Math 104 Pre-Calculus
Sample Final
Solve for x:
1. −3|𝑥 + 2| = 8 |y| can’t be <0, No solution
2.
√𝑥 + 7 + √𝑥 + 12 = 5
√𝑥 + 7 = −√𝑥 + 12 + 5; square both sides x+7 = x+12 +25 -10√𝑥 + 12
7-12-25 = -30 = -10√𝑥 + 12; square both sides √𝑥 + 12 = 3, x+12 = 9, x = -3 ; check
solution for extraneous root, is OK
3. −|2𝑥 − 1| + 4 > 3
-|2x-1|> -1 or |2x – 1| < 1. Gives two inequalities: 2x-1 < 1 and 2x – 1 > -1
x < 1 and x >0; 0 <x < 1
4.
2
3
≤
5−3𝑥
−2
≤
3
4
Left side: -4 ≥15-9x, x≥19/9
Right side: -10 + 6x ≤ 3, x ≤13/6
19/9 ≤x ≤ 13/6; Make sure there is an overlap: 19/9 is 1/9 larger than 2, and 13/6 is 1/6
larger than 2. Note that 1/9 is less than 1/6, so is a value between the two fractions.
Consider the functions 𝑓(𝑥) =
𝑥+2
𝑥−2
and 𝑔(𝑥) = 𝑥 − 3.
5. Find (𝑓 ∘ 𝑔)(𝑥); give its domain
𝑥−3+2 𝑥−1
substitute x-3 for x in f(x): 𝑥−3−2 = 𝑥−5; domain x ≠5
6. Find f -1(x); give its range and domain
y = (x+2)/(x-2),; solve for x: (x-2)y=x+2, x(y-1)=2+2y, x = (2+2y)/(y-1);
rewrite f-1(x) = (2+2x)/(x-1) ; domain x≠ 1, range is the domain of the original: y≠ 2
7. The hypotenuse of a right triangle is 15 cm. If one of the legs of the triangle has length
𝑥, find a function that expresses the perimeter of the triangle as a function of x.
Pythagoras: c2 = a2 + b2; Let x be one side and y the other and solve for y.
15(15) = x(x) + y(y), y(y) = 225 – x(x), y = ±√225 − 𝑥 2 , Perimeter P, is the sum of the
lengths of the sides: P = 15 + x + √225 − 𝑥 2
8. Find the coordinates of the point on the graph of 𝑦 = √𝑥 + 3 that is closest to the
point (3, 0).
distance between the point and a point x, y on the graph, d: d2 = (x-3) 2 + (y-0)2; subst. for y:
(x-3) 2 +x+ 3 = x 2 – 6x + 9 + x + 3 = x 2 – 5x +12; Minimum at x = 5/2, the vertex
(5/2, √11/2)
9. Consider the function 𝑓(𝑥) = 5𝑥 2 − 3. Find and simplify the difference quotient given
below.
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
Just substitute in for f(x) and f(x+h): [5x2 + 10xh + 5h2 – 3 –( 5x2 – 3)]/h = 10x + 5h
10. Graph 𝑓(𝑥) = √𝑥 2 − 1 + 2 and state its domain and range.
What kind of graph is this? y-2 = √𝑥 2 − 1, or (y-2)2 – x2 =-1 or x2 - (y-2)2 = 1; it is a
portion of a hyperbola. However, x must be 1, because of the square root and y must
be 2, since the square root is positive. Thus we have just a portion of a hyperbola
11. What is the largest possible area for a rectangle with a perimeter of 80 cm?
Let L = length, W = width. Area, A = LW, Perimeter, P = 2L +2W = 80, which gives
L = 40-W; A = W(40-W) = W2 -40W; Maximum is at axis of the parabola W2-40W +400,
which is at W=20; If W = 20, L = 20 from the value of the perimeter, so the area is 400.
In problems 12 – 15, solve the given equation. Simplify your answers, leaving
them in logarithm terms only when necessary.
12. log 5 (𝑥 + 6) + log 𝑥 (𝑥) = 5
log 𝑥 (𝑥) = 1, so log 5 (𝑥 + 6) + 1 = 5, log 5 (𝑥 + 6) = 4; making both sides exponential,
with 5 as the base, gives x+6 = 54 = 625, x = 619
13. log 4 (𝑥) + log 4 (𝑥 + 3) = 1
combining the logs, since log a + log b = log ab, we have log4(x(x+3)) = 1, exponentiation
of the log gives us x(x+3) = 4; x2 +3x – 4=0, factoring, (x-1)(x+4) = 0;
x = 1, x = -4; Reject x = -4, gives log of a negative number.
14. 42𝑥+1 = 5𝑥
Take logs (any base) of both sides: log 42𝑥+1 = log5𝑥 ; (2x+1)log4 = x log 5. Rearranging:
2x log 4 – x log 5 = -log 4; x = -log4 / (2log4 – log5)
15. 4𝑥−2 = 16√2
4𝑥−2 = 24 21/2 = 29/2 since 16 = 24 and √2 =21/2. Additionally, 4(x – 2) = 2(2x – 4)Taking logs
gives 2x-4 = 9/2, 2x= 17/2 or x = 17/4
16. Use Cramer’s rule to solve the following system of equations: 2x + 3y = 5, 4x – 3y = -1
2 3
5
3
2 5
|
|=-6 - 12 = -18 = D, |
|= -15 + 3 = -12 = Dx, |
|= -2 – 20 = -22 = Dy
4 −3
−1 −3
4 −1
x = 2/3, y = 11/9
2 −1 −1
1 1
17. Given that the inverse of the matrix ( 1
0 −1) is (0 2
−2 1
2
1 0
1
1), solve the following
1
system of equations: 2x – y – z = 4, x - z = 8, -2x + y + 2z = 9
1 1
(0 2
1 0
1 4
1) (8) = 4+8+9, 0+16+9, 4+0+9 or x = 21, y = 25, z = 13
1 9
18. Solve the following system of nonlinear equations.
y = log2 (x+1)
y = 5 – log2 (x-3)
equating y’s, log2 (x+1) = 5 - log2 (x-3); log a + log b = log ab, so we have
log2 (x+1)(x-3) = 5; exponentiation both sides gives (x+1)(x-3) = 25 = 32;
x2 - 2x – 35 = 0
x = -5, x = 7 ; however, x = -5 gives us the log of a negative number. So x = 7; calculate
y=log2 (7+1) = 3, so the solution is (7,3)
19.
Find the center and radius of the following circle.
5𝑥 2 + 5𝑦 2 + 10𝑥 + 10𝑦 + 25 = 20
5(𝑥 2 + 2x) + 5(𝑦 2 + 2y) = -5; 5(x+1)2 + 5(y+1) 2 = 15 + 10 = 5
(x+1) 2 + (y+1) 2 = 1; center is (-1, -1), radius 1
to check, this means that the point x = 0 y = -1 is on the circle
Plugging in, 0 + 5 - 10 + 25 = 20 which checks
20.
Find the coordinates of the foci of the following hyperbola.
9𝑥 2 − 16𝑦 2 − 72𝑥 − 32𝑦 = 16
9(x2 – 8x) – 16(y2 – 2x) = 16
9(x-4) 2 – 16(y+1) 2 = 16(1+9 - 1) = 9(16) = 144
equation is (x-4) 2 /16 - (y+1) 2 /9 = 1
center at (4, -1), c2 = 9 + 16 =25 or c = 5. Since this is an east-west hyperbola, the
axis is at y = 1 and the foci at (9, -1) and (-1, -1)
21.
Find the vertices of the following ellipse: 4𝑥 2 + 9𝑦 2 − 8𝑥 − 18𝑦 = 62.
4(x2-2) + 9(y2-2) = 68
4(x-1) 2 + 9(y-1) 2 = 68 + 4 +9 = 81
(x-1) 2/(9/4)+ (y-1) 2 /9 since 9 > 9/4, the axis is parallel to the y axis and has
length 6. Center is at (1,1), vertices are at (1, 1+3) = (1, 4) and (5/2, -2)
22.
Find all roots of the following polynomial function. State the multiplicity if it’s
greater than 1.
𝑓(𝑥) = 𝑥 4 + 𝑥 3 − 25𝑥 2 − 𝑥 + 24
Possible rational roots are 1, 2, 4, 6, 12, 24; 1 is a root
1| 1 1 -25 -1 24 x3 + 2x2 – 23x – 24=0; -1 is a root
1 2 -23 -24
1 2 -23 -24 0
-1| 1 2 -23 -24 x2 + x – 24=0; x = (-1 √97 )/2
-1 -1 24
1 1 -24
23.
For the function f(x) as given in problem 22, find the solutions to the inequality
y = f(x)<0
Roots were found in problem 22. Check in-between them. Smallest is (-1 -√97 )/2. A
number smaller than the root is -10, so try -6. 64 -63 -25(36) +12 + 24 = 36(36-6) +
36(-25 + 1)>0. -1 is next smallest. Try -2; 16 -8 – 100 + 2 + 24 <0; next try 0, 24>0.
Now need something between 1 and (-1 +√97 )/2, try x = 2; 16 + 8 – 100 + 2 + 24 <0
For x large and positive, f(x) is positive.
In summary, holds for (-1 -√97 )/2<x<-1, 1<x<(-1 +√97 )/2
24.
If a polynomial of degree 4 has roots 𝑥 = 1 − 𝑖 and 𝑥 = 2 + 3𝑖, how many real roots
does it have? None
In problems 25 – 26, evaluate the given series.
25.
∑20
𝑛=1(2𝑛 − 8) sum = 20/2(a1 + an) = 10(-6 + 32) = 260
26.
∑∞
𝑛=0 3 (− 5) sum = a/(1-r) = 3/(1+2/5) = 3/(7/5) = 15/7
27.
Prove the following statement using Mathematical Induction.
2 𝑛
5 + 9 + 13 … + (4𝑛 + 1) = 𝑛(2𝑛 + 3)
n = 1: 5 = 1(2+3) = 5
LHS: n+1: n(2n+3) + (4(n+1)+1) = 2n2 + 3n + 4n + 5 = n2 + 7n + 5
RHS: (n+1)(2(n+1)+3) = (n+1)(2n + 2 + 3) = (n+1)(2n+5) = n2 + 7n + 5
28. Find the inverse function f-1(x) for f(x) = log4(3x+1); give its domain and range.
y = log4 (3x+1); exponentiation gives 4y = 3x+1; x =(4y -1)/3, or f-1(x) = (4x -1)/3; domain is
all reals, the range is the range of the original function, x ≥ -1/3
29. Find the 9th term in the expansion of (3𝑎 + 𝑏)12 . Leave your result in exponential
and/or factorial form, if appropriate.
power on the first term, 3a, is 12-9+1=4, power on the second term, b = 9-1=8
(
12!
12∙11∙10∙9
12
)(3a)4b8 = 8!4!(3a)4b8 = 4∙3∙2 (3a)4b8
8