Math 110 Midterm 1 Study Guide October 14, 2013

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Augustus Lewis
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1 Name: For more practice exercises, do the study set problems in sections: , 4.1, and Find the domain of f, and express the solution in interval notation. (a) f(x) = x 6 D = (, ) or D = R (b) f(x) = 4 3x D = (, 4 3 ] (c) f(x) = 3x 3x 4 D = (, 4 3 ) ( 4 3, ) or D = R { 4 3 } (d) f(x) = x 2 + 2x 1 D = (, ) or D = R (e) f(x) = x 2 3x + 3 (f) f(x) = 16 x 2 D = [ 4, 4] x (g) f(x) = x 2 D = (2, ) D = (, ) or D = R 2. Sketch the graph of f. Find the domain D and the range R of f, and express the solution in interval notation. (a) f(x) = 3x + 3 D = (, ) or D = R R = (, ) or R = R (b) f(x) = 5x + 7 D = (, ) or D = R R = (, ) or R = R (c) f(x) = x 2 6x + 3 D = (, ) or D = R R = (, 12] (d) f(x) = 9 x 2 D = [ 3, 3] R = [ 3, 0] (e) f(x) = 9 x 2 D = [ 3, 3] R = [0, 3] 3. Explain how the graph of g is obtained from the graph of f. (a) f(x) = x 2 g(x) = x Shifted 4 units up (b) f(x) = x 2 g(x) = 3x Shifted 2 units up, stretched vertically by factor of 3 (c) f(x) = x 2 g(x) = (x 1) Reflected about x-axis, shifted 1 unit to the right, shifted 1 unit up (d) f(x) = x 2 g(x) = (x + 2) 2 3 Shifted 2 units to the left, shifted 3 units down (e) f(x) = x g(x) = x + 1 Shifted 1 unit to the left (f) f(x) = x g(x) = x + 5 Shifted 5 units up (g) f(x) = x 3 g(x) = x Shifted 8 units up 4. Express the quadratic function in standard form. (a) f(x) = 2x x 2 f(x) = (x 1) (b) f(x) = x 2 + 2x 1 f(x) = (x + 1) 2 2 (c) f(x) = x 2 3x + 3 f(x) = (x ) (d) f(x) = 3x 2 12x + 13 (e) f(x) = 1 x x 2

2 5. Find the maximum or mininum value of each function. (a) f(x) = x 2 + x (b) f(x) = 3x 2 + 4x 7 (c) f(x) = x 2 + 4x + 3 (d) f(x) = 3x 2 6x + 1 max: min: min: f( 2) = 1 min: 6. Use f(x) = 3x 5 and g(x) = 2 x 2 to evaluate the expression. (a) f(g(x)) = 3(2 x 2 ) 5 = 6 3x 2 5 = 3x f(g(0)) = = 1 (b) g(f(x)) = 2 (3x 5) 2 = 2 (3x 5)(3x 5) = 2 (9x 2 30x + 25) = 9x x 23 g(f(3)) = = 14 (c) f(f(x)) = 3(3x 5) 5 = 9x 20 f(f( 2)) = = Find the domain of (f g)(x) and (g f)(x). (a) f(x) = x 15 g(x) = x 2 + 2x f(g(x)) = x 2 + 2x 15 The domain of g is R since g is a quadratic funciton. Find the domain of f(g(x)) = x 2 + 2x 15 x 2 + 2x 15 0 Page 2

3 (x 3)(x + 5) 0 In order to find all the values of x such that (x 3)(x + 5) 0, we need to make a sign chart. The factors x 3 and x + 5 are zeros at 3 and 5, respectively. The intervals are (, 5], [ 5, 3], and [3, ). Then the solution of (x 3)(x + 5) 0 are the values of x for which the product of the factors is positive or equal to 0 that is where the resulting sign is positive. This corresponds to the intervals (, 5] [3, ). So the domain of f g is (, 5] [3, ). Answer: The domain of f g is (, 5] [3, ). f(x) = x 15 g(x) = x 2 + 2x g(f(x)) = ( x 15) x 15 = x x 15 The domain of f is [15, ). Find the domain of g(f(x)) = x x 15. The domain of g f is [15, ). Answer: The domain of g f is D = [15, ). (b) f(x) = x g(x) = x f(g(x)) = x + 1 Page 3

4 The domain of g is [0, ). Find the domain of f(g(x)) = x + 1. The domain of f g is R since f g is a linear function. Answer: The domain of f g is [0, ). f(x) = x g(x) = x g(f(x)) = x The domain of f is R since f is a quadratic function. Find the domain of g(f(x)) = x The domain of g f is [ 1, 1]. Answer: The domain of g f is [ 1, 1]. (c) f(x) = x 2 g(x) = 1 x 3 f(g(x)) = ( ) 2 1 x 3 = 1 x 6 The domain of g is R {0}. Page 4

5 Find the domain of f(g(x)) = 1 x 6 The domain of f g is R {0}. Answer: The domain of f g is R {0}. f(x) = x 2 g(x) = 1 x 3 g(f(x)) = 1 (x 2 ) 3 = 1 x 6 The domain of f is R since f is a quadratic function. Find the domain of g(f(x)) = 1 x 6 The domain of g f is R {0}. Answer: The domain of g f is R {0}. (d) f(x) = 1 x 1 g(x) = x 1 f(g(x)) = 1 x 1 1 = 1 x 2 The domain of g is R since g is a linear function. Page 5

6 Find the domain of f(g(x)) = 1 x 2. The domain of f g is (, 2) (2, ) or R {2}. Answer: The domain of f g is R {2}. f(x) = 1 x 1 g(x) = x 1 g(f(x)) = 1 x 1 1 = 1 x 1 x 1 x 1 = 1 x + 1 = x + 2 x 1 x 1 The domain of f is R {1}. Find the domain of g(f(x)) = x + 2 x 1. The domain of g f is D = R {1}. Answer: The domain of g f is R {1}. 8. Use the Intermediate Value Theorem to show that f has a zero between a and b. (a) f(x) = 2x 3 + 6x 2 3 a = 3 b = 2 f( 3) = 2( 3) 3 + 6( 3) 2 3 = 3 < 0 f( 2) = 2( 2) 3 + 6( 2) 2 3 = 5 > 0 (b) f(x) = x 3 4x 2 + 3x 2 a = 3 b = 4 Page 6

7 f(x) = (3) 3 4(3) 2 + 3(3) 2 = 2 < 0 f(x) = (4) 3 4(4) 2 + 3(4) 2 = 10 > 0 (c) f(x) = 2x 4 + 3(4) 2 a = 0 b = 1 f(x) = 2(0) 4 + 3(0) 2 = 2 < 0 f(x) = 2(1) 4 + 3(1) 2 = 3 > 0 9. Find all the values of x such that f(x) > 0 and all the values of x such that f(x) < 0, and sketch the graph of f. (a) f(x) = x 3 + 2x 2 + 8x f(x) > 0 for all x in (, 2) (0, 4) f(x) < 0 for all x in ( 2, 0) (4, ) (b) f(x) = 1 6 (x + 2)(x 3)(x 4) f(x) > 0 for all x in ( 2, 3) (4, ) f(x) < 0 for all x in (, 2) (3, 4) (c) f(x) = x 3 + 2x 2 4x 8 f(x) > 0 for all x in (2, ) f(x) < 0 for all x in (, 2) ( 2, 2) 10. Use synthetic division to show that c is a zero of f. (a) f(x) = 4x 3 9x 2 8x 3 c = 3 (b) f(x) = 4x 3 6x 2 + 8x 3 c = 1 2 (c) f(x) = 27x 4 9x 3 + 3x 2 + 6x + 1 c = Use synthetic division to find the quotient and remainder if the first polynomial is divided by the second. (a) 3x x 2 7x + 8 x + 4 (b) 5x 3 18x 2 15 x 4 (c) x 4 2x 3 + 3x 36 x 3 (d) x 3 + x 2 11x + 10 x 2 Page 7

Review all the activities leading to Midterm 3. Review all the problems in the previous online homework sets (8+9+10).

Review all the activities leading to Midterm 3. Review all the problems in the previous online homework sets (8+9+10). MA109, Activity 34: Review (Sections 3.6+3.7+4.1+4.2+4.3) Date: Objective: Additional Assignments: To prepare for Midterm 3, make sure that you can solve the types of problems listed in Activities 33 and