1324 Math Exam 2

f(x) = x − 2 2x − 1

f −1 (x) = x − 2 2x − 1

f(x) = 1 − 4 + 3x 5

f −1 (x) = − 5 /3 x + 1/ 3

. f(x) = 3 + 7x 5 − 2x

f(x) = 3 + 7x 5 − 2x Domain: (−∞, 5 2 ) ∪ ( 5 2 , ∞) Vertical asymptote: x = 5 2 As x → 5 2 − , f(x) → ∞ As x → 5 2 + , f(x) → −∞ No holes in the graph Horizontal asymptote: y = − 7 2 As x → −∞, f(x) → −7 2 + As x → ∞, f(x) → −7 2 −

f(x) = 3x 2 + 2x + 10

f(x) = 3x 2 + 2x + 10 = 3 x − − 1 3 + √ 29 3 i x − − 1 3 − √ 29 3 i Zeros: x = − 1 3 ± √ 29

f(x) = x 2 − 2x + 5

f(x) = x 2 − 2x + 5 = (x − (1 + 2i))(x − (1 − 2i)) Zeros: x = 1

6. f(x) = x 3 + 1 x 2 − 1

f(x) = x 3 + 1 x 2 − 1 = x 2 − x + 1 x − 1 Domain: (−∞, −1) ∪ (−1, 1) ∪ (1, ∞) Vertical asymptote: x = 1 As x → 1 −, f(x) → −∞ As x → 1 +, f(x) → ∞ Hole at (−1, − 3 2 ) Slant asymptote: y = x As x → −∞, the graph is below y = x As x → ∞, the graph is above y = x

11. f(x) = x 3 + 2x 2 + x x 2 − x − 2

f(x) = x 3 + 2x 2 + x x 2 − x − 2 = x(x + 1) x − 2 Domain: (−∞, −1) ∪ (−1, 2) ∪ (2, ∞) Vertical asymptote: x = 2 As x → 2 −, f(x) → −∞ As x → 2 +, f(x) → ∞ Hole at (−1, 0) Slant asymptote: y = x + 3 As x → −∞, the graph is below y = x + 3 As x → ∞, the graph is above y = x +

p(x) = x^4, c= 3/2

p (3/2) = 73/16

p(x) + 4x^2- 33x-180, C=12

p(12)= 0, p(x) = (x-12) (4x+15)

x + √ 3x + 10 = −2

x = −3

√ x − 2 + √ x − 5 = 3

x=6

f(x) = 4x 4 − 4x 3 + 13x 2 − 12x + 3

√ 11 6 i 37. f(x) = 4x 4 − 4x 3 + 13x 2 − 12x + 3 = x − 1 2 2 4x 2 + 12 = 4 x − 1 2 2 (x + i √ 3)(x − i √ 3)

f(x) = x 2 , g(x) = 2x + 1

(g ◦ f)(0) = 1 (f ◦ g)(−1) = 1 (f ◦ f)(2) = 16 (g ◦ f)(−3) = 19 (f ◦ g) 1 2 = 4 (f ◦ f)(−2) = 1

x x^-1 > 0

(-1,0) U (1,oo)

1 x+2 > 0

(-2,oo)

x^2 + 5x +6 x^2- 1 > 0

(-oo, -3) U (-2,-1) U (1,oo)

(2x^3-x+1) / (x^2 + x +1)

(2x-2) + (-x+3)

(9x^3 + 5) / (2x-3)

(2x-3) (9/2x^2 + 27/4x + 81/8) + 283/8

f(x) = √ 3 − x, g(x) = x^2 + 1

(g ◦ f)(0) = 4 (f ◦ g)(−1) = 1 (f ◦ f)(2) = √ 2 (g ◦ f)(−3) = 7 (f ◦ g) 1 2 = √ 7 2 (f ◦ f)(−2) = p 3 −

f(x) = |x|, g(x) = √ 4 − x

(g ◦ f)(x) = p 4 − |x|, domain: [−4, 4] (f ◦ g)(x) = | √ 4 − x| = √ 4 − x, domain: (−∞, 4] (f ◦ f)(x) = ||x|| = |x|, domain: (−∞, ∞)

1 1 x^2-3 x + 3 + x − 3 = x 2 − 9

-1

x 5x + 4 = 3

-6/7

2x − 1 /3 (x − 3) 1 /3 + x 2/3 (x − 3)− 2/3 ≥ 0

. (−∞, 0) ∪ [2, 3) ∪ (3, ∞)

f(x) = x 2x + 1 2x + 1 , g(x) = x

. For f(x) = 3x x−1 and g(x) = x x−3 (g ◦ f)(x) = x, domain: (−∞, 1) ∪ (1, ∞) (f ◦ g)(x) = x, domain: (−∞, 3) ∪ (3, ∞) (f ◦ f)(x) = 9x 2x+1 , domain: −∞, − 1 2 ∪ − 1 2 , 1 ∪ (12222

10 − √ x − 2 ≤ 11

. [2, ∞)

f(x) = 3√ x − 1 − 4

. f −1 (x) = 1 9 (x + 4)2 + 1, x ≥ −4

f(x) = 3 4 − x

. f −1 (x) = 4x − 3 x

14. f(x) = −x 3 + 4x x 2 − 9

. f(x) = −x 3 + 4x x 2 − 9 = −x 3 + 4x (x − 3)(x + 3) Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) Vertical asymptotes: x = −3, x = 3 As x → −3 −, f(x) → ∞ As x → −3 +, f(x) → −∞ As x → 3 −, f(x) → ∞ As x → 3 +, f(x) → −∞ No holes in the graph Slant asymptote: y = −x As x → −∞, the graph is above y = −x As x → ∞, the graph is below y = −

. 5 − (4 − 2x) 2/ 3 = 1

. x = −2, 6

f(x) = 4 x + 2

1

. Sqroot over (−25)(−4)

10

f(x) = 3x 2 − 5x − 2 x^2 − 9

10

Simplify. Sqroot over (−9)(−16)

12

. f(x) = x^3 + 2x^2 + x x^2 − x − 2

13

f(x) = x x^2 + x − 12

6

f(x) = x 3 − 4x 2 − 4x − 5 x 2 + x + 1

8. f(x) = x 3 − 4x 2 − 4x − 5 x 2 + x + 1 = x − 5 Domain: (−∞, ∞) No vertical asymptotes No holes in the graph Slant asymptote: y = x − 5 f(x) = x − 5 everywhere

x^4 -9x^2 - 4x +12

All of the real zeros lie in the interval [−13, 13] Possible rational zeros are ±1, ±2, ±3, ±4, ±6, ±12 There are 2 or 0 positive real zeros; there are 2 or 0 negative real zeros

3x^3+ 3x^2-11x-10

All of the real zeros lie in the interval − 14 3 , 14 3 Possible rational zeros are ± 1 3 , ± 2 3 , ± 5 3 , ± 10 3 , ±1, ±2, ±5, ±10 There is 1 positive real zero; there are 2 or 0 negative real zeros

-2^3+19x^2-49x+20

All of the real zeros lie in the interval − 51 2 , 51 2 Possible rational zeros are ± 1 2 , ±1, ±2, ± 5 2 , ±4, ±5, ±10, ±20 There are 3 or 1 positive real zeros; there are no negative real zeros

f(x)= x^4 +2x^3 -12 X^2-40x-32

All real zeros lie in interval [-41,41] Possible rational zeros are +1.+2+4+8+16+32 There is 1 positive Zero, there are 3 or 1 negative zeros.

f(x) = 5x √3 x 3 + 8

Domain: (−∞, −2) ∪ (−2, ∞) (+) −2 (−) 0 0 (+) Vertical asymptote x = −2 Horizontal asymptote y = 5 No unusual steepness or cuspsl

f(x) = x 3 /2 (x − 7) 1/ 3

Domain: [0, ∞) 0 0 (−) 7 0 (+) No asymptotes Unusual steepness at x = 7 No cusps

f(x) = x √ 1 − x ^2

Domain: [−1, 1] −1 0 (−) 0 0 (+) 1 0 No asymptotes Unusual steepness at x = −1 and x = 1 No cusps

f(x) = 3x − 5, g(x) = √x

For f(x) = 3x − 5 and g(x) = √ x (g ◦ f)(x) = √ 3x − 5, domain: 5 3 , ∞ (f ◦ g)(x) = 3√ x − 5, domain: [0, ∞) (f ◦ f)(x) = 9x − 20, domain: (−∞, ∞)

f(x) = 6 − x − x^2 , g(x) = x √ x + 10

For f(x) = 6 − x − x 2 and g(x) = x √ x + 10, (g ◦ f)(0) = 24 (f ◦ g)(−1) = 0 (f ◦ f)(2) = 6 (g ◦ f)(−3) = 0 (f ◦ g) 1 2 = 27−2 √ 42 8 (f ◦ f)(−2) = −14

f(x) = √ 2x + 5, g(x) = 10x x 2 + 1

For f(x) = 6 − x − x 2 and g(x) = x √ x + 10, (g ◦ f)(0) = 24 (f ◦ g)(−1) = 0 (f ◦ f)(2) = 6 (g ◦ f)(−3) = 0 (f ◦ g) 1 2 = 27−2 √ 42 8 (f ◦ f)(−2) = −14

z = 1 − i √ 3, w = −1 − i √ 3

For z = 1 − i √ 3 and w = −1 − i √ 3 z + w = −2i √ 3 zw = −4 z 2 = −2 − 2i √ 3 1 z = 1 4 + √ 3 4 i z w = 1 2 + √ 3 2 i w z = 1 2 − √ 3 2 i z = 1 + i √ 3 zz = 4 (z) 2 = −2 + 2i √

z = i, w = −1 + 2i

For z = i and w = −1 + 2i z + w = −1 + 3i zw = −2 − i z 2 = −1 1 z = −i z w = 2 5 − 1 5 i w z = 2 + i z = −i zz = 1 (z) 2 = −1