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MATH 105 Practice Final
1) The function A described by A(r) = r^2
√ 3 4 gives the area of an equilateral triangle with side r. Find the area when the side is 6 in.
Find A( 6 ).
A( 6 ) = ( 6 )^2
2) Write an equation of the line containing the point (3, − 2 ) and
(a) Parallel to the line 3x + 4 y = 5 (b) Perpendicular to the line 3x + 4 y = 5
First find the slope of 3x + 4 y = 5 by putting it into slope-intercept form.
3 x + 4 y = 5 4 y = − 3 x + 5
y = −
x +
So the slope m = − 34.
(a) Parallel lines have the same slope so use slope-point form to get the equation.
y − (− 2 ) = −
(x − 3 )
y = −
x +
y = −
x +
(b) Perpendicular lines have slopes which are negative reciprocals of one another. So the slope of a perpendicular line is 43.
y − (− 2 ) =
(x − 3 )
y =
x − 4 − 2
y =
x − 6
3) For f (x) = (^) x^2 +x 1 , g(x) = 2 x + 5, determine the domain of (^) gf.
f (x) is undefined when x + 1 = 0 so x 6 = −1. g(x) is defined for all real numbers. (^) gf is
undefined when either f or g is undefined or when g(x) = 2 x + 5 = 0 so x 6 = − 52 and x 6 = −1. In set-builder notation:
{x|x is real and x 6 = − 52 and x 6 = − 1 }
In interval notation: (−∞, − 52 ) ∪ (− 52 , − 1 ) ∪ (−1, ∞)
4) Solve 2 x 3
3 y 4
x 3
7 y 18
Multiply the second equation by −2, add and then solve.
2 x 3 +^
3 y 4 =^
11 12 − 23 x − 79 y = − 1 0 + 34 y − 79 y = − 121
27 y − 28 y = − 3 y = 3
7) Solve | − 5 x − 3 | ≥ 10
Consider the two cases where − 5 x − 3 ≥ 0 and where − 5 x − 3 < 0. In the first case | − 5 x − 3 | = − 5 x − 3. In the second | − 5 x − 3 | = 5 x + 3 giving the two inequalities below.
− 5 x − 3 ≥ 10 5 x + 3 ≥ 10 − 5 x ≥ 13 5 x ≥ 7 x ≤ −
x ≥
So the solution in set-builder notation is {x|x ≤ − 135 or x ≥ 75 }. In interval notation: (−∞, − 135 ] ∪ [ 75 , ∞).
8) Multiply (a + b + 1 )(a + b − 1 )
[(a + b) + 1 ][(a + b) − 1 ] (a + b)^2 − 1 a^2 + 2 ab + b^2 − 1
9) Factor completely: (x + 3 )^2 − 2 (x + 3 ) − 35
First set u = x + 3 and substitute.
u^2 − 2 u − 35 (u + 5 )(u − 7 )
Substitute x + 3 back in for u
[(x + 3 ) + 5 ][(x + 3 ) − 7 ] (x + 8 )(x − 4 )
10) Factor completely: p^6 − q^6
Solve using the identities a^2 − b^2 = (a + b)(a − b), a^3 − b^3 = (a − b)(a^2 + ab + b^2 ), and a^3 + b^3 = (a + b)(a^2 − ab + b^2 ).
p^6 − q^6 (p^3 )^2 − (q^3 )^2 (p^3 + q^3 )(p^3 − q^3 ) (p + q)(p^2 − pq + q^2 )(p − q)(p^2 + pq + q^2 )
11) Solve (x + 2 )(x − 5 ) = 8
(x + 2 )(x − 5 ) = 8 x^2 + 7 x − 10 = 8 x^2 + 7 x − 18 = 0 (x − 6 )(x + 3 ) = 0
x = 6 or x = − 3
12) Divide and simplify: a^3 + 4 a a^2 − 16
÷
a^2 + 8 a + 15 a^2 + a − 20
Flip and multiply, then simplify.
a^3 + 4 a a^2 − 16
a^2 + a − 20 a^2 + 8 a + 15 a(a^2 + 4 )(a − 4 )(a + 5 ) (a − 4 )(a + 4 )(a + 3 )(a + 5 ) a(a^2 + 4 ) (a + 3 )(a + 4 )
14) Simplify
x y
y x x^2 y − y
x y
y x x^2 y −^ y^
xy xy x^2 y y
xy^2 x x^3 y y − xy^2
x^2 − y^2 x^3 − xy^2 x^2 − y^2 x(x^2 − y^2 ) 1 x
15) Solve y + 5 y + 1
y y + 2
4 y + 15 y^2 + 3 y + 2
y + 5 y + 1
y y + 2
4 y + 15 (y + 2 )(y + 1 )
(y + 2 )(y + 1 )
y + 5 y + 1
y y + 2
4 y + 15 (y + 2 )(y + 1 ) (y + 2 )(y + 1 )
(y + 2 )(y + 5 ) − y(y + 1 ) = 4 y + 15 y^2 + 7 y + 10 − y^2 − y = 4 y + 15 2 y = 5 y = (^52)
16) Russ drives 72 miles, has a flat tire and no spare, and walks 4 miles to a gas station. His driving rate was 12 times faster than his walking rate. If Russ spent 2^12 hours driving and walking, at what rate did he walk?
Set his walking rate to w, then his driving rate is 12w. Since his time walking is thus (^) w^4 and his time driving is (^1272) w , this gives the following equation since the sum of the two times is 2.5. 4 w
12 w
w
w
2 w ·
w
w
· 2 w
8 + 12 = 5 w 5 w = 20 w = 4
So his walking rate is 4 mph.
17) Use Polynomial division to find the indicated function value (synthetic division is preferred). f (x) = 6 x^4 + 15 x^3 + 28 x + 6; find f (− 3 ).
The Remainder Theorem tells us that f (− 3 ) is the remainder when dividing f (x) by x + 3. Using synthetic division. − 3 6 15 0 28 6 − 18 9 − 27 − 3 6 − 3 9 1 | 3
So the remainder is 3 and thus f (− 3 ) = 3.
18) Solve for q:
p
q
f
pq f
p
q
f
· pq f
q f + p f = pq p f = pq − q f p f = (p − f )q p f p − f = q
q = p f p − f
Checking: √ 20 −
True
So x = 809 is a solution.
22) Divide and Simplify. Write in the form a + bi: 5 + 3 i 7 − 4 i
5 + 3 i 7 − 4 i
7 + 4 i 7 + 4 i 35 + 20 i + 21 i − 12 49 + 16 23 + 41 i 65 23 65
i
23) Completely solve x^2 + 9 = 4 x
x^2 − 4 x = − 9 x^2 − 4 x + 4 = − 9 + 4 (x − 2 )^2 = − 5 x − 2 = ±i
x = 2 ± i
24) Two hoses are connected to a swimming pool. Working together, they can fill the pool in 4 hr. The larger hose, working alone, can fill the pool in 6 hr less time than the smaller one. How long would the smaller take, working alone, to fill the pool?
Set the time the smaller hose would take to be s, then the time the larger hose takes is s − 6.
1 s − 6
s
4 s + 4 (s − 6 ) = s(s − 6 ) 4 s + 4 s − 24 = s^2 − 6 s 0 = s^2 − 14 s + 24 0 = (s − 2 )(s − 12 )
s = 2 or s = 12
But if s = 2 then the time the larger hose takes would be -4, so the only solution is s = 12. Thus the smaller hose takes 12 hrs to fill the pool.
25) Solve y
1 (^3) − y
1 (^6) − 6 = 0
Substitute u = y
1 (^6) into the equation and solve.
u^2 − u − 6 = 0 (u − 3 )(u + 2 ) = 0
u = 3 or u = − 2
Substitute y
1 (^6) = u back in.
y
1 (^6) = 3 or y
1 (^6) = − 2 y = 36 or y=(− 2 )^6 y = 729 or y = 64
Checking y = 729
729
1 (^3) − 729
1 (^6) − 6 = 0 9 − 3 − 6 = 0 0 = 0 True
This gives us everything we need to plot the graph.
27) Find a formula for the inverse of f (x) = (x − 2 )^3
Set f (x) = y. y = (x − 2 )^3
Swap x and y. x = (y − 2 )^3
Solve for y. √ (^3) x = y − 2
y = 3
x + 2
Set y = f −^1 (x). f −^1 (x) = 3
x + 2
28) Solve for x: loga 125 = 3
a^3 = 125 a = 5
29) If loga x = 2, loga y = 3, and loga z − 4, what is loga
√ (^3) x (^2) z √ (^3) y (^2) z− 2
Switch to rational exponent notation from radical notation.
loga x (^23) z (^13)
y (^23) z−^ (^23)
loga x (^23) z y (^23)
loga x
(^23)
(^23) 2 3 loga x^ +^ loga z^ −^
2 3 loga y 2 3 ·^2 +^4 −^
2 3 ·^3 4 3 +^4 −^2 10 3
