = 0
x=0
0
+ | +
no extrema
51) y = xe2x
y' = 2xe2x + e2x = 0
e2x (2x + 1) = 0
-
x=-
1
2
1
2
- |
+
absolute min at x = -
1
2
Use the Concavity Test to find the intervals where the graph of the function is concave up.
52) y = x3 - 3x2 - 9x + 3
y' = 3x2 - 6x - 9
y'' = 6x - 6 = 0
x=1
1
- | +
concave up for (1, )
9
53) y =
4
x
y' = y'' =
4
x2
8
x3
= 0 or Und
0
- | +
x=0
concave up for (0, )
Find the points of inflection.
54) y = x3 + 10x + 2
y' = 3x2 + 10
y'' = 6x = 0
x=0
0
- | +
inflection point at x = 0
55) y =
1 4
x - 2x3 + 12
2
y' = 2x3 - 6x2
y'' = 6x 2 - 12x = 0
6x (x - 2) = 0
x = 0, 2
0
2
+ | - | +
inflection points at x = 0, 2
Use the given derivative of the function to find the local extrema of the function.
56) y' = (x - 3)2(x + 1)
y' = 0 at x = -1, 3
3
-1
- | + | +
min at x = -1
Solve the problem analytically.
57) Of all numbers whose difference is 16, find the two that have the minimum product.
Let x and y be the two numbers, where x > y. Find the min of xy
x - y = 16
x = 16 + y
P = xy = (16 + y)(y) = 16y + y2
P' = 16 + 2y = 0
y = -8
-8
- | +
Pmin at y = -8, x = 8
10
Solve the problem.
58) A company wishes to manufacture a box with a volume of 24 cubic feet that is open on top and is twice as long
as it is wide. Find the width of the box that can be produced using the minimum amount of material. Round to
three decimal places.
Let L be the length of the box, W be the width and H be the height. Find W at the minimum of LW + 2LH + 2WH.
L = 2W
V = LWH = 2W2 H = 24
H=
12
W2
S = LW + 2LH + 2WH = 2W2 + 4WH + 2WH = 2W2 + 6WH = 2W2 + 6W(
S' = 4W -
72
=0
W2
W=
3
18
12
72
) = 2W2 +
2
W
W
2.621
59) The strength S of a rectangular wooden beam is proportional to its width times the square of its depth. Find the
dimensions of the strongest beam that can be cut from a 11-in.-diameter cylindrical log. (Round answers to
three decimal places.)
11"
121 = w2 + d2
d2 = 121 - w2
S = wd2 = w(121 - w2) = 121w - w3
S' = 121 - 3w2 = 0
w=
121
3
6.351
d = 8.981
11
Use the graph of f to estimate where f' is 0, positive, and negative.
60)
f' = 0 at x = -1, 1
f' > 0 for ( , -1)
f' < 0 for (-1, 1)
(1, )
12