Name:
Student number:
Math 110-003
Quiz 3 (Mar 3)
Write full solutions to the questions below. It is to your own benefit to write as clearly and
legibly as possible.
1. Let f (x) = (x + 1)5 − 5x − 2. Find
(a) the intervals of concavity and
(b) the (local) maxima and minima.
Hint: you may use either the first or second derivative test, but you can save time by
using the second derivative test.
Solution.
(a) We have f 0 (x) = 5(x + 1)4 − 5 and f 00 (x) = 20(x + 1)3 . Thus,
f 00 (x) = 0 ⇒ 20(x + 1)3 = 0
⇒ (x + 1)3 = 0
⇒x+1=0
⇒ x = −1.
(1)
(2)
(3)
(4)
Since f 00 (−2) < 0 and f 00 (0) > 0, the funtion f is concave down on (−∞, −1) and
concave up on (−1, ∞).
(b) We have
f 0 (x) = 0 ⇒ 5(x + 1)4 = 5
⇒ (x + 1)4 = 1
⇒ x + 1 = ±1
⇒ x = −2, 0.
(5)
(6)
(7)
(8)
As determined in (a), f 00 (−2) < 0 and f 00 (0) > 0, so there is a local maximum at
x = −2 and a local minimum at x = 0.
Name:
Student number:
2. Let g(x) =
Math 110-003
Quiz 3 (Mar 3)
2x2 + x − 1
. Find
x2 + x − 2
(a) the horizontal asymptote(s) and
(b) the vertical asymptote(s),
if any exist. You may use l’Hôpital’s rule if you wish but it is not needed for this
question.
Solution.
(a) For x 6= 0, we have
2x2 + x − 1
2x2 + x − 1 x2
= 2
÷ 2
x2 + x − 2
x +x−2
x
2 + x1 − x12
.
=
1 + x1 − x22
(9)
(10)
As x → ±∞,
1
1
→ 0 and 2 → 0,
x
x
so
2+
2x2 + x − 1
= lim
2
x→±∞ x + x − 2
x→±∞ 1 +
lim
1
x
1
x
−
−
1
x2
2
x2
=
(11)
2+0−0
= 2.
1+0−2×0
(12)
Thus, y = 2 is the only horizontal asymptote of g(x).
(b) Since g(x) is a rational function, its vertical asymptotes (if any) must occur at
zeros of the denominator. By the quadratic formula,
√
−1 ± 1 + 8
−1 ± 3
2
x +x−2=0⇒x=
=
= −2, 1,
(13)
2
2
so the vertical asymptotes (if any) must occur at x = −2, 1. Since the numerator
is finite for both these values of x, they are both indeed vertical asymptotes.
Name:
Student number:
Math 110-003
Quiz 3 (Mar 3)
√
3. Bonus. Find all horizontal asymptote(s) of h(x) =
9x6 − x
.
x3 + 1
Solution.
By definition,
√
x6 =
p
(x3 )2 = |x3 |.
(14)
Thus,
√
h(x) =
q
x6 (9 −
1
)
−x
x5
=
x3 + 1
x3 (1 + x13 )
q
3
|x | 9 − x15
= 3
.
x (1 + x13 )
9x6
(15)
(16)
For x > 0, we have |x3 | = x3 and so
q
9−
q
x 9−
3
h(x) =
x3 (1 +
1
x5
1
)
x3
=
1+
1
x5
1
x3
(17)
and so
√
9−0
= 3.
lim h(x) =
x→∞
1+0
Thus, y = 3 is a horizontal asymptote of h(x).
On the other hand, for x < 0, we have |x3 | = −x3 , so
q
q
1
3
− 9 − x15
−x 9 − x5
=
h(x) = 3
x (1 + x13 )
1 + x13
(18)
(19)
and so
√
− 9−0
lim h(x) =
= −3.
x→−∞
1+0
Thus, y = −3 is a horizontal asymptote of h(x).
(20)