c
Math 151 WIR, Spring 2010, Benjamin
Aurispa
Math 151 Week in Review 3
Sections 2.3, 2.5, 2.6
1.
18
5
2. (a)
4
3
(b) 8
(c)
1
25
(d) − 163√5
(e) DNE There is a vertical asymptote at x = −1 and the limit does not exist since lim f (x) = ∞,
x→−1−
but
(f)
D
lim f (x) = −∞.
x→−1+
− 21 , −1
E
(g) DNE since
lim f (x) = 6, but
x→−3−
lim f (x) = −6.
x→−3+
(h) −5
(i) 0 (Use Squeeze Theorem)
3. −1 (Use Squeeze Theorem)
4. (a) Not continuous at x = −4 or x = 4 since f is not defined here. (These x-values are not in the
domain.)
Details:
At x = 4, the function has a removable discontinuity and lim f (x) = 89 . The function is not
x→4
continuous from the left or right at x = 4.
At x = −4, the function has a vertical asymptote (infinite discontinuity) and is not continuous
from the left or the right.
(b) Not continuous at x = −3 or x = 2 since f is not defined at x = −3 and the limit does not exist
at x = 2.
Details:
At x = −3 the function has a vertical asymptote and is not continuous from the left or the right.
At x = 2, lim f (x) = 11, but lim f (x) = 14, so the limit does not exist and there is a jump
x→2−
x→2+
discontinuity here. Since f (2) = 14, the function is continuous from the right at x = 2.
(c) Not continuous at x = 1 or x = 5 since the limit does not exist at x = 1 and, although the limit
exists at x = 5, it does not equal f (5).
Details:
At x = 1, lim f (x) = −6, but lim f (x) = −9. The limit does not exist and there is a jump
x→1−
x→1+
discontinuity. Since f (1) = −6, the function is continuous from the left at x = 1.
At x = 5, lim f (x) = 15 (from both sides), but f (5) = 10. There is a removable discontinuity
x→5
and the function is not continuous from the left or the right.
5. (a) This function is continuous everywhere since it is a polynomial.
(b) There is a removable discontinuity at x = −3.
lim f (x) = 60, but f (−3) is not defined.
x→−3
g(x) = (x − 2)(x − 9) = x2 − 11x + 18
(c) This function does not have a removable discontinuity. (It does have a vertical asymptote or
infinite discontinuity at x = 6.)
1
c
Math 151 WIR, Spring 2010, Benjamin
Aurispa
6. c = 1, d = 7
7. f (x) = x3 − 3x2 + 1 is continuous everywhere since it is a polynomial. f (2) = −3 and f (3) = 1. Since
f (2) < 0 < f (3), then by the Intermediate Value Theorem, there must exist a number c in (2, 3) such
that f (c) = 0. Thus, there must be a root to the equation x3 − 3x2 + 1 = 0 on the interval (2, 3).
8. g is continuous everywhere since it is a polynomial. g(1) = 2 and g(2) = 15. Since g(1) < 9 < g(2),
then by the Intermediate Value Theorem, there must exist a number c in (1, 2) such that g(c) = 9.
9. (a)
5
6
(b) 0
(c)
5
4
(d) −2
(e)
7
2
10. (a) Vertical Asymptote: x = − 19
Horizontal Asymptote: y = − 29
(b) Vertical Asymptote: x = 7
No Horizontal Asymptotes since lim f (x) = ∞ and lim f (x) = −∞
x→∞
(c) No Vertical Asymptotes
Horizontal Asymptotes: y =
√1 ,
5
x→−∞
y = − √15
2