SSEA Math 41 Track
1
Final Exam Solutions
September 1st, 2011
(12 points)
Compute the following limits if they exist, or write DNE if they do not
exist. Write ∞ or −∞ to indicate an infinite limit.
(a)
t2 − 9
lim 2
.
t→3 t + t − 12
Solution: This equals
t+3
6
(t − 3)(t + 3)
= lim
= .
t→3 t + 4
t→3 (t − 3)(t + 4)
7
lim
(b)
x3 + 2x − 1
.
x→1
x−1
Solution: The numerator approaches 2, and the denominator approaches zero, so
we have an infinite limit. Checking the signs: the numerator is positive and the
denominator is negative, so the limit is −∞.
lim−
(c)
lim cos y.
y→∞
Solution: cos y oscillates back and forth as y goes to infinity, so the limit does not
exist.
(d)
(x2 + 1)2 (x − 3)
.
x→∞
(x − 5)6
lim
Solution: The numerator is a degree-5 polynomial, and the denominator is a
degree-6 polynomial, so the limit as x → ∞ is zero. We don’t even need to
compare the coefficients.
SSEA Math 41 Track
2
Final Exam Solutions
September 1st, 2011
(16 points) Compute the first derivative of each of the following functions: (problems
worth 3/3/3/3/4 points)
(a)
f (t) = t2 sin t.
Solution: By the Product Rule, the answer is:
2t sin t + t2 cos t.
(b)
f (x) = cos(3e2x ).
Solution: By the Chain Rule, the answer is:
− sin(3e2x ) · 3e2x · 2 = −6xe2x sin(3e2x ).
(c)
Z
x
ln(t + 4) dt.
g(x) =
−3
Solution: By the Fundamental Theorem of Calculus, the answer is ln(x + 4).
(d)
arctan x
.
2x + 1
Solution: By the Quotient Rule, the answer is:
f (x) =
2x+1
1+x2
− 2 arctan x
.
(2x + 1)2
(e) Compute f 0 (x) directly from the limit definition of the derivative:
f (x) = x2 + 3.
Solution: We want
(x + h)2 + 3 − (x2 + 3)
2xh + h2
= lim
= lim (2x + h) = 2x.
h→0
h→0
h→0
h
h
lim
SSEA Math 41 Track
3
Final Exam Solutions
September 1st, 2011
(12 points)
Sketch the graph of a function f (x) which satisfies each of the following
conditions (your answer to this problem should be a single graph satisfying all twelve
conditions). You will get one point for each correct condition.
• limx→−∞ f (x) = −1, limx→∞ f (x) = 1, limx→−1− f (x) = 1, limx→−1+ = −∞;
limx→0 f (x) = −1, f (0) = −2;
• f (x) is differentiable everywhere except for x = −1, x = 0, and x = 3, and is
continuous everywhere except for x = −1 and x = 0;
• If x < 3 and f 0 (x) is defined, f 0 (x) ≥ 0, but f 0 (x) = 0 only at x = 2;
• f (x) has an inflection point at (−2, 0) and a local maximum at (3, 2).
SSEA Math 41 Track
4
Final Exam Solutions
4
2
September 1st, 2011
(12 points)
Sketch the graph of f (x) = x − 2x + 2. You should find the intervals of
increase and decrease, find the intervals of concavity, and label all local and absolute
maxima and minima as well as all inflection points.
SSEA Math 41 Track
5
Final Exam Solutions
September 1st, 2011
(12 points)
(a) (6 points) Find the equation of the tangent line to f (x) = ex cos x at x = 0, and
then use linear approximation to estimate e0.1 cos(0.1).
Solution: The derivative is f 0 (x) = −ex (sin x) + ex (cos x) = ex (cos x − sin x), so
f 0 (0) = 1(1 − 0) = 1. And f (0) = 1 also. So the equation of the tangent line is
L(x) = x + 1.
From this, we know that e0.1 cos(0.1) ∼ L(0.1) = 1.1.
(b) (6 points) Find the absolute maximum and minimum values of f (x) = ex cos x on
the interval [0, π/2]. Hint: it may help to remember that tan(π/4) = 1.
Solution: We already computed the derivative, and it is defined everywhere on the
interval. So set it equal to zero; ex cannot be zero, so we must have cos x = sin x,
i.e. tan x = 1. This means (in this interval) that x = π/4. So π/4 is the
only critical number. This means we have √
to check the points 0, π/4, and π/2.
π/4
f (0) = 1. f (π/2) = 0. And f (π/4) = e / 2. The absolute minimum is clearly
at π/2 and the
√ value is zero. As for the absolute maximum, it’s not obvious
whether eπ/4 / 2 or 1 is bigger. But checking the sign of f 0 (x) between 0 and π/4
tells us that f 0 (x) > 0 there, so f (x) is increasing between
0 and π/4. Hence the
√
absolute maximum is at π/4 and the value is eπ/4 / 2.
SSEA Math 41 Track
6
Final Exam Solutions
September 1st, 2011
(10 points)
A comet is making a near approach to Earth. The comet travels along
the graph of the function y = 5 − 2x, where Earth is located at the origin of our
coordinate system and x and y are measured in millions of kilometers. (Of course,
comets do not travel in straight lines, but it’s a reasonable approximation when a
comet is in a small region of the solar system.)
(a) (4 points) Draw a diagram and write an expression for the distance from the
comet to Earth. Your answer should only depend on x, the x-coordinate of the
comet.
p
√
Solution: The distance is D(x) = (x2 +(5−2x)2 ) = 5x2 − 20x + 25. Diagram
below:
(b) (6 points) At what point is the comet closest to Earth, and how far away is the
comet at that point?
Solution: We want to minimize the distance. The derivative
10x − 20
D0 (x) = √
.
2 5x2 − 20x + 25
Setting this equal to zero gives that D0 (x) = 0 exactly when x = 2. Moreover,
checking the signs shows that D0 (x) > 0 when x > 2 and D0 (x) < 0 when x < 2.
So D(x) has a global √
minimum at x = 2, and the comet is closest
to Earth at the
√
point (2, 1). D(2) = 5, so the comet is at a distance of 5 million miles from
Earth, or roughly 2.2 million miles.
SSEA Math 41 Track
7
(8 points)
Final Exam Solutions
September 1st, 2011
Consider the integral
Z
8
√
1 + x dx.
3
(a) Write an approximation to this integral, using the Left Endpoint Rule with 5
subintervals.
√
√
√
√
√
Solution: 1 + 3 + 1 + 4 + 1 + 5 + 1 + 6 + 1 + 7.
(b) Do the same with the Right Endpoint Rule and 10 subintervals.
√
√
√
√
√
√
√
√
√
√
Solution: (0.5)( 4.5 + 5 + 5.5 + 6 + 6.5 + 7 + 7.5 + 8 + 8.5 + 9).
(c) Do the same with the Midpoint Rule and 5 subintervals.
√
√
√
√
√
Solution: 4.5 + 5.5 + 6.5 + 7.5 + 8.5.
(d) Are your answers to a) and b) overestimates or underestimates? Explain.
√
Solution: 1 + x is increasing (obvious, or can take the derivative), so the Left
Endpoint Rule gives an underestimate and the Right Endpoint Rule gives an
overestimate.
(e) (Extra Credit - 2 points) Is your answer to c) an overestimate or underestimate?
Explain.
Solution: The graph of f (x) is concave down - the second derivative is −(1/4)(x+
1)−3/2 , which is negative. See Figure 5 on page 404 of Stewart for the picture that
should convince you that the Midpoint Rule gives an overestimate in this case.
SSEA Math 41 Track
8
(16 points)
Final Exam Solutions
September 1st, 2011
Evaluate the following integrals:
(a)
Z
1
(x3 − 2x2 + x − 5) dx.
0
Solution:
x4
4
−
1 1 2 1
2x3 x2
59
+
− 5x = − + − 5 = − .
3
2
4 3 2
12
0
(b)
Z
π 1/3
x2 sin(x3 ) dx.
0
Solution: Make the substitution u = x3 ; then du = 3x2 dx, so x2 dx = (1/3) du.
The limits of integration become 0 and π. We get:
Z π
π 2
1
1
sin u du = (− cos u)0 = .
3
3
0 3
(c)
Z
x cos x dx.
Solution: Integrate by parts with f (x) = x and g 0 (x) = cos x;then f 0 (x) = 1 and
g(x) = sin x, so by the integration by parts formula:
Z
Z
x cos x dx = x sin x − sin x dx = x sin x + cos x + C.
(d) If
F (x) =
evaluate
Z
(sin x)9 (cos x)2001
,
1+x
π/4
F 0 (x) dx.
0
Solution: By the Fundamental Theorem of Calculus, the answer is F (π/4)−F (0).
F (0) = 0 since sin 0 = 0, so the answer is F (π/4), which is
√
√
(1/ 2)9 (1/ 2)2001
1
= 1005
.
1 + (π/4)
2 (1 + (π/4))