COLLEGE ALGEBRA II (MATH 006)
FALL 2013
——PRACTICE FOR EXAM I——
SOLUTIONS
1. Find all rational zeros of the polynomial, and write the
polynomial in factored form.
P (x) = x3 + 2x2 − 5x − 6
Solution:
p
p
q
q
±1, ±2, ±3, ±6 ±1 ±1, ±2, ±3, ±6
−1 ) 1
2 −5 −6
−1 −1
6
1
1 −6
0
P (x) = (x + 1)(x2 + x − 6) = (x + 1)(x + 3)(x − 2)
Zeros: −3, −1, and 2
2. Find all rational zeros of the polynomial, and write the
polynomial in factored form.
P (x) = 2x4 − 7x3 + 3x2 + 8x − 4 Solution:
p
p
q
q
±1, ±2, ±4 ±1, ±2 ±1, ±2, ±4, ± 21
−1 ) 2 −7 3
8 −4
−2 9 −12
4
2 −9 12 −4
0
Note: −1 is a lower bound for the zeros of P (x).
P (x) = (x + 1)(2x3 − 9x2 + 12x − 4)
1
)
2 −9 12 −4
2 −7
5
2 −7
5
1
2
)
1
2 −9
12 −4
4 −10
4
2 −5
2
0
P (x) = (x + 1)(x − 2)(2x2 − 5x + 2)
P (x) = (x + 1)(x − 2)(2x − 1)(x − 2)
P (x) = (2x − 1)(x + 1)(x − 2)2
Zeros: −1, 12 , and 2
3. Find all the real zeros of the polynomial. Use the quadratic
formula if necessary.
P (x) = x3 − x2 − 3x + 3
Solution:
p
p
q
q
±1, ±3 ±1 ±1, ±3
−1
)
1 −1 −3 3
−1
2 1
1 −2 −1 4
1
)
1 −1 −3
3
1
0 −3
1
0 −3
0
2
P (x) = (x
√− 1)(x
√ − 3) = (x − 1)(x +
Zeros: − 3, 1, 3
√
3)(x −
√
3)
4. Use Descartes’ Rule of Signs to determine how many positive and how many negative real zeros the polynomial can
have. Then determine the possible total number of real zeros.
P (x) = −4x7 + x3 − x2 + 2
Solution:
P (x) = −4x7 |{z}
+ x3 |{z}
− x2 |{z}
+ 2
There are either three or one positive real zeros.
P (−x) = 4x7 |{z}
− x3 − x2 |{z}
+ 2
There are two or zero negative real zeros.
Possible total real zeros: five, three, or one.
2
5. Show that the given values for a and b are lower and upper
bounds for the real zeros of the polynomial.
P (x) = 8x3 + 10x2 − 39x + 9; a = −3, b = 2
Solution:
−3
)
10 −39
9
−24
42 −9
8 −14
3
0
8
2
Alternating (or zero).
)
10 −39 9
16
52 26
8 26
13 35
8
All entries are nonnegative.
6. Find the intercepts, asymptotes, domain, and range, and
then sketch the graph of each rational function.
(a) s(x) =
(b) r(x) =
(c) f (x) =
x−1
x2 −4
2x−1
x+1
x2 −x−6
x+1
Solution:
x−1
(a) s(x) = xx−1
2 −4 = (x−2)(x+2)
−1
x−int: x = 1, y−int: y = −4
= 14
Horiz. Asymp: y = 0, Vert. Asymp: x = ±2
y
5
−5
5
x
−5
Figure 1: Domain: {x : x 6= ±2}, Range: (−∞, ∞)
3
(b) r(x) = 2x−1
x+1
x−int: 2x − 1 = 0, x = 12 , y−int: y =
Horiz. Asymp: y = 21 = 2
Vert. Asymp: x + 1 = 0, x = −1
y
−1
1
=1
5
−5
5
x
−5
Figure 2: Domain: {x : x 6= −1}, Range: {y : y 6= 2}
2
−x−6
(c) f (x) = x x+1
= (x−3)(x+2)
x+1
x−int: (x − 3)(x + 2) = 0, x = 3, and x = −2
y−int: y = −6
1 = −6
Horiz. Asymp: None, Vert. Asymp: x + 1 = 0, x = −1
Oblique/Slant: y = x − 2(Long Division)
y
5
−5
5
x
−5
Figure 3: Domain: {x : x 6= −1}, Range: (−∞, ∞)
4
7. Express the equation in exponential form.
(a) log8 2 = 13
(b) log2 ( 81 ) = −3
Solution:
1
(b) 2−3 = 18
(a) 8 3 = 2
8. Express the equation in logarithmic form.
(a) 53 = 125
(b) 10−4 = 0.0001
Solution:
(a) log5 125 = 3
(b) log 0.0001 = −4
9. Use the definition of the logarithmic function to find x.
(a) log2 x = 5
(b) log2 16 = x
Solution:
(a) x = 25 = 32
(b) 2x = 16 = 24 , x = 4
10. Find the domain of the function.
g(x) = log3 (x2 − 1)
Solution:
g(x) is defined whenever x2 − 1 = (x − 1)(x + 1) > 0.
To solve this nonlinear inequality, we set (x − 1)(x + 1) = 0
with solutions x = ±1.
We proceed as follows:
A
B
-1
C
1
Region
A
B
C
Interval
(−∞, −1) (−1, 1) (1, ∞)
Test Point(t)
−2
0
2
2
t −1
3
−1
3
Inequality satisfied?
Yes
No
Yes
Domain of g(x):(−∞, −1) ∪ (1, ∞)
5
11. Use the Laws of Logs to expand each expression.
2
3 4
√ +1) )
(a) log( xzy6 )
(b) log2 ( x(x
x2 −1
Solution:
3 4
(a)
log( xzy6 ) = log(x3 y 4 ) − log z 6
= log x3 + log y 4 − log z 6
= 3 log x + 4 log y − 6 log z
√
2
x(x
+1)
(b) log2 ( √x2 −1 ) = log2 (x(x2 + 1)) − log2 x2 − 1
1
= log2 x + log2 (x2 + 1) − log2 (x2 − 1) 2
= log2 x + log2 (x2 + 1) − 21 log2 (x2 − 1)
12. Use the Laws of Logs to combine each expression.
(a) log2 A + log2 B − 2 log2 C
(b) −4 log x − 13 log(x2 + 1) + 2 log(x − 1)
Solution:
(a) log2 A + log2 B − 2 log2 C = log2 (AB) − log2 C 2
= log2 ( AB
C2 )
(b) −4 log x − 13 log(x2 + 1) + 2 log(x − 1)
1
= log x−4 + log(x2 + 1)− 3 + log(x − 1)2
= log( 4 21 1 ) + log(x − 1)2
= log(
x (x +1) 3
(x−1)2
1
x4 (x2 +1) 3
)
13. Use the change of base formula and a calculator to evaluate
each logarithm.
(a) log2 5
(b) log3 16
Solution:
5
(a) log2 5 = log
log 2 = 2.322
16
(b) log3 16 = log
log 3 = 2.524
6
14. Find the solution to the exponential equation.
5x = 4x+1
Solution:
5x = 4x+1
ln 5x = ln 4x+1
x ln 5 = (x + 1) ln 4
x ln 5 = x ln 4 + ln 4
x(ln 5 − ln 4) = ln 4
ln 4
x = ln 5−ln
4 = 6.21
15. Solve the equation.
e2x − 3ex + 2 = 0
Solution:
Let r = ex , so that e2x − 3ex + 2 = r2 − 3r + 2.
r2 − 3r + 2 = 0
(r − 2)(r − 1) = 0
r = 2 or r = 1
ex = 2 or ex = 1
ln ex = ln 2 or ln ex = ln 1
x ln e = ln 2 or x ln e = ln 1
x = ln 2 or x = ln 1
x = .6931 or x = 0
16. Solve the logarithmic equation for x.
log5 (x + 1) − log5 (x − 1) = 2
Solution:
log5 (x + 1) − log5 (x − 1) = 2
log5 ( x+1
x−1 ) = 2
x+1
2
x−1 = 5
(x + 1) = 25(x − 1)
x + 1 = 25x − 25
7
−24x = −26
13
x = 26
24 = 12
17. The population of a certain city was 112, 000 in 2006, and
the observed doubling time for the population is 18 years.
(a) Find an exponential model n(t) = n0 ert for the population t years after 2006.
(b) Estimate when the population will reach 500, 000.
Solution:
(a) We need to evaluate r using the doubling time, n(18) =
224, 000.
224000 = 112000e18r
2 = e18r
ln 2 = ln e18r
ln 2 = 18r ln e
r = ln182 = .0385
n(t) = 112000e.0385t
(b) 500000 = 112000e.0385t
125
.0385t
28 = e
.0385t
ln( 125
28 ) = ln e
ln( 125
28 ) = .0385t ln e
1
ln( 125
t = .0385
28 ) = 38.9(or the year 2045)
8
18. The half-life of radium-226 is 1600 years. Suppose we have
a 22 mg sample.
(a) Find an exponential model m(t) = m0 e−rt for the mass
remaining after t years.
(b) How much of the sample will remain after 4000 years?
(c) After how long will only 18 mg of the sample remain?
Solution:
ln 2
(a) r = lnh2 = 1600
= .000433
−.000433t
m(t) = 22e
(b) m(4000) = 22e−.000433(4000) = 3.89
(c) 18 = 22e−.000433t
9
−.000433t
11 = e
9
ln( 11
) = ln e−.000433t
9
ln( 11
) = −.000433t ln e
1
9
t = − .000433
ln( 11
) = 463.4
9
19. A hot bowl of soup starts to cool according to Newton’s
Law of Cooling. Its temperature at time t is given by
T (t) = 65 + 145e−0.05t
where t is measured in minutes and T is measured in o F .
(a) What is the initial temperature of the soup?
(b) What is the temperature after 10 minutes?
(c) After how long will the temperature be 100o F ?
Solution:
(a) T (0) = 65 + 145e−0.05(0) = 210
(b) T (10) = 65 + 145e−0.05(10) = 152.9
(c) 100 = 65 + 145e−0.05t
35 = 145e−0.05t
7
−0.05t
29 = e
7
ln( 29
) = ln e−0.05t
7
ln( 29
) = −0.05t ln e
7
1
ln( 29
) = 28.4
t = − .05
10