MATH 101 HOMEWORK 1 – SOLUTIONS
1. Find all values of k such that the line y = 3x − 4 is tangent to the graph of y = kx3 .
If the line is tangent at (a, ka3 ) then the slope of the curve at that point is 3: 3ka2 = 3,
ka2 = 1. Also, the point (a, ka3 ) lies on the line y = 3x − 4, so that 3a − 4 = ka3 = a. We
get a = 2, k = 1/a2 = 1/4.
2. A ladder 12 ft long rests against a vertical wall. If the bottom of the ladder slides away
from the wall at a rate of 1 ft/s. how fast is the top of the ladder sliding down the wall
when the bottom of the ladder is 4 ft from the wall?
We have x2 + h2 = 12, so that 2xx0 + 2hh0 = 0. When x = 4, h =
Therefore at that time
h0 = −
√
122 − 44 =
√
128.
1
xx0
4·1
= − √ ≈ .35355.
= −√
h
128
8
3. A box with an open top is to be constructed from a rectangular piece of cardboard of size
3 ft × 2 ft by cutting out a square from each of the four corners and bending up the sides.
Find the largest volume that such a box can have.
The volume of the box is V (x) = x(3 − 2x)(2 − 2x) = 6x − 10x2 + 4x3 , with 0 ≤ x ≤ 1. At
the endpoints, V (0) = V (1) = 0. We look for critical points: V 0 (x) = 6 − 20x + 12x2 = 0,
6x2 − 10x + 3 = 0, i.e.
√
√
10 ± 100 − 4 · 6 · 3
5± 7
x=
=
.
2·6
6
The larger of these values is outside the interval (0, 1). For x =
ft3 . This is the maximum value.
1
√
5− 7
6
we have V ≈ 1.0563
4. Sketch the graph of the function y = x2x−4 . Include the following information: local minima and maxima, intervals of increase and decrease, intervals of concavity and inflection
points, asymptotes, limits at infinity.
The function f (x) =
x
x2 −4
is odd: f (−x) = −f (−x), and defined for all x 6= ±2. We have
lim
x→2−
lim
x→−2−
x
x
=
−∞,
lim
= ∞,
x2 − 4
x→2+ x2 − 4
x2
x
x
= −∞, lim
= ∞,
2
+
−4
x→−2 x − 4
x
lim 2
= 0.
x→±∞ x − 4
Thus the function has asymptotes x = ±2, y = 0. We also have
f 0 (x) =
(x2 − 4) − x · 2x −x2 − 4
,
(x2 − 4)2
(x2 − 4)2
which is not defined for x = ±2 and negative for all other x. Thus the function is decreasing
on each of the intervals (−∞, −2), (−2, 2), (2, ∞). There are no local minima or maxima.
Now we check the concavity:
f 00 (x) =
−2x(x2 − 4)2 − (−x2 − 4) · 2(x2 − 4) · 2x
2x(x2 + 12)
=
,
(x2 − 4)4
(x2 − 4)3
not defined when x = ±2. This is positive (f concave up) on (−2, 0) and (2, ∞), and
negative (f concave down) on (−∞, −2) and (0, 2). At x = 0 f 00 changes sign, so f has an
inflection point.
2