140 Version EXAM 1 Radin (58305) This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001
6 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6
1
while at x0 = 3,
x 3-
lim f (x) = -4 =
x 3+
lim f (x) = 0 .
10.0 points
Consequently, on (-7, 7) the function f is discontinuous only at x = 3, -3 . 002 10.0 points
Below is the graph of a function f .
4 2 -6 -4 -2 -2 -4
-8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7
2
4
6
Use this graph to determine all of the values of x on (-7, 7) at which f is discontinuous. 1. none of the other answers 2. x = 3 3. no values of x 4. x = 3 , -3 correct 5. x = -3 Explanation: Since f (x) is defined everywhere on (-7, 7), the function f will be discontinuous at a point x0 in (-7, 7) if and only if
x x0
Determine where f is continuous, expressing your answer in interval notation. 1. (-, -2) (4, ) 2. (-, -2) (-2, 4) (4, ) 3. (-, ) correct 4. (-, -2) (-2, ) 5. (-, 4) (4, ) Explanation: The function is piecewise continuous, so we have to check the left and right hand limits at the points where the definition of f changes, i.e., at x = -2 and x = 4. Now at x = -2 lim f (x) =
x-2- x-2-
A function f is defined by x -2, 2 - x, x2 , -2 < x < 4, f (x) = 12 + x, x 4.
lim f (x) = f (x0 )
or if
x x0 -
lim 2 - x = 4 , lim x2 = 4 ,
lim
f (x) =
x x0 +
lim
f (x).
lim f (x) =
x-2+
x-2+
As the graph shows, the only possible candidates for x0 are x0 = -3 and x0 = 3. But at x0 = -3, f (-3) = 0 =
x -3
hence, the function is continuous at the point x = -2. On the other hand, at x = 4 lim f (x) = lim x2 = (4)2 = 16 ,
x4- x4-
lim f (x) = 4,
lim f (x) = lim 12 + x = 16 .
x4+ x4+
Version 140 EXAM 1 Radin (58305) Thus, the function is also continuous at the point x = 4. 003 10.0 points 5. limit =
2
Let f be a continuous function on [-3, 1] such that f (-3) = -1 , f (1) = 7 .
1 3 Explanation: After rationalizing the numerator we see that (x + 2) - 9 x+2-3 = x+2+3 = Thus 1 x+2-3 = x-7 x+2+3 x-7 . x+2+3
Which of the following is a consequence of the Intermediate Value Theorem without further restrictions on f ? 1. f (c) = 1 for some c in (-3, 1) 2. f (0) = 0 3. f (c) = 1 for some c in (-3, 1) correct 4. f (c) = 0 for some c in (-3, 0) 5. -1 f (x) 7 for all x in (-3, 1) Explanation: Because f is continuous on [-3, 1] the Intermediate Value Theorem ensures that for each M, -1 < M < 7, there exists at least one choice of c in (-3, 1) for which f (c) = M . In particular, f (c) = 1 for some c in (-3, 1) . But without further restrictions on f , none of the other properties need hold. 004 Determine
x7
for all x = 7. Consequently, limit = lim
x7
1 1 = . 6 x+2+3
005
10.0 points
Determine if the limit
x0
lim
1+
sin(-5x) 7x
exists, and if it does, find its value. 1. limit = - 2. limit = - 3. limit = 4. limit = 2 7 2 5
10.0 points
2 correct 7 2 5
lim
x+2-3 . x-7
5. limit doesn't exist Explanation: Since sin - = - sin ,
x0
1. limit =
1 correct 6
2. limit = 3 3. limit doesn't exist 4. limit = 6
lim
1+
sin(-5x) 7x
x0
= lim
1-
sin 5x 7x
.
Version 140 EXAM 1 Radin (58305) On the other hand,
x0
3
lim
sin ax = a. x
using, say, a graphing calculator. They look like
Consequently, lim 1+ sin(-5x) 7x = 2 . 7
x0
006 10.0 points Determine the value of
x3
lim f (x) (3, 5) g: h:
when f satisfies the inequalities 5 f (x) x2 - 6x + 14 on [-5, 3) (3, 5]. 1. limit = 5 correct 2. limit = 3 3. limit = 6 4. limit = 4 5. limit does not exist 6. limit = 2 Explanation: Set g(x) = 5, h(x) = x - 6x + 14 .
2
(not drawn to scale), so the graphs of g and h `touch' at the point (3, 5) while the graph of f is `sandwiched' between these two graphs. Thus again we see that
x3
lim f (x) = 5 .
007
10.0 points
When f is the function defined by f (x) = determine if
x 2-
2x - 1,
x < 2, x 2,
5x - 8,
Then, by properties of limits,
x3
lim g(x) = lim 5 = 5 ,
x3
lim f (x)
while
x3
exists, and if it does, find its value.
x3
lim h(x) = lim ( x2 - 6x + 14) = 5 .
1. limit = 4 2. limit = 1 3. limit = 3 correct 4. limit = 5
By the Squeeze Theorem, therefore,
x3
lim f (x) = 5 .
To see why the Squeeze theorem applies, it's a good idea to draw the graphs of g and h
5. limit does not exist
Version 140 EXAM 1 Radin (58305) 6. limit = 2 Explanation: The left hand limit
x 2-
4 10.0 points
009
Determine if the limit 8 -8 lim x + 1 x0 x exists, and if it does, find its value. 1. limit = -9 2. limit = 8 3. limit = -8 correct 4. limit = 9 5. limit does not exist Explanation: After bringing the numerator to a common denominator and rearranging, we obtain, 8 -8 8 - 8(x + 1) x+1 = x x(x + 1) = - Now
x0
lim f (x)
depends only on the values of f for x > 2. Thus
x 2-
lim f (x) =
x 2-
lim 2x - 1 .
Consequently, limit = 2 2 - 1 = 3 . 008 10.0 points
Find the value of x-5-1 lim . x0 3(x - 4) 1. limit does not exist correct 2. limit = 1 3
3. limit = 1 4. limit = 1 6
8x 8 = - . x(x + 1) x+1 8 x+1
lim
2 5. limit = 3 Explanation: Since the domain of f (x) = x-5
exists, so the limit 8 -8 x+1 lim x0 x exists and limit = - lim 010 8 x+1 = -8 .
consists of the interval [5, ), the function x-5-1 3(x - 4) is not defined near x = 0. Consequently, limit does not exist .
x0
10.0 points 2 + 3 . x3
Find the derivative of f when f = (x) 6x2 +
Version 140 EXAM 1 Radin (58305) 1. f (x) = 6 2x5 - 1 x4 3. x = 0 4 3 4 5. x = - , 0 correct 3 1 6. x = 3 2 7. x = - , 0 3 Explanation: By the quotient rule, 4. x = - y (x) = 9x2 + 12x 6x(3x + 2) - 9x2 = . (3x + 2)2 (3x + 2)2
5
correct
2. none of the other answers 3. f (x) = 6 4. f (x) = 3 5. f (x) = 3 2x5 + 1 x4 2x4 + 1 x3 1 - 2x5 x4
1 - 2x5 6. f (x) = 6 x4 2x4 - 1 7. f (x) = 3 x3
Now the tangent line is horizontal to the graph when y (x) = 0, i.e., when 9x2 + 12x = 3x(3x + 4) = 0. Hence the tangent line is horizontal when x = 0, - 012 4 3 .
Explanation: Since
d r (x ) = rxr-1 dx holds for all r, we see that f (x) = 12x - Consequently, f (x) = 6 2x5 -1 . 6 . x4
10.0 points
Find the y-intercept of the tangent line at the point P (2, f (2)) on the graph of f (x) = x2 + 3x + 2 . 1. y-intercept = -1
x4
keywords: derivatives, negative powers 011 10.0 points
2. y-intercept = 6 3. y-intercept = 1 4. y-intercept = -2 correct 5. y-intercept = 2 6. y-intercept = -6
Find all values of x at which the tangent line to the graph of 3x2 y = 3x + 2 is horizontal. 1. x = 1 , 0 3 2 2. x = - 3
Explanation: The slope, m, of the tangent line at the point P (2, f (2)) on the graph of f is the value of the derivative f (x) = 2x + 3
Version 140 EXAM 1 Radin (58305) at x = 2, i.e., m = 7. On the other hand, f (2) = 12. Thus by the point-slope formula, an equation for the tangent line at P (2, f (2)) is y - 12 = 7(x - 2) , which after simplification becomes y = 7x - 2 . Consequently, the tangent line at P has y-intercept = -2 013 10.0 points . Determine the derivative of f when f (x) = x2 sin x + 2x cos x . 1. f (x) = (x2 + 2) cos x correct 2. f (x) = -(x2 + 2) sin x 3. f (x) = (x2 + 2) sin x 4. f (x) = -(x2 + 2) cos x 5. f (x) = -(x2 - 2) sin x 6. f (x) = (x2 - 2) cos x Explanation: By the Product Rule, f (x) = 2x sin x + x2 cos x + 2 cos x - 2x sin x . Consequently, f (x) = (x2 + 2) cos x .
6
Find the derivative of f when f (x) = 5 sec x + cos x . 1. f (x) = sin x 1 - 5 sec2 x 2. f (x) = sin x 5 sec2 x - 1 correct 3. f (x) = cos x 5 csc2 x + 1 4. f (x) = cos x 1 - 5 csc2 x 5. f (x) = sin x 5 sec2 x + 1 6. f (x) = cos x 5 csc2 x - 1 Explanation: Since d (sec x) = sec x tan x = sin x sec2 x dx while we see that f (x) = sin x 5 sec2 x - 1 . 014 10.0 points d (cos x) = - sin x , dx
015
10.0 points
Determine f (2) (x) when f (x) = 2 sin x - 3 cos x . 1. f (2) (x) = -2 sin x - 3 cos x 2. f (2) (x) = -3 cos x + 2 sin x 3. f (2) (x) = -2 cos x - 3 sin x 4. f (2) (x) = -2 sin x + 3 cos x correct 5. f (2) (x) = 3 sin x + 2 cos x 6. f (2) (x) = -2 cos x + 3 sin x
Version 140 EXAM 1 Radin (58305) Explanation: Since d sin x = cos x, dx we see that f (x) = 2 cos x + 3 sin x . Consequently, f (2) (x) = -2 sin x + 3 cos x . 016 10.0 points 4. d cos x = - sin x , dx
7
3.
correct
Determine which one of the following is the graph of f (x) = 3x . x-2 5.
1.
6. 2.
Explanation:
Version 140 EXAM 1 Radin (58305) The graph of f will have a vertical asymptote at x - 2 = 0, i.e., to the right of the y-axis. This eliminates two of the possible graphs. In addition, the graph of f must pass through the origin, eliminating a third graph. To decide which of the remaining three graphs is that of f we look at the behaviour of f near the asymptote:
x 2-
8
What is her speed after 9 minutes, and in what direction is she heading at that time? 1. away from RLM at 20 yds/min. 2. away from RLM at 25 yds/min. 3. towards RLM at 20 yds/min. 4. towards RLM at 30 yds/min. 5. away from RLM at 30 yds/min. 6. towards RLM at 25 yds/min. correct Explanation: The graph is linear on [8, 10], so the student's speed at time t = 9 is the (absolute value of the) slope of this line. Hence 100 - 150 = -25 . 10 - 8
lim f (x) = - , lim f (x) = + .
while
x 2+
Consequently, the graph of f must be
slope = keywords: limit, infinity 017 10.0 points
The fact that her distance from RLM is decreasing at t = 9 indicates that she is walking towards RLM at that time. 018 10.0 points
A Calculus student leaves the RLM building and walks in a straight line to the PCL Library. Her distance from RLM after t minutes is given by the graph yards 500 400 300 200 100 2 4 6 8 10 mins 12
If f is a function having
6 5 4 3 2 1 0 -1 -2 -3 -4 -5 -6 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5
4 2 -4 -2 -2 -4 2 4
Version 140 EXAM 1 Radin (58305) as its graph, which of the following is the graph of the derivative f of f ?
5 4 1.
9
4 3 2 2 1 0 -1 -4 -2 -2 -2 -3 -4 -4 -5 -6 6 5 -6 -5 -4 -3 -2 -1 0 4 2. 4 3 2 2 1 0 -1 -4 -2 -2 -2 -3 -4 -4 -5 -6 5 -6 -5 -4 -3 -2 -1 0 4 3. 4 3 2 2 1 0 -1 -4 -2 -2 -2 -3 -4 -4 -5 -6 5 -6 -5 -4 -3 -2 -1 0 4 4. 4 3 2 2 1 0 -1 -4 -2 -2 -2 -3 -4 -4 -5 -6 6 5 -6 -5 -4 -3 -2 -1 0 4 5. 4 3 2 2 1 0 -1 -4 -2 -2 -2 -3 -4 -4 -5 -6 -6 -5 -4 -3 -2 -1 0
2
4
The tangent line to the graph of f is horizontal at (-2, f (-2)) and (1, f (1)), so the graph of f must have x-intercepts at x = -2 and x = 1; in particular, the graph of f cannot be a straight line. This eliminates three of the possible graphs. By observing the slope of the graph of f near x = -2 and x = 1 we see that the graph of f is
6 5 4
1 2 3 4 5
3
4 2 -4 -2 -2 -4
-6 -5 -4 -3 -2 -1 0 1 2 3 4 5
correct
2 1 0 -1
2
4
2
4
-2 -3 -4 -5
1 2 3 4 5
-6
2
4
1 2 3 4 5
2
4
1 2 3 4 5
2
4
1 2 3 4 5
Explanation: