MATH171-15c
Time Allowed: 2 hours
MATH 171, FALL SEMESTER 2015
Version A
Name (print):
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Section No:
GUIDELINES
1. In Part 1 (Problems 1–19), mark your responses on your ScanTron form using a
No: 2 pencil. For your own record, mark your choices on the examination paper as
well . ScanTrons will be collected at the conclusion of the examination; they will not
be returned.
2. Calculators should not be used throughout the examination.
3. In Part 2 (Problems 19–26), present your solutions in the space provided. Show
all your work neatly and concisely, and indicate your final answer clearly. You
will be graded, not merely on the final answer, but also on the quality and correctness
of the work leading up to it.
4. Be sure to write your name, section number, and version letter of the
examination on the ScanTron form.
1
Part 1 – Multiple Choice ( 57 points)
1. Find the slope of the tangent line to the graph of x3 + y 3 = 6xy − 1 at the point (2, 3).
(a)
6
21
(b)
2. Compute
−12
21
lim + 3
x
2+x
(c)
−4
5
(d)
4
5
(e)
2
X
5
.
x→−2
(a) 3
(b) ∞
(c) −∞
(d)
3. Compute cot sin−1 (3/4) .
√
(a) 3/5
(b) 3/ 7
(c)
√
√
3
7/3 X
(e) 0 X
(d) 5/3
(e) 7/3
.
4. 250 + 250 =
(a) 2100
(b) 22500
(c) 450
5. If f (15) (0) = 2 and g(x) = f
(a) 214
x
2
(b) Non of these
d) 251 X
(e) 2101
, find g (15) (0).
(c)
1
X
214
(d) 215
6. Which of the following points x are points of inflection of
f (x) =
x2
+ sin x,
4
π
2π
and x =
3
3
π
5π
x = and x =
X
6
6
π
x = only
6
π
x = only
3
f does not have inflection points
(a) x =
(b)
(c)
(d)
(e)
2
0 ≤ x ≤ π.
(e)
1
215
7. Given that f (x) =
√
1 − x, the range of the inverse is
(b) [1, ∞)
(a) (−∞, 1] X
(c) [0, ∞)
(d) [−1, 1]
(e) (−∞, ∞)
√
8. Find the liner approximation L(x) to the function f (x) = sin(x2 ) at a = π
√
√
√
(a) L(x) = 2 π − 2x
(b) L(x) = −2 πx
(c) L(x) = 2π − 2 πx X
√
√
(d) L(x) = −2π + 2 πx
(e) L(x) = π − x
9. The domain of sin−1 (x − 1) is
(a) [0, 2] X
10.
(b) [−2, 0]
(d) [−π/2, π/2]
(e) (−∞, ∞)
Solve the equation ln(x + e) + ln(x − e) = 2 + ln 3
(a) x = 3e only
(b) x = 1 and x = 3e (c) x = 2e only X
(d) x = 2e and x = −2e
11.
(c)[−1, 1]
(e) No solution
√
Find the slope of the tangent line to the curve r(t) = ( t, t2 + 5) at t = 2
√
1
1
1
(a) 8 2 X
(b)
(d)
(c) √
(e) 32
32
2
8 2
12. For 0 < x < π/2, the derivative of f (x) = xtan x is
(a) (tan x)xtan x−1
sec x
(b) xsec x
+ ln x sec x tan x
x
tan x
(c) xtan x
+ ln x sec2 x X
x
(d) (sec x)xsec x−1
sec2 (x) (e) xtan x
x
ex − 1 − x
=
x→0
cos x
(a) −∞
(b) −1
13. lim
(c) 0 X
(d) 1
3
(e) ∞
14. Assume the derivative of a function f (x) is
f 0 (x) = x3 − 2x2 .
Which of the following statements is NOT true:
(a) f (x) has an absolute minimum at x = 2,
(b) f (x) has a local extreme point at x = 0, X
(c) f (x) is increasing on [2, ∞),
(d) f (x) is concave up on (−∞, 0),
(e) f (x) is concave up on (4/3, ∞).
15. Let f be a differentiable function, and let G be the function defined by
2
G(x) = [f (f (x))] . Compute the derivative of G with respect to x.
(a) 2[f (f (x))]f 0 (f (x))
(c) 2f 0 (f (x))
(b) 2[f (f (x))]f 0 (f (x))f 0 (x) X
(d) 2f 0 (f (x))f 0 (x)
(e) 2f (x)f 0 (x)
16. Suppose that f is a function defined on (0, ∞), and its derivative is given by f 0 (x) =
1 − ln(x)
, 0 < x < ∞. Determine on which interval f is concave upwards.
x2
(a) (e3/2 , ∞) X
(b) (0, e3/2 )
(d) (3/2, ∞)
(c) (0, 3/2)
(e) (0, e)
17. If f (x) = x3 + x2 and g(x) is the inverse of f (x). Find g 0 (2).
(a) 1/3
(b) 1/5 X
1 −
=
18. lim
x→0 x
sin x
(a) −∞
(b) −1
(c) 1/10
(d) 1/16
(e) 1/24
1
19.
d
dx
(a)
sin(e)
x
(c) 0 X
(d) 1
(e)∞
=
x
cos(e))x sin(e)
(b) ln(sin(e)) sin(e)
x
(d) cos(e) ln(sin(e)) sin(e)
x
X
x
(e) ln(cos(e)) cos(e)
4
x−1
(c) x sin(e)
Part 2 – Work Out (57 points)
In this part you need to show your work leading to the answer: NO WORK
(understandable to the reader), NO CREDIT.
20. (7 pts.)A spectator watches a helicopter which flies at a constant altitude of 80m at a
speed of 10m/s. At what rate is the angle of elevation from the feet of the spectator
to the helicopter (see figure) changing 6 seconds after the helicopter passes directly
above the spectator.
Solution: x: horizontal distance from spectator to helicopter. The vertical distance is 80.
It follows therefore that
x
(tan(θ) = 80/x also works).
cot(θ) =
80
Differentiating gives:
− csc2 (θ)
1 dx
10
1
dθ
=
=
= .
dt
80 dt
80
8
After 6 seconds we find for x:
x(s) =
p
802 + (6 · 10) = 100.
and
csc(θ) =
5
x
= .
80
4
Thus after 6 seconds we have
dθ
−1 4 2
2
=
) = − rad/s.
dt
8 5
25
2
The answer 25 (which one obtains taking the outside angle is also acceptable).
21. (7 pts.) A Norman window is constructed by adjoining semicircle to the top of an
ordinary rectangular window. Find the dimensions of a Norman window of maximum
area given that the total perimeter is 16 feet.
Solution: Perimeter: 16 = πr + 2r + 2h, it follows therefore h = 16 − 12 (πr + 2r)
Area of Window:
h
i
π
π
π
1
A = r2 + 2rh = r2 + 2r 8 − (πr + 2r) = 8r −
+ 4 r2 .
2
2
2
2
Domain of A: [0, 16/(π + 2)] Then
A0 = 16 −
an thus A0 = 0 when
r=
00
Since A = −
π
2
π
2
π
2
+ 2 2r,
8
16
=
.
+2
π+4
+ 2 2 < 0 this must be an absolute Maximum.
5
22. (5 pts.) State the Extreme Value Theorem.
23. (10) For f (x) = x4 − 6x2 + 4
a) Find the intervals where f is increasing and decreasing.
b) Find the local maxima and minima.
c) Find the intervals where f is concave up and down.
Solution:
f 0 (x) = 4x3 −
= 4x(x2 − 3).
√ 12x √
Thus x = 0, 3, − 3 are the critical points.√
√
Intervals for which function is increasing: ( 3, ∞)√
and (− √3, 0),
Intervals for which
is decreasing: (−∞, − 3), (0, 3)
√ function
√
Local minima: − 3, 3; and local Maxima 0
f 00 (x) = 12x2 − 12. Thus f 00 (x) = 0 for x = ±1
Intervals for which function is concave up: (−∞ − 1) and (1−, ∞)
Intervals for which function is concave down : (−1, 1)
6
24. (12) a) State the Mean Value Theorem.
b) Assume that f (x) is continuous on [0, 1], and twice differentiable on (0, 1). Show
that if f (0) = f (1) and if f (x) is concave down then f (x) has exactly one local maximum
which is also the absolute maximum on [0, 1].
Solution: By Rolle’s Theorem there is a c ∈ (0, 1) so that f 0 (0) = 0. Now since f (x) is
concave down, it follows that f 0 (x) is decreasing, which means that f 0 (x) > f 0 (c) = 0 for
all x < c, and f 0 (x) < f 0 (c) = 0 for all x > c.
But this implies that there is no other critical point, and moreover that c is a a local
maximum. Since f 0 (x) > 0 on the interval [0, c] it follows that f (x) is increasing on [0, c]
and thus f (c) > f (0) = f (1).
This implies that the end points cannot be absolute maxima. So the only choice is
that c is the absolute maximum.
25. (6) Compute
1
lim+ (tan2 (x)) ln x .
x→0
Show all your steps clearly.
ln(tan2 (x))
1 Solution: We first consider ln (tan2 (x)) ln x =
ln(x)
which is for x → 0+ of the undetermined form −∞
−∞ . Thus using L’Hospital’s rule
ln(tan2 (x))
lim
= lim+
ln(x)
x→0
x→0+
2 tan(x) sec2 (x)
tan2 (x)
1
x
2x sec2 (x)
2x cos(x)
= lim+
= lim+
= 2.
tan(x)
x→0
x→0 cos2 (x) sin(x)
It follows therefore that
1
lim+ (tan2 (x)) ln x = e2 .
t→0
7
26. (10) (a) Write down the formula for the derivative of the inverse of a function f (x).
(b) using part (a) derive the formula
d
−1
arccos(x) = √
dx
1 − x2
(the inverse function of cos(x) is denoted by arccos(x)).
Solution: a) if g = f −1 then
d
1
g(x) = 0
dx
f (g(x)
b)
dx
1
−1
−1
arccos(x) =
=p
=√
.
2
dx
− sin(arccos(x))
1 − x2
1 − cos (arccos(x)
8
Rough(Scratch) Work
This blank page is meant for rough work only; work presented here will not be graded.
9
QN
PTS
1–19
20
21
22
23
24
25
26
TOTAL
10