1 Chapter P: Preparing for Calculus P.1: Functions and Their Graphs Concepts and Vocabulary 1. If f is a function defined by y = f(x), then x is called the independent variable and y is the dependent variable. 2. True . The independent variable is sometimes referred to as the argument of the function. 3. False . If no domain is specified for a function f, then the domain of f is taken to be the largest set of real numbers for which the value f(x) is defined and is a real number. 3(x2−1) 4. False . The domain of the function f(x)= is the set {x|x(cid:54)=1}. x−1 5. False . A function can have at most one y-intercept; otherwise, there would be more than one y-value corresponding to x=0. 6. A set of points in the xy-plane is the graph of a function if and only if every vertical line intersects the graph in at most one point. 7. If the point (5,−3) is on the graph of f, then f( 5 )= −3 . 8. Let f(x) = ax2 +4. For the point (−1,2) to be on the graph of f, we must have f(−1) = 2. But f(−1)=a(−1)2+4=a+4, so we must have a+4=2, or a=−2 . 9. A function f is (a) increasing on an interval I if, for any choice of x and x in I, with x < x , 1 2 1 2 then f(x )0. The solution of this inequality is {x|x>1}, so the domain of g is the set of real numbers {x|x>1} , or in interval notation (1,∞) . 3 21. Because division by zero is not defined, x3−4x=x(x2−4)=x(x−2)(x+2) cannot be zero; that is, x(cid:54)=0,x(cid:54)=2,andx(cid:54)=−2. Thus,thedomainofhisthesetofrealnumbers {x|x(cid:54)=−2,x(cid:54)=0,x(cid:54)=2} . 22. Because the square root of a negative number is not a real number, the value of t + 1 must be nonnegative. The solution of the inequality t+1 ≥ 0 is {t|t ≥ −1}. Moreover, because division by zero is not defined, t−5 cannot be 0; that is, t cannot be equal to 5. The intersection of the sets {t|t≥−1} and {t|t(cid:54)=5} is the set {t|−1≤t<5}∪{t|t>5}. The domain of s is therefore the set of real numbers {t|−1≤t<5}∪{t|t>5} , or in interval notation [−1,5)∪(5,∞) . 23. Let f(x)=−3x+1. Then f(x+h)−f(x) −3(x+h)+1−(−3x+1) −3x−3h+1+3x−1 −3h = = = = −3 . h h h h 1 24. Let f(x)= . Then x+3 f(x+h)−f(x) 1 − 1 (x+h+3)(x+3) = x+h+3 x+3 · h h (x+h+3)(x+3) x+3−(x+h+3) −h 1 = = = − . h(x+h+3)(x+3) h(x+h+3)(x+3) (x+h+3)(x+3) √ 25. Let f(x)= x+7. Then √ √ √ √ f(x+h)−f(x) x+h+7− x+7 x+h+7+ x+7 = · √ √ h h x+h+7+ x+7 x+h+7−(x+7) h 1 = √ √ = √ √ = √ √ . h( x+h+7+ x+7) h( x+h+7+ x+7) x+h+7+ x+7 2 26. Let f(x)= √ . Then x+7 √ √ f(x+h)−f(x) √ 2 − √2 x+h+7 x+7 = x+h+7 x+7 · √ √ h h x+h+7 x+7 √ √ √ √ 2 x+7−2 x+h+7 x+7+ x+h+7 = √ √ · √ √ h x+h+7 x+7 x+7+ x+h+7 2(x+7)−2(x+h+7) = √ √ √ √ h x+h+7 x+7( x+7+ x+h+7) −2h = √ √ √ √ h x+h+7 x+7( x+7+ x+h+7) 2 = −√ √ √ √ . x+h+7 x+7( x+7+ x+h+7) 27. Let f(x)=x2+2x. Then f(x+h)−f(x) (x+h)2+2(x+h)−(x2+2x) x2+2xh+h2+2x+2h−x2−2x = = h h h 2xh+h2+2h h(2x+h+2) = = = 2x+h+2 . h h 4 28. Let f(x)=(2x+3)2. Then f(x+h)−f(x) (2(x+h)+3)2−(2x+3)2 4(x+h)2+12(x+h)+9−(4x2+12x+9) = = h h h 4x2+8xh+4h2+12x+12h+9−4x2−12x−9 = h 8xh+4h2+12h h(8x+4h+12) = = = 8x+4h+12 . h h 29. This graph fails the vertical line test (for example, the vertical line x = 2 intersects the graph in two points), so it does not represent a function . 30. This graph passes the vertical line test, and so it does represent a function . (a) From the graph, the domain of the function is the set of all real numbers or the interval (−∞,∞) , and the range of the function is the set of all positive real numbers or the interval (0,∞) . (b) The graph has no x-intercepts, and the y-intercept is 1. Thus, the only intercept is (0,1) . (c) The graph is not symmetric with respect to the x-axis, the y-axis, or the origin. 31. This graph passes the vertical line test, and so it does represent a function . (a) From the graph, the domain of the function is the set {x|−π ≤x≤π} or the closed interval [−π,π] ,andtherangeofthefunctionistheset {y|−1≤y ≤1} ortheclosedinterval [−1,1] . π (cid:16) π (cid:17) (b) Thegraphhasx-interceptsof± anday-interceptof1,sotheinterceptsare ± ,0 and (0,1) . 2 2 (c) The graph is symmetric with respect to the y-axis but not symmetric with respect to the x-axis or the origin. 32. This graph fails the vertical line test (for example, the vertical line x = 2 intersects the graph in two points), so it does not represent a function . 33. (a) f(−1)=−1+3=2; f(0)=0+3=3; f(1)=5; f(8)=−8+2=−6 (b) The graph of f consists of three pieces corresponding to each equation in the definition. On the interval [−2,1), the graph is the line y = x+3, and on the interval (1,∞), the graph is the line y =−x+2. The graph also contains the point (1,5). 5 4 3 2 1 -2 -1 1 2 3 -1 5 (c) The individual components of f have domains of {x|−2 ≤ x < 1}, {x|x = 1}, and {x|x > 1}. Thedomainoff istheunionofthesethreesets;thatis, {x|x≥−2} insetnotationor [−2,∞) in interval notation. Moreover, the individual components of f have ranges of {y|1 ≤ y < 4}, {y|y =5},and{y|y <1}. Therangeoff istheunionofthesethreesets;thatis, {y|y =5 or y <4} in set notation or (−∞,4)∪{5} in interval notation. The x-intercept is 2, and the y-intercept is 3, so the intercepts are (2,0) and (0,3) . 34. (a) f(−1)=2(−1)+5=3; f(0)=−3; f(1)=−5(1)=−5; f(8)=−5(8)=−40 (b) The graph of f consists of three pieces corresponding to each equation in the definition. On the interval [−3,0), the graph is the line y =2x+5, and on the interval (0,∞), the graph is the line y =−5x. The graph also contains the point (0,−3). 5 -3 -2 -1 1 2 3 -5 -10 -15 (c) The individual components of f have domains of {x|−3 ≤ x < 0}, {x|x = 0}, and {x|x > 0}. Thedomainoff istheunionofthesethreesets;thatis, {x|x≥−3} insetnotationor [−3,∞) in interval notation. Moreover, the individual components of f have ranges of {y|−1 ≤ y < 5}, {y|y =−3}, and {y|y <0}. The range of f is the union of these three sets; that is, {y|y <5} in 5 set notation or (−∞,5) in interval notation. The x-intercept is − , and the y-intercept is −3, 2 (cid:18) (cid:19) 5 so the intercepts are − ,0 and (0,−3) . 2 35. (a) f(−1)=1+(−1)=0; f(0)=02 =0; f(1)=12 =1; f(8)=82 =64 (b) The graph of f consists of two pieces corresponding to each equation in the definition. On the interval (−∞,0), the graph is the line y = 1+x, and on the interval [0,∞), the graph is the parabola y =x2. 6 8 6 4 2 -5 -4 -3 -2 -1 1 2 3 -2 -4 (c) The individual components of f have domains of {x|x < 0} and {x|x ≥ 0}. The domain of f is the union of these two sets; that is, all real numbers or the interval (−∞,∞) . Moreover, the individual components of f have ranges of {y|y <1} and {y|y ≥0}. The range of f is the union of these two sets; that is, all real numbers or the interval (−∞,∞) . The x-intercepts are −1 and 0, and the y-intercept is 0, so the intercepts are (−1,0) and (0,0) . 1 √ √ √ 36. (a) f(−1)= =−1; f(0)= 30=0; f(1)= 31=1; f(8)= 38=2 −1 (b) The graph of f consists of two pieces corresponding to each equation in the definition. On the √ interval (−∞,0), the graph is y = 1, and on the interval [0,∞), the graph is y = 3x. x 4 2 -4 -2 2 4 6 8 10 -2 -4 -6 -8 (c) The individual components of f have domains of {x|x < 0} and {x|x ≥ 0}. The domain of f is the union of these two sets; that is, all real numbers or the interval (−∞,∞) . Moreover, the individual components of f have ranges of {y|y <0} and {y|y ≥0}. The range of f is the union of these two sets; that is, all real numbers or the interval (−∞,∞) . The x-intercept is 0, and the y-intercept is 0, so the only intercept is (0,0) . 37. Because the graph of f includes the point (0,3), f(0)=3 ; because the graph also includes the point (−6,−3), f(−6)=−3 . 38. Because the graph of f lies above the x-axis at x=3, f(3) is positive . 7 39. Because the graph of f lies below the x-axis at x=−4, f(−4) is negative . 40. Because the graph of f includes the points (−3,0), (6,0), and (10,0), f(x) = 0 for x=−3, x=6, and x=10 . 41. Because the graph of f lies above the x-axis for −3 < x < 6 and for 10 < x ≤ 11, f(x) > 0 for −3

Instructor's Solution Manuals to Calculus Early Transcendentals

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