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of 5 (7) The graph of y = (x 1)° + 2 is the graph of y = x° shifted (A) 2- units right, 1- unit up (C) 2- units right, 1- unit down (B) 1 unit right, 2 units up (D)1 unit right, 2 units down

of 5 (7) The graph of y = (x 1)° + 2 is the graph of y = x° shifted (A) 2- units right, 1- unit up (C) 2- units right, 1- unit down (B) 1 unit right, 2 units up (D)1 unit right, 2 units down

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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2 of 5 (7) The graph of y = (x - 1)° + 2 is the graph of y = x' shifted (A) 2- units right, 1- unit up (C) 2- units right, 1- unit down %3D (B) 1 unit right, 2 units up (D)1 unit right, 2 units down (8) To verify that (x+1) is a factor of P(x) = x° + 2x² – 5x – 6, which of the following is correct? %3D P: The graph of P(x) has an x-intercept at x = 1 Q: The remainder R=0 when dividing P(x) by (x+1) R: To check P(-1) = 0 (A) P and Q (B) P and R (C) Q and R (D) P, Q, and R (9) When a certain drug is taken orally, the concentration of the drug in the patient's bloodstream after t minutes is given by C(t) = 0.024t-0.0002t, where 0 < t < 120, and the concentration is measured in mg/Lt. When is the maximum serum concentration reached, and what is that maximum concentration? %3D (A) C Max = 0.72 mg/L at t = 60min. (C) CMax = 0 mg/L at t = 120min. (B) C Max = 0.64 mg/L at t = 40min. (D) CMax = 0.64 mg/L at t = 80 min. %3D Questions #9-13 are based on the following information: Let f(x) = x? – 1 and g(x) = 2x - 4 be two functions. (10) Find the inverse function for g(x) = 2x - 4 %3D 1 (A) x = ½ y + 2 (B) y = ½ x + 2 (C) y = ½ x - 4 %3D 2х -4 (11) Find fo g1 (2) =? 1 (A) f•g* (2) = f( 2) (B) f•g° (2) = 3 - 3 = 9; f(2) = 3, gʻ (2) = 2 + 4)/2 = 3 (B) f•g" (2) = f('/g2)= undefined; g(2) = 0 (D)f•g' (2) = f(3) = 8; g" (2) = ( * 4)/2 = 3 = undefined; g(2) = 0 g(2) %3D Find (fg ")(2) =? (A) (fg )(2) = f( 2)– (B) (fg")(2) = 3 -3= 9; f(2) = 3, g' (2) = 2 + 4)/2 = 3 (12) 1 = undefined; g(2) = 0 (B) (fg')(2) = f('/g2)= undefined; g(2) = 0 g(2) (D) (f g')(2) = f(3) = 8; g" (2) = (2 + 4)/2 = 3 %3D %3D Are (fog)(x) = (gof)(x)? (13) 1. Yes. (f °g)(x) = (g°f)(x) = f(x) • g(x) = g • f(x) = (x² – 1) (2x- 4) II. Yes. (fog)(x) = (2x – 4) – 1 = 2x – 5 and (gof)(x) = 2x? – 1 – 4 = 2x² – 5 III. No. (fog)(x) = (2x – 4)² – 1 = 4x² – 16x +15, but (gof)(x) = 2(x² – 1) – 4) = 2x² – 6 (A) II only %3D %3D %3D (B) III only (C) I and II (D) None is true (14) If f(x) = x² – 1, x 2 0, and f'(x) = Vx + 1 are graphed on the same xy-plane, which transformation %D would relate them onto one another? (A) Reflection about the x-axis (C) Reflection about the origin (B) Reflection about the y-axis (D) Reflection about the line y = x х2-2х -3 (15) Rational function r(x) = has a vertical asymptote at x = 4. The asymptotic behavior х- 4 around x = 4 could be described as (B) y → 0, as x →4° & y → - ∞, as x →4* (D) y → - 0, as x →4 & y → - ∞, as x →4* (А) у —> - оо, as x ->4' & y 0, as x >4* (C) y → 0, as x →4° & y → ∞, as x →4*
Transcribed Image Text:2 of 5 (7) The graph of y = (x - 1)° + 2 is the graph of y = x' shifted (A) 2- units right, 1- unit up (C) 2- units right, 1- unit down %3D (B) 1 unit right, 2 units up (D)1 unit right, 2 units down (8) To verify that (x+1) is a factor of P(x) = x° + 2x² – 5x – 6, which of the following is correct? %3D P: The graph of P(x) has an x-intercept at x = 1 Q: The remainder R=0 when dividing P(x) by (x+1) R: To check P(-1) = 0 (A) P and Q (B) P and R (C) Q and R (D) P, Q, and R (9) When a certain drug is taken orally, the concentration of the drug in the patient's bloodstream after t minutes is given by C(t) = 0.024t-0.0002t, where 0 < t < 120, and the concentration is measured in mg/Lt. When is the maximum serum concentration reached, and what is that maximum concentration? %3D (A) C Max = 0.72 mg/L at t = 60min. (C) CMax = 0 mg/L at t = 120min. (B) C Max = 0.64 mg/L at t = 40min. (D) CMax = 0.64 mg/L at t = 80 min. %3D Questions #9-13 are based on the following information: Let f(x) = x? – 1 and g(x) = 2x - 4 be two functions. (10) Find the inverse function for g(x) = 2x - 4 %3D 1 (A) x = ½ y + 2 (B) y = ½ x + 2 (C) y = ½ x - 4 %3D 2х -4 (11) Find fo g1 (2) =? 1 (A) f•g* (2) = f( 2) (B) f•g° (2) = 3 - 3 = 9; f(2) = 3, gʻ (2) = 2 + 4)/2 = 3 (B) f•g" (2) = f('/g2)= undefined; g(2) = 0 (D)f•g' (2) = f(3) = 8; g" (2) = ( * 4)/2 = 3 = undefined; g(2) = 0 g(2) %3D Find (fg ")(2) =? (A) (fg )(2) = f( 2)– (B) (fg")(2) = 3 -3= 9; f(2) = 3, g' (2) = 2 + 4)/2 = 3 (12) 1 = undefined; g(2) = 0 (B) (fg')(2) = f('/g2)= undefined; g(2) = 0 g(2) (D) (f g')(2) = f(3) = 8; g" (2) = (2 + 4)/2 = 3 %3D %3D Are (fog)(x) = (gof)(x)? (13) 1. Yes. (f °g)(x) = (g°f)(x) = f(x) • g(x) = g • f(x) = (x² – 1) (2x- 4) II. Yes. (fog)(x) = (2x – 4) – 1 = 2x – 5 and (gof)(x) = 2x? – 1 – 4 = 2x² – 5 III. No. (fog)(x) = (2x – 4)² – 1 = 4x² – 16x +15, but (gof)(x) = 2(x² – 1) – 4) = 2x² – 6 (A) II only %3D %3D %3D (B) III only (C) I and II (D) None is true (14) If f(x) = x² – 1, x 2 0, and f'(x) = Vx + 1 are graphed on the same xy-plane, which transformation %D would relate them onto one another? (A) Reflection about the x-axis (C) Reflection about the origin (B) Reflection about the y-axis (D) Reflection about the line y = x х2-2х -3 (15) Rational function r(x) = has a vertical asymptote at x = 4. The asymptotic behavior х- 4 around x = 4 could be described as (B) y → 0, as x →4° & y → - ∞, as x →4* (D) y → - 0, as x →4 & y → - ∞, as x →4* (А) у —> - оо, as x ->4' & y 0, as x >4* (C) y → 0, as x →4° & y → ∞, as x →4*
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