Precalculus I: College Algebra
Review Problems
Final Exam solutions
MTH 119
November 28, 2006
1. If f (x) = 7x3 + 3x2 − x + C and f (−2) = 1, what is the value of C?
f (−2) = 7(−2)3 + 3(−2)2 − (−2) + C = 1
− 56 + 12 + 2 + C = 1
− 42 + C = 1
C = 43
a. C = −13
b. C = 43
c. C = 39
d. C = −45
2. What is the2domain of the2 rational function
2(x −2)
2(x2 −2)
−4
f (x) = 3x22x
+6x−45 = (3(x2 +2x−15) = 3(x+5)(x−3) → x 6= −5 and x 6= 3
a. {x|x 6= 3, x 6= −3, x 6= −5}
b. x ∈ R
c. {x|x 6= −3, x 6= 5}
d. {x|x =
6 3, x 6= −5}
3. Find the domain of the following functions:
(a) f (x) =
2t−1
t2 +3t
√
here, t2 + 3t 6= 0 → t(t + 3) 6= 0 → t 6= 0, t 6= −3
2x2 + 3x + 1 here, 2x2 + 3x + 1 ≥ 0 → (2x + 1)(x + 1) ≥ 0
x = −1 once
zeroes:
x = − 21 once
From the sign chart:
(b) h(x) =
Domain must be x ≤ −1 ∪ x ≥ − 21
(c) k(x) = 3 log(3 − x)
here 3 − x > 0 → 3 > x
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4. Based on the graphs of f (x) and g(x) answer the following:
a.
c.
e.
g.
Domain of f : − 6 ≤ x ≤ 6
f (−4) = 3
(f + g)(2) = f (2) + g(2) = 3 + 1 = 4
(f ◦ g)(−2) = f (g(−2)) = f (−6) = −2
b. Range of g : − 6 ≤ y ≤ 2
d. for what value(s) of x is f (x) = 3 → x = −4, x ≈ .5, x = 2?
f. (g − f )(−4) = g(−4) − f (−4) = −5 − 3 = −8
h. (f ◦ f )(6) = f (f (6)) = f (6) = 6
5. Use the properties of logarithms to find the exact value of the expression:
log17 34 − log17 2 = log17 34
2 = log17 17 = 1
a. 6
b. 1
c. 34
d. 17
√
6. For the given functions,√f (x) = x and g(x) = 3x + 8. Find f ◦ g and state its domain.
f (g(x)) = f (3x + 8) = 3x + 8 here 3x + 8 ≥ 0 → x ≥ − 38
7. If f (x) = 2 + 3x2 ,
(a) f (−4) = 2 + 3(−4)2 = 2 + 3(16) = 50
(b) f (−2x) = 2 + 3(−2x)2 = 2 + 3(4x2 ) = 12x2 + 2
(c)
f (a+h)−f (a)
h
=
2+3(a+h)2 −[2+3a2 ]
h
2+3(a2 +2ah+h2 )−2−3a2
h
=
=
2+3a2 +6ah+3h2 −2−3a2
h
= 6a + 3h
8. Perform the indicated operation
based on the given functions:
√
f (x) = 2x2 − 3 g(x) = x − 8 h(x) = 2+x
x
1
1
(a) (f + h)(−3) = f (−3) + h(−3) = 2(−3)2 − 3 + 2−3
−3 = 15 + 3 = 15 3
√
√
(b) (f ◦ g)(x) = f (g(x)) = f ( x − 8) = 2( x − 8)2 − 3 = 2x − 19 here x ≥ 8
(c)
f (x)−f (−1)
x+2
=
2x2 −3−(−1)
x+2
=
2x2 −2
x+2
=
2(x+1)(x−1)
x+2
9. Find the average
rate of change of f (x) =
√
√
f (8)−f (1)
16− 9
=
= 17
8−1
7
√
x + 8 from x = 1 to x = 8.
10. Given a base function, be able to graph a function using knowledge of shifts and reflections.
11. The volume V of a square based pyramid with base sides s and height h is V = 13 s2 h.
If the height is half the length of a base side, express the volume V as a function of h.
We want V in terms of h only. need to replace s in terms of h, where h = 21 s → s = 2h
V = 13 (2h)2 h = 43 h3
2
12. Given a rectangular box with a square base of side length x where the height of the
box, h, is twice the length of one side of the square base. Express the surface area of
the box in terms of x.
Surface area of a rectangular box with a square base of side length x and height of the
box, h is:
S = 2x2 + 4xh
We want this in terms of x only, need to replace h in terms of x using h = 2x
S(x) = 2x2 + 4x(2x) = 10x2
13. In the picture below find the area of the rectangle inside the circle.
From the picture one side of rectangle is x = 6. By using the equation of the circle the
height of the rectangle y can be found:
62 + y 2 = 100 → y 2 = 64 → y = 8
The area of the rectangle in question is (6)(8) = 48 square units.
14. A stone is thrown upward from the top of a building. Its height (in feet) above the
ground after t seconds is given by h(t) = −16t2 + 48t + 32. What maximum height does
it attain and at what time does this happen?
b
48
= − 2(−16)
= 48
Maximum time at t = − 2a
32 = 1.5 → 1.5 seconds.
2
Maximum height is: h(1.5) = −16(1.5) + 48(1.5) + 32 = 68 → 68 feet.
15. Find the quadratic function that has vertex (4, 2) and passes through the point (0, −2)
Use the standard quadratic function: f (x) = a(x − h)2 + k where (h, k) = (4, 2) is the
vertex.
f (x) = a(x − 4)2 + 2
Use the second point (0, −2) → f (0) = −2 to find a.
f (0) = a(−4)2 + 2 = −2
16a = −4
a = − 41 → therefore, f (x) = − 41 (x − 4)2 + 2
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16. Use the following information to sketch a possible graph of a polynomial function P (x).
P (x) is a 4th degree polynomial
Give the table of values:
x
-4
-3
-2
0
1
P (x)
0
2
0
3
0
A sign chart for P (x) is given as:
17. Sketch the graph of the polynomial function: f (x) = (x + 2)(x2 − 2x − 8)
factoring completely: f (x) = (x + 2)(x − 4)(x + 2) = (x + 2)2 (x − 4)
y-intercept: f (0) = (2)(−8) = −16 → (0, −16)
x = −2 twice appears quadratic
xintercept(s):
x = 4 once appears linear
A sign chart for f (x) is given as:
2
18. Find the y-intercept of f (x) = x −3x+7
6x
Set x = 0 this is undefined due to the zero in the denominator, thus there is NO
y-intercept.
19. For the following rational function; R(x) =
R(x) =
14−2x
x2 −x−20
=
2(7−x)
(x−5)(x+4)
(a) y-intercept: R(0) =
14
−20
14−2x
x2 −x−20 ,
find the following:
7
= − 10
(b) x-intercept(s): Set Numerator = 0. So, x = 7
(c) Vertical Asymptote(s): Set Denominator = 0. So, x = 5 and x = −4
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(d) Horizontal Asymptote: y = 0 since the degree(top) < degree(bottom)
20. What is the horizontal asymptote of f (x) =
x2 −4x3
2x3 +8
Since the degree(top)=degree(bottom) then the horizonal asymptote is the ratio of the
leading coefficients.
y = −2 is the horizontal asymptote.
21. Solve the non-linear inequality:
(x−1)(3−x)
(x−2)2
≤0
The zeroes and undefined values are needed to help form the intervals for the sign chart.
x = 1 (once)
zeroes:
x = 3 (once)
undefined: quad x = 2 (twice)
sign chart:
Want intervals that are less than or equal to 0: x ≤ −1 ∪ x ≥ 3
22. Find the inverse function of the following:
(a) f (x) = 3 + 4 ln(x − 1)
y = 3 + 4 ln(x − 1)
x = 3 + 4 ln(y − 1)
x − 3 = 4 ln(y − 1)
x−3
4 = ln(y − 1)
x−3
e 4 =y−1
x−3
x−3
e 4 − 1 = y → f −1 (x) = e 4 − 1
√
(b) f (x) =
8 3 x2 + 2
√
3
y = 8 px2 + 2
x = 8p3 y 2 + 2
3
x
y2 + 2
8 =
x 3
( 8 ) = y2 + 2
( x8 )3 − 2 = y 2
p x
p
( 8 )3 − 2 = y → f −1 (x) = ( x8 )3 − 2
23. Solve the following equations:
(a) log2 (1 − x) = 4 → 42 = 1 − x → x = −15
(b) 32x+1 = 52 → log3 32x+1 = log3 (52) → 2x + 1 = log3 (52) → 2x = log3 (52) − 1
log (52)−1
x= 3 2
(c) 82x = 42x−1 Write these with common base 2
(23 )2x = (22 )2x−1
26x = 24x−2 set exponents equal 6x = 4x − 2 → 2x = −2 → x = −1
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24. State the domain of the equation: log4 (x + 3) + log4 (x − 3) = 2, and solve the equation
for all real valued solutions.
For solutions to exist, we must have x > 3
log4 (x + 3)(x − 3) = 2
42 = (x + 3)(x − 3)
16 = x2 − 9
25 = x2
x = ±5
Only real valued solution is x = 5
25. A lab culture initially contains 3000 bacteria. After 3 hours the bacterial count is 4500.
How many bacteria are there after 7 hours? How long will it take for the bacterial count
to reach 20000?
Let N (t) = N0 ert represent the number of bacteria after t hours.
N0 = 3000 initial amount at t = 0
Model becomes: N (t) = 3000ert
Use N (3) = 4500 and N (3) = 3000e3r to find r.
≈ 0.1352
4500 = 3000e3r → 1.5 = e3r → ln(1.5) = 3r → r = ln(1.5)
3
Model for the amount of bacteria for any time t in hours becomes: N (t) = 3000e0.1352t
After 7 hours, there are N (7) = 3000e0.1352(7) ≈ 7729 bacteria
To find the time when 20000 bacteria are present, set N (t) = 20000 and solve for t.
ln( 20
0.1352t
3 )
→ ln( 20
20000 = 3000e0.1352t → 20
3 = e
3 ) = 0.1352t → t = 0.1352 ≈ 14 hours
26. In 1992, the population of bluetopia was estimated at 5 million. For any subsequent
year, the population, P (t) (in millions), can be modeled using the function,
250
P (t) = 5+44.99e
−0.0208t where t is the number of years since 1992. Determine the year
when the population will be 39 million.
Set 39 = P (t) and solve for t.
250
39 = 5+44.99e
−0.0208t
39[5 + 44.99e−0.0208t] = 250
5 + 44.99e−0.0208t = 250
39
44.99e−0.0208t ≈ 6.41 − 5
1.41
e−0.0208t ≈ 44.99
− 0.0208t ≈ ln(0.03135)
t ≈ ln(0.03135)
−0.0208 ≈ 166.5 years.
27. A thermometer is taken from a room at 71◦ F to the outdoors where the temperature
is 14◦ F. Determine what the reading on the thermometer will be after 5 minutes, if
the reading drops to 45◦ after 1 minute. Assume the cooling follows Newton’s Law of
Cooling: T (t) = Ts + (T0 − Ts )ekt where Ts and T0 represent the temperature of the
surrounding and initial temperature of the object respectively. (Round your answer to
two decimal places).
In this case we have Ts = 14 (temperature outside) and T0 = 71 (initial temperature).
Model becomes: T (t) = 14 + 57ekt which represents the temperature of the object after
t minutes
Use T (1) = 45 and T (1) = 14 + 57ek to find k.
31
k
45 = 14 + 57ek → 31 = 57ek → 31
57 = e → k = ln( 57 ) ≈ −0.6091
New model for the temperature at any time t in minutes: T (t) = 15 + 57e−0.6091t
After 5 minutes, the temperature is T (5) = 14 + 57e−0.6091(5) ≈ 16.71◦F.
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28. A tour group is split into two groups when waiting in line for food at the zoo snack
shack. The first group bought 7 slices of pizza and 5 sodas for $28.50. The second group
bought 6 slices of pizza and 4 sodas for $23.98. How much does one slice of pizza cost?
Let P represent the cost of 1 slice of pizza.
Let S represent the cost of 1 soda.
Total cost for group one: 7P + 5S = 28.50
Total cost for group two: 6P + 4S = 23.98
Multiply to top equation by (-4) and bottom equation by 5, this forms opposite coefficients on the S variable ,then add these equations
−28P − 20S = −114
30P + 20S = 119.9
2P = 5.9
P = 2.95
The cost of one slice is : $2.95
3x2 − y 2 = 11
x2 + 4y 2 = 8
Multiply the top equation by 4, this gives opposite coefficients on the y-variable.
29. Find all solution of the following system of equations.
12x2 − 4y 2 = 44
x2 + 4y 2 = 8
13x2 = 52
x2 = 4 → x = ±2
Still need to solve for y by substituting the value for x2 into either equation.
x2 + y 2 = 8
4 + 4y 2 = 8
4y 2 = 4
y 2 = 1 → y = ±1
The solutions are : (±2, ±1)
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