MATH 100, Section 110 (CSP)
Weeks 10 and 11: Marked Homework Solutions
2010 Nov 25
1. [16] f (x) =
x2 +3
2(x−1)
is defined for all x 6= 1.
(i) Critical numbers, local maxima and minima, intervals where f is increasing
or decreasing:
2
+3)(2)
(x+1)(x−3)
−4x−6
−2x−3
f 0 (x) = (2x)(2x−2)−(x
= 2x4(x−1)
= x2(x−1)
.
2
2 =
(22 (x−1)2
2(x−1)2
0
f (x) is not defined iff x = 1, but f (x) is not defined at x = 1.
Critical numbers: f 0 (x) = 0 iff x = −1 or x = 3. Note that the tangent line is horizontal at
both critical numbers.
On (−∞, −1): f 0 (x) > 0, f is increasing.
On (−1, 1): f 0 (x) < 0, f is decreasing; local maximum value f (−1) = −1.
On (1, 3): f 0 (x) < 0, f is decreasing. (Note that f (1) is not defined.)
On (3, ∞): f 0 (x) > 0, f is increasing; local minimum value f (3) = 3.
2
2
(ii) Inflection points, intervals where the graph is concave upward or downward:
2
2
−2x−3)4(x−1)
4
= (x−1)
f 00 (x) = (2x−2)2(x−1)22 −(x
3.
(x−1)4
00
f (x) is not defined iff x = 1, but f (x) itself is not defined at x = 1.
Inflection points: none, since f 00 (x) 6= 0 for all x in the domain of f .
On (−∞, 1): f 00 (x) < 0, the graph of f is concave downwards.
On (1, ∞): f 00 (x) > 0, the graph of f is concave upwards.
(iii) Asymptotes (horizontal, vertical, slant):
We should suspect a vertical asymptote x = 1 since f (x) is a rational function whose denominator is zero when x = 1 but the numerator is not. We also suspect a slant asymptote
because the degree of the numerator is exactly one greater than the degree of the denominator.
Vertical asymptote:
x2 +3
limx→1− 2(x−1)
= −∞ (numerator is near 4, denominator is arbitrarily small and negative
e.g. if x = 0.999).
x2 +3
limx→1+ 2(x−1)
= +∞ (numerator is near 4, denominator is arbitrarily small and positive
e.g. if x = 1.001).
So x = 1 is indeed a vertical asymptote, and we get some useful information for the sketch.
Slant asymptote:
By long division,
2
,
f (x) = 12 x + 21 +
x−1
which is in the form
f (x) = mx + b + (something which has a limit of 0 as x → ±∞),
using
with m = 21 and b = 21 , so y = 21 x + 21 is the expected slant asymptote.
To verify this
1
1
2
the slant asymptote definition, we calculate limx→∞ f (x) − 2 x + 2 = limx→∞ x−1 = 0,
1
2
= 0. So we have now verified that
and similarly limx→−∞ f (x) − 21 x + 21 = limx→−∞ x−1
1
1
y = 2 x + 2 is indeed a slant asymptote (for both x → +∞ and x → −∞).
Sketch of graph:
√
2. [16] f (x) = x 3 − x.
(a) Domain:
√
For 3 − x to be defined, we need 3 − x ≥ 0, i.e. x ≤ 3. The domain is
(−∞, 3].
(b) The x-coordinates of the local maxima and minima (if any) and intervals
where f (x) is increasing or decreasing:
By the Product Rule, f 0 (x) = 1 · (3 − x)1/2 + x 21 (3 − x)−1/2 (−1), and after simplifying we
get
3(2 − x)
f 0 (x) = √
.
2 3−x
f 0 (x) is not defined at x = 3, f 0 (x) = 0 iff x = 2. Therefore x = 2 is a critical number (we
don’t count x = 3 as a critical number because it is an endpoint of the domain).
On (−∞, 2): f 0 (x) > 0, f is increasing.
On (2, 3): f 0 (x) < 0, f is decreasing; there must be a local maximum value attained at x = 2
(according to the way the textbook defines local extrema, there is not a local minimum
attained at x = 3 because that point is an end point of the domain, but it’s ok if you said
there was a local minimum attained at x = 3).
(c) Intervals where f (x) is concave upwards or downwards, and the x-coordinates
of inflection points (if any):
2
By the Quotient Rule, f 00 (x) =
1
−1/2 (−1)
1/2
3 (−1)(3−x) −(2−x) 2 (3−x)
2
(3−x)
f 00 (x) =
and after simplifying we get
3(x − 4)
.
4(3 − x)3/2
Note that x = 4 is outside the domain, f 00 does not change sign in the domain.
On (−∞, 3): f 00 (x) < 0, the graph of f is concave downwards.
(d) Vertical tangent:
√
= −∞ (e.g. x = 2.9999, f 0 (x) ≈
Since limx→3− f 0 (x) = limx→3− 23(2−x)
3−x
tangent at x = 3, i.e. at the point (3, 0).
−3
),
+0
there is a vertical
(e) Behaviour at infinity:
√
√
√
For a very negative number (e.g. x = 1000) 3 − x ≈ −x = |x|1/2 , so f (x) = x 3 − x ≈
x|x|1/2 = −|x|3/2 and we guess a = 3/2. Now we must verify (recall that if x < 0, then
|x| = −x):
√
√
x 3−x
x 3−x
= lim
lim
x→−∞ |x| |x|1/2
x→−∞ |x|3/2
√
x 3−x
√
= lim
x→−∞ −x
−x
q
= lim (−1) − x3 + 1
x→−∞
= −1.
(f) Sketch of graph:
3. [10] Mimimize the total travel time T =
domain 0 ≤ x ≤ 3, where x is in km.
3−x
8
3
+
√
x2 +4
3
=
3
8
− 18 x + 31 (x2 + 4)1/2 hr, over the
= − 18 + 13 · 21 (x2 + 4)−1/2 · 2x = − 18 + 3√xx2 +4 . dT
= 0 iff 3√xx2 +4 = 18
Find critical points: dT
dx
dx
√
√
iff 8x = 3 x2 + 4, 64x2 = 9(x2 + 4), 55x2 =
√ 36, x = ±6/ 55. Possible x-values where T
could attain a minimum value are x = 0,√6/ 55 ≈ 0.81, 3, and
the corresponding T -values
√
√
55
13
25
3
are T (0) = 24 ≈ 1.04, T (6/ 55) = 8 + 12 ≈ 0.99, T (3) = 3 ≈ 1.20. (If a calculator is
not used, the First Derivative Test can justify the minimum value is attained.) To minimize
the time required to reach the cabin, the hiker should walk 3 − √655 ≈ 2.2 km down the road
before setting off through the forest straight for the cabin.
4. [10] We want to minimize L = L1 + L2 (see the left side of the figure below). Since
8
= sin θ and L12 = cos θ, we have
L1
L = L(θ) =
8
1
+
= 8 csc θ + sec θ,
sin θ cos θ
with the domain
0<θ<
π
.
2
The derivative is L0 (θ) = −8 csc θ cot θ + sec θ tan θ = −8(cos θ/ sin2 θ) + (sin θ/ cos2 θ), or
sin3 θ − 8 cos3 θ
.
L (θ) =
sin2 θ cos2 θ
0
Since the denominator is always positive on the domain 0 < θ < π/2, L0 (θ) = 0 iff
sin3 θ − 8 cos3 θ = 0, or equivalently, tan3 θ = 8, θ = tan−1 2.
On (0, tan−1 2): L0 (θ) < 0, L is decreasing.
On (tan−1 2, π/2): L0 (θ) > 0, L is increasing.
By the First Derivative Test for Absolute Extreme Values, L(tan−1 2) is the absolute minumum value of L on (0, π/2). This absolute minimum
value √is the length of the shortest
√
√
2
8 1+22
−1
−1
−1
ladder: L(tan 2) = 8 csc(tan 2) + sec(tan 2) = 2 + 1+2
= 5 5 ≈ 11.18 ft.
1
This problem can also be done without trigonometric
functions, using similar triangles: see
p
y
2
the right side of the figure above. L = y + (1 + x)2 and by similar triangles, 1+x
= x8 .
This gives y = 8(1 + x−1 ), which can be substituted into L. Rather than minimizing L, it is
equivalent and easier to minimize
f (x) = L2 = 64(1 + x−1 )2 + (1 + x)2 .
4
The domain is 0 < x < ∞. Then f 0 (x) = −128(1+x−1 )x−2 +2(1+x) and f 0 (x) = 0 iff 1+x =
64(1+x−1 )x−2 , or (multiplying by x3 to get rid of the negative powers) x3 (1+x) = 64(x+1),
thus x3 = 64, x = 4. Writing f 0 (x) = 2x−3 (x + 1)(x3 − 64), we see that on (0, 4) we have
f 0 (x) < 0, f is decreasing, and on (4, ∞) we have f 0 (x) > 0, f is increasing. Therefore by
the First Derivative Test For Absolute Extreme Values, f (4) =√125 is the
√ absolute minimum
2
value of f = L , and the length of the shortest ladder is L = 125 = 5 5 ft.
5. [10] The circumference of the circular disk was originally 2πR, and after the sector is
cut out, what remains from the circumference has length 2πR − 2παR = 2π(1 − α)R. This
becomes the circumference of the base of the cone, and therefore the radius of the base of
the cone is r = (1 − α)R.
p
√
2 − r2 = R
By
Pythagoras’
Theorem,
the
height
of
the
cone
is
h
=
R
1 − (1 − α)2 =
√
R 2α − α2 , so the volume of the cup is
√
1
1
V = πr 2 h = πR3 (1 − α)2 2α − α2 ,
3
3
√
1
3
=
πR
−2(1
−
α)
·
with 0 < α < 1. The derivative is dV
2α − α2
dα
3
1−α
2
2
+(1 − α)2 · 21 (2α − α2 )−1/2 (2 − 2α) = 31 πR3 √2α−α
2 [−2(2α − α ) + (1 − α) ], or
dV
1
1−α
= πR3 p
(3α2 − 6α + 1).
dα
3
α(2 − α)
√
= 0 iff 3α2 − 6α + 1 = 0 iff α = 6± 36−12
=
Since 31 πR3 √ 1−α > 0 for all 0 < α < 1, dV
dα
6
α(2−α)
q
1 ± 23 . Since 3α2 − 6α + 1 is a quadratic polynomial with a positive coefficient of α 2 , its
graph is a parabola opening upwards, withq
two α-axis intercepts.
q
q
Therefore 3α2 − 6α + 1 > 0 if α < 1 − 23 , 3α2 − 6α + 1 < 0 if 1 − 23 < α < 1 + 23 ,
q
and 3α2 − 6α + 1 > 0 if 1 + 23 < α. But we need 0 < α < 1, so 3α2 − 6α + 1 > 0 if
q
q
0 < α < 1 − 23 , and 3α2 − 6α + 1 < 0 if 1 − 23 < α < 1.
q q 2
dV
Therefore on 0, 1 − 3 , dα > 0 and V is increasing, and on 1 − 23 , 1 , dV
< 0 and
dα
V is decreasing. By the First Derivative Test For Absolute
Extreme Values, V attains its
q
absolute maximum value on 0 < α < 1 when α = 1 −
2
3
≈ 0.1835.
Another way to justify this is to observe that the domain could be extended to the closed
interval 0 ≤ α ≤ 1 and V is still defined and continuous. Then use the q
Closed Interval
Method. The absolute maximum value must be attained at α = 0 or 1 − 23 or 1. Since
q
V = 0 at α = 0 and at α = 1, and V > 0 at α = 1 − 23 , the maximum value must be
q
attained at α = 1 − 23 .
It is also possible, and in fact the algebra is much easier, if the volume V is expressed
as
√
a function of the height h of the cone: V = 31 π(R2 − h2 )h. The value of h = R/ 3 that
5
gives the maximum value of V is found by setting dV /dh = 0, and justified either by the
First Derivative Test For Absolute Extreme Values, or by extending the domain to the closed
interval 0 ≤ h ≤ R and using the Closed Interval Method, evaluating
V at the
p end points
√
2
2
and the critical number. Then the corresponding value
p of r = R − h = R 2/3 can be
found, the finally the value of α = 1 − (r/R) = 1 − 2/3.
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