Download Solutions for Review Problems for the Calculus I/Precalculus

Solutions for Review Problems
for the Calculus I/Precalculus Placement Test - Fall, 2000
I. Basic Algebra Skills - 4. Polynomials
Polynomial Equations:
1. Determine if x = ?2 is a solution of the equation Ýx ? 1ÞÝx 2 + 4ÞÝ2x ? 1ÞÝx + 3Þ = 0.
Since x + 2 is not a factor of the left-hand side of the equation, x = ?2 is not a solution.
2. Solve the equation Ý3x + 4ÞÝx 2 ? x ? 1ÞÝ2x 2 + 3x ? 1Þ = 0.
Ý3x + 4ÞÝx 2 ? x ? 1ÞÝ2x 2 + 3x ? 1Þ = 0 if and only if 3x + 4 = 0, x 2 ? x ? 1 = 0, or 2x 2 + 3x ? 1 = 0.
3x + 4 = 0, x = ? 43 ;
x 2 ? x ? 1 = 0, x =
2x 2 + 3x ? 1 = 0, x
Solutions: x = ? 43 ,
?Ý?1Þ± Ý?1Þ 2 ?4Ý1ÞÝ?1Þ
= 1±2 5
2
3± 3 2 ?4Ý2ÞÝ?1Þ
=
= 3± 417 .
2Ý2Þ
1+ 5 1? 5
, 3+ 417 , 3? 417 .
2
2
;
3. Solve each equation by factoring.
(a) 81x 6 ? 64 = 0
81x 6 ? 64 = ÝÝ9x 3 Þ 2 ? 8 2 Þ = Ý9x 3 ? 8ÞÝ9x 3 + 8Þ = 0
9x 3 ? 8 = 0, x 3 = 89 , x = 3 89 = 32 ;
9
9x 3 + 8 = 0, x 3 = ? 89 , x =
Solutions: x =
2
39
3
? 89 = ? 32 .
, x = ? 32 .
9
9
(b) x 4 + 2x 3 ? 8x ? 16 = 0
x 4 + 2x 3 ? 8x ? 16 = x 3 Ýx + 2Þ ? 8Ýx + 2Þ = Ýx + 2ÞÝx 3 ? 8Þ
x + 2 = 0, x = ?2; x 3 ? 8 = 0, x = 3 8 = 2.
Solutions: x = ?2, x = 2.
(c) x 4 + 5x 2 ? 36 = 0
x 4 + 5x 2 ? 36 = Ýx 2 + 9ÞÝx 2 ? 4Þ = Ýx 2 + 9ÞÝx ? 2ÞÝx + 2Þ
x 2 + 9 ® 0, no solution from this equation.
x ? 2 = 0, x = 2; x + 2 = 0, x = ?2.
Solutions: x = 2, x = ?2.
Polynomial Functions and Graphs:
1. Based on the graph of y = x 3 , sketch the graph of the function gÝxÞ = Ýx + 1Þ 3 ? 2 without using
calculator.
200
150
50
-4
0
-2
2 x
4
-50
-100
– y=x 3 , -. y=(x+1) 3 ? 2
2. Based on the graph of y = x 4 , sketch the graph of the function gÝxÞ = ?Ýx ? 2Þ 4 + 1 without using
calculator.
4
2
-2
-1
1
2
3
4
-2
-4
- y=x 4 , -. y=-(x-2) 4 ? 1
3. Determine the right-hand and left-hand behavior of the graph of the polynomial function without
sketching the graph.
(a) fÝxÞ = 2x 5 ? 5x + 7. 5
Since the coefficient of x 5 is 2 > 0, lim x¸+K fÝxÞ = +K and lim x¸?K fÝxÞ = ?K.
(b) gÝxÞ = ? 14 Ý3x 4 ? 2x + 5Þ
gÝxÞ = ? 34 x 4 + 12 x ?
lim x¸?K gÝxÞ = ?K.
5
4
. Since the coefficient of x 4 is ? 34 < 0, lim x¸+K gÝxÞ = ?K and
4. The graph of a polynomial PÝxÞ = ax 3 + bx 2 + cx + d is given below. (1) Determine if the constant a
is positive or negative; and (2) find all zeros of PÝxÞ.
4
2
-3
-2
-1
0
1
x
2
3
-2
-4
y=P(x)
(1) a < 0.
(2) zeros: x = ?2, x = ?1, and x = 1.
5. Find all zeros of the polynomial function.
(a) fÝxÞ = x 5 + x 3 ? 6x
x 5 + x 3 ? 6x = xÝx 4 + x 2 ? 6Þ = xÝx 2 + 3ÞÝx 2 ? 2Þ = 0 if and only if x = 0, x 2 + 3 = 0 or x 2 ? 2 = 0.
zeros: x = 0, x = ± 2 .
(b) gÝxÞ = x 3 ? 4x 2 ? 2x + 8
x 3 ? 4x 2 ? 2x + 8 = x 2 Ýx ? 4Þ ? 2Ýx ? 4Þ = Ýx ? 4ÞÝx 2 ? 2Þ = 0 if and only if x ? 4 = 0, or x 2 ? 2 = 0.
zeros: x = 4, and x = ± 2 .
6. Find a polynomial function that has zeros: ?2, ? 1, 0, 3, 4.
PÝxÞ = Ýx + 2ÞÝx + 1ÞxÝx ? 3ÞÝx ? 4Þ.
7. An open box is to be made from a square piece of material, 36 centimeters on a side, by cutting
equal squares with a size x from the corners and turning up the sides.
(a) Write the volume VÝxÞ of the box in x.
The sizes of the box is: l = w = 36 ? 2x and h = x. So, VÝxÞ = xÝ36 ? 2xÞ 2 = 4xÝ18 ? xÞ 2 .
(b) Find the domain of VÝxÞ.
The domain of VÝxÞ: 0 < x < 18 (18 ? x > 0)
(c) Evaluate the volume of the box when x = 4 centimeters.
VÝ4Þ = 16Ý14Þ 2 = 3136 cm 3 .
(d) Estimate x at which the volume V is maximized.
3500
3000
2500
2000
1500
1000
500
0
2
4
6
8
10
x
12
14
16
18
y=V(x)
From the graph of y = VÝxÞ, x u 6 cm.
20