MATH 101 HOMEWORK 1 – SOLUTIONS
√
√
1. First of all, for the limit to be finite we must have lim ax + b − 4 = 0, ie. b = 4,
x→0
b = 16. Also, from l’Hopital’s rule we have
√
ax + 16 − 4
= lim
lim
x→0
x→0
x
So we must have a/8 = 1, ie. a = 8.
√ 1
2 ax+16
1
·a
a
a
= √ = .
8
2 16
2. (a) At (t, t2 ), the tangent has slope y 0 = 2t, hence the equation of the tangent is
y − t2 = 2t(x − t). Rewrite this as y = 2tx − t2 and set m = 2t, then the equation becomes
y = mx − m2 /4 as required.
(b) The slope of the tangent at x = t is y 0 = 2at + b, so the equation of the tangent is
y − (at2 + bt + c) = (2at + b)(x − t),
which simplifies to y = (2at + b)x − at2 + c. Let m = 2at + b, then t = (m − b)/2a, so that
m−b 2
(m − b)2
f (m) = −at2 + c = −a(
) +c−
+ c.
2a
4a
3. (a) Let |AB| = x, |AC| = y, |AD| = d, where AD is the bisector. The area of the
triangle ABC is
√
1
xy 3
◦
xy sin(120 ) =
.
2
4
This area is also equal to the sum of areas of the triangles ABD and ADC, which is
√
1
3
1
◦
◦
xd sin(60 ) + yd sin(60 ) =
(x + y)d.
2
2
4
This gives us an equation for d: xy = (x + y)d. We solve it and plug in xy = 1:
1
1
x
xy
=
=
= 2
.
d=
x+y
x+y
x + 1/x
x +1
(b) We need to maximize the function d(x) found in (a), so we compute
x2 + 1 − x · 2x 1 − x2
d0 (x) =
.
(x2 + 1)2 (x2 + 1)2
1
This is 0 when x = 1, positive when 0 < x < 1, and negative when x > 1 (remember that
x is the sidelength of a triangle, so it must be positive). Hence d(x) is maximized when
x = 1, and its maximal value is d(1) = 1/2.
4. The function is defined for all x 6= −1. We have
lim −
x→±∞
1
= 0,
x+1
lim −
x→−1−
1
= ∞,
x+1
lim −
x→−1+
1
= −∞,
x+1
so that
lim y = e0 = 1,
x→±∞
lim y = ∞,
x→−1−
lim y = 0.
x→−1+
Thus there is a horizontal asymptote y = 1 and a vertical asymptote x = −1.
Next, we compute
1
y 0 = e−1/(x+1) ·
> 0 for all x,
(x + 1)2
hence y is increasing on each of the intervals (−∞, −1) and (−1, ∞). There are no local
maxima or minima (note that y approaches 0 as x → −1+ , but does not actually reach
it.) Finally,
y 00 = e−1/(x+1) ·
1
−2
−2x − 1 −1/(x+1)
+ e−1/(x+1) ·
=
e
.
4
3
(x + 1)
(x + 1)
(x + 1)4
This is 0 when x = −1/2, positive when x < −1/2, and negative when x > −1/2. Thus
y is concave up on (−∞, −1) and (−1, −1/2), concave down on (−1/2, ∞), and has an
inflection point at (−1/2, e−2 ).
The graph is below.
2