MATH 113 (B1) - SUMMER 2008
Assignment 3 Solutions
Question 1. (10 Points) Estimate the value of
√
3
27.2 by using linear approximation.
1
Solution: Let f (x) = x 3 , then
1
f (x) = x 3
1 −2
f 0 (x) =
·x 3
3
so that,
1
f (27) = (27) 3 = 3
−2
1
1
f 0 (27) =
· (27) 3 = 3
3
3
Therefore, we have:
√
3
so that
√
3
x ≈ f (27) + f 0 (27)(x − 27) = 3 +
27.2 ≈ 3 +
1
(x − 27)
33
1
0.2
812
(27.2 − 27) = 3 +
=
(≈ 3.01)
3
3
27
270
Question 2. (10 Points) Let f (x) = x3 − 2x + 3.
(a) Compute its differential at x = 2.
(b) Use it to estimate f at x = 2.1.
Solution: (a) Since dy = f 0 (x)dx = (3x2 − 2)dx, we have the differential at x = 2 is
dy = f 0 (2)dx = (3 · 22 − 2)dx = 10dx.
(b) At x = 2.1, dx = ∆x = 2.1 − 2 = 0.1 and thus dy = 10 · 0.1 = 1.
Use f (2.1) = f (2) + ∆y ≈ f (2) + dy to get
f (2.1) ≈ 23 − 2 · 2 + 3 + 1 = 8.
(Notice the above estimate formula is the linear approximation formula : f (2.1) ≈ f (2) +
f 0 (2)(2.1 − 2).)
Question 3. (20 Points) Find the absolute maximum and minimum values of f on the
given interval:
(a) f (x) =
x2
x+1
, [−1, 3]
+x+1
4
(b) f (x) = x 5 (x − 1)2 , [−1, 32]
Solution:
(a) Step 1. Find the critical numbers
y0 =
1 · (x2 + x + 1) − (2x + 1)(x + 1)
−x2 − 2x
−x(x + 2)
=
= 2
.
2
2
2
2
(x + x + 1)
(x + x + 1)
(x + x + 1)2
¡
¢2
Since x2 + x + 1 = x + 12 + 34 , the critical numbers comes from f 0 (x) = 0.
f 0 (x) = 0 =⇒ x = −2, x = 0.
Notice that x = −2 is a critical number, however it is not inside the interval [−1, 3], so
we are not interested in it.
Step 2. Calculate f (−1), f (0), f (3)
f (0) = 1, f (−1) = 0, and f (3) =
4
4
=
9+3+1
13
Thus the absolute maximum value is f (0) = 1 and the absolute minimum value is f (−1) =
0.
(b) Step 1. Find the critical numbers
y 0 = 45 x
−1
5
4
(x − 1)2 + x 5 [2(x − 1)] = 25 x
−1
5
(x − 1)[2(x − 1) + 5x] = 25 x
−1
5
(x − 1)(7x − 2).
The critical numbers comes from f 0 (x) = 0 and f 0 (x) is undefined.
2
f 0 (x) = 0 =⇒ x = , x = 1
7
f 0 (x) is undefined =⇒ x = 0.
Step 2. Calculate f (−1), f (0), f (1), f
f (−1) = 1 · (−2)2 = 4,
¡ ¢ ¡ ¢ 4 ¡ ¢2
,
f 72 = 72 5 · −5
7
¡2¢
7
, f (32)
f (0) = 0,
f (1) = 0,
f (32) = 24 · (−31)2 = 24 · 312 = 15376.
Thus, the absolute maximum value is f (32) = 15376 and the absolute minimum value is
f (0) = f (1) = 0.
Question 4. (10 points) Show the equation 3x4 + 12x = 1 has exactly 2 solutions.
Solution: Let f (x) = 3x4 + 12x − 1.
Step 1: Use IVT to show f (x) has at least 2 roots.
Since f (x) is a polynomial, it is continuous on R. This means that we can use the
Intermediate Value Theorem (IVT). Since
f (0) = 0 + 0 − 1 = −1 < 0
and
f (1) = 3 + 12 − 1 = 14 > 0,
the IVT implies that there is at least a c1 ∈ (0, 1) such that f (c1 ) = 0. Moreover, since
f (−2) = 3 · (−2)4 + 12 · (−2) − 1 = 3 · 16 − 24 − 1 = 23 > 0
and
f (0) = −1 < 0,
the IVT implies that there is at least a c2 ∈ (−2, 0) such that f (c2 ) = 0. Clearly c1 6= c2 .
Therefore, f (x) has at least 2 roots.
Step 2: Use Rolle’s Theorem to show f (x) has at most 2 roots.
Notice that f 0 (x) = 12x3 + 12 = 12(x3 + 1) = 12(x + 1)(x2 − x + 1). Since the
polynomial
µ
¶2
1
3
2
x −x+1= x−
+ >0
2
4
has no real roots, x = −1 is the unique real root of f 0 (x).
Suppose f (x) had three distinct real roots, say x1 < x2 < x3 . Since it is a polynomial,
f (x) is continuous on [x1, x2] and differentiable on (x1 , x2 ). Moreover, since f (x1 ) =
f (x2 ) = 0, the Rolle’s Theorem implies the existence of a number c1 , x1 < c1 < x2 , such
that f 0 (c1 ) = 0. Similarly, the Rolle’s Theorem allows us to deduce the existence of c2 ,
x2 < c2 < x3 , such that f 0 (c1 ) = 0. Therefore, the hypothesis f (x) has at least three
distinct real roots leads to the conclusion that f 0 (x) has at least 2 distinct real roots.
However this contradicts our observation that f 0 (x) has only one real root. Therefore,
f (x) has at most two distinct real roots.
From Step 1 and Step 2, we can conclude that f (x) = 3x4 + 12x − 1 has exactly 2
roots, that is, the equation 3x4 + 12x = 1 has exactly 2 solutions.
Question 5. (20 Points) Identify the intervals on which f is increasing and decreasing.
(Notice that you should already have their derivatives from Question 3.)
(a) f (x) =
x2
x+1
+x+1
4
(b) f (x) = x 5 (x − 1)2
Solution: (a) We need to investigate when the derivative,
f 0 (x) =
1 · (x2 + x + 1) − (2x + 1)(x + 1)
−x2 − 2x
−x(x + 2)
=
= 2
2
2
2
2
(x + x + 1)
(x + x + 1)
(x + x + 1)2
is positive or negative.
The critical numbers are x = −2 and 0, so we need to examine the three intervals
(−∞, −2) ,
(−2, 0) ,
(0, ∞)
as in the table below.
Interval
x
(−∞, −2) −
x(x + 2)
+ x + 1)2
x+2
1
(x2 +x+1)2
−
+
−
f 0 (x) = -
(x2
(−2, 0)
−
+
+
+
(0, ∞)
+
+
+
−
Therefore, for f 0 (x) we have
+
−
−2
−
0
Thus, f (x) is decreasing on the interval (−∞, −2), increasing on the interval (−2, 0), and
decreasing on the interval (0, ∞).
(b) We need to investigate when the derivative,
4
2 −1
4 −1
f 0 (x) = x 5 (x − 1)2 + x 5 [2(x − 1)] = x 5 (x − 1)(7x − 2),
5
5
is positive or negative.
The critical numbers are x = 0, x = 1 and x = 72 , so we need to examine the four
intervals
µ
¶
¶
µ
2
2
(−∞, 0) ,
0,
,1 ,
and
(1, ∞)
,
7
7
as in the table below.
Interval
(−∞, 0)
x
−1
5
x − 1 7x − 2
f 0 (x) = 25 x
−1
5
(x − 1)(7x − 2)
−
−
−
−
+
−
−
+
¡2
¢
,
1
7
+
−
+
−
(1, ∞)
+
+
+
+
¡
0, 27
¢
Therefore, for f 0 (x) we have
+
−
0
+
−
1
2
7
Thus, f (x) is decreasing on the interval (, −∞, 0 ), increasing on the interval ( 0, 27 ),
decreasing on the interval ( 27 , 1 ), and increasing on the interval ( 1, ∞ ).
Question 6. (20 Points) Sketch the graph of f (x) =
x
.
(x − 1)2
Solution: Find the first two derivatives,
f 0 (x) =
f 00 (x) = −
1 · (x − 1)2 − x · 2(x − 1)
x+1
=−
,
4
(x − 1)
(x − 1)3
(1) · (x − 1)3 − (x + 1) · 3(x − 1)2
2(x + 2)
=
,
(x − 1)6
(x − 1)4
(a) Find the domain of f.
dom(f ) = {x ∈ R | x 6= 1}
for all x 6= 1,
for all x 6= 1.
(b) Find the x-intercept and y-intercept of the graph of f.
The x-intercept is x = 0 and the y-intercept is y = 0.
(c) Find any symmetries of the graph of f, for example, is f an even function? an odd
function? a periodic function?
There are no symmetries
(d) Find all horizontal asymptotes and all vertical asymptotes to the graph of f.
Note that
x
lim f (x) = lim
= lim µ
x→±∞
x→±∞ (x − 1)2
x→±∞
1
x ¶ = 0 = 0,
2
1
1
1−
x
and the line y = 0 is a horizontal asymptote to the graph.
Also note that
lim− f (x) = lim−
x→1
x→1
x
= +∞
(x − 1)2
and
lim+ f (x) = lim+
x→1
x→1
x
= +∞,
(x − 1)2
and the line x = 1 is a vertical asymptote to the graph.
(e) Find the intervals where f is increasing and the intervals where f is decreasing.
Since f 0 (x) =
−(x + 1)
, the critical numbers are x = −1 and x = 1. For y 0 we have
(x − 1)3
+
−
−1
−
1
Thus, f (x) is decreasing on the interval ( −∞, −1 ), increasing on the interval
( −1, 1 ), and decreasing on the interval ( 1, ∞ ).
(f) Find all local maxima and all local minima of the function f.
Since 1 is not in the domain of f , we only need to discuss if we have a local maximum,
a local minimum, or neither at the critical point x = −1.
Since
f 0 changes from ª to ⊕
the first derivative test implies that y has a local minimum at x = −1.
(g) Find all intervals where the graph of f is concave up (∪) and all intervals where the
graph of f is concave down (∩). Find all inflection points of the graph of f.
Since f 00 (x) =
2(x + 2)
,
(x − 1)4
f 00 (x) = 0 =⇒ x = −2
f 00 (x) is undefined =⇒ x = 1
For f 00 (x) we have
+
−
+
1
−2
Thus, f (x) is concave down on the interval ( −∞, −2 ), concave up on the interval
( −2, 1 ), and concave up on the interval ( 1, ∞ ). Since the concavity changes at
x = −2, the point (−2, −2/9) is an inflection point of the graph of y.
(h) Sketch the graph of f.
y
−2
−1
0
1
x
Question 7. (10 Points) Find the points on the ellipse 4x2 + y 2 = 4 that are farthest
away from the point (1, 0).
Solution: The distance between the point (x, y) and (1, 0) is
p
d = (x − 1)2 + (y − 0)2 .
The distance d has a an absolution maximum if and only if d2 = (x − 1)2 + y 2 also has
an absolution maximum.
Since the point (x, y) is on the ellipse, we have,
f (x) = d2 = (x − 1)2 + 4 − 4x2 , with − 1 ≤ x ≤ 1
Differentiating,
f 0 (x) = 2(x − 1) − 8x
and
f 00 (x) = 2 − 8 = −6 < 0
for −1 < x < 1. The only critical point occurs when −2 − 6x = 0 ⇒ x = 13 .
By the Second D-Test, this is an absolute maximum. Therefore,
on the
´
³ the √points
1 ±4 2
2
2
.
ellipse 4x + y = 4 that are farthest away from the point (1, 0) are 3 , 3
Question 8. (10 Points) A Norman window has the shape of a rectangle surmounted by
a semicircle. Thus the diameter of the semicircle is equal to the width of the rectangle.
If the perimeter of the window is 30 ft, find the dimensions of the window so that the
greatest possible amount of light is admitted.
Solution: In the figure below, the perimeter is
πx ³
π´
30 = x + 2y +
= 1+
x + 2y,
2
2
so that
y=
³
1h
π´ i
30 − 1 +
x .
2
2
y
y
x
Therefore, the area of the Norman window is
³ x ´2
1
A(x) = xy + · π
2
2
µ 2¶
h
³
πx
1
π´ i 1
= x·
30 − 1 +
x + ·
2
2
2
4
µ 2¶
³
´
1
π 2 1
πx
= 15x −
1+
x + ·
2
2
2
4
·³
µ 2 ¶¸
´
1
π 2
πx
= 15x −
1+
x −
2
2
4
³
´
1
π 2
= 15x −
1+
x
2
4
and differentiating
A0 (x) = 15 −
4+π
x=0
4
and
A00 (x) = −
when
x=
60
,
4+π
4+π
< 0 for all x ≥ 0.
4
From the second derivative test, the area is an absolute maximum when x =
in this case
·
¸
³
π ´ 60
30
1
30 − 1 +
=
.
y=
2
2 4+π
4+π
60
, and
4+π