04_ch03b_pre-calculas12_wncp_solution.qxd 5/17/12 7:49 PM Page 39 Home Quit Lesson 3.5 Exercises, pages 243–249 A 4. For each graph below, sketch its image after a reflection in the line y = x. a) b) y 4 2 y f(x) y f(x) 4 y 4 2 x 2 0 4 4 2 2 0 2 x f(y) 4 c) 2 x f(y) x 4 4 d) x f(y) y y 4 y f(x) 4 y f(x) x f(y) 2 2 x x 2 0 2 4 0 2 4 6 2 5. Determine an equation of the inverse of each function. a) y = 2x + 9 x ⴝ 2y ⴙ 9 x ⴚ 9 ⴝ 2y yⴝ xⴚ9 2 An equation of the inverse xⴚ9 is: y ⴝ b) y = x2 + 5 x ⴝ y2 ⴙ 5 x ⴚ 5 ⴝ y2 √ yⴝ— xⴚ5 An equation of the inverse is: √ yⴝ— xⴚ5 2 c) y = 5x - 3 4 xⴝ 5y ⴚ 3 4 4x ⴝ 5y ⴚ 3 4x ⴙ 3 ⴝ 5y yⴝ 4x ⴙ 3 5 An equation of the inverse is: y ⴝ 4x ⴙ 3 5 ©P DO NOT COPY. d) y = 2x2 - 3 x ⴝ 2y2 ⴚ 3 x ⴙ 3 ⴝ 2y2 y2 ⴝ xⴙ3 2 yⴝ— √x ⴙ 3 2 √x ⴙ 3 An equation of the inverse is: yⴝ— 2 3.5 Inverse Relations—Solutions 39 04_ch03b_pre-calculas12_wncp_solution.qxd 5/17/12 7:49 PM Page 40 Home Quit B 6. For each function below: i) Determine an equation of its inverse. ii) Sketch the graphs of the function and its inverse. iii) Is the inverse a function? Explain. b) y = -x 2 + 3 a) y = 4x - 1 i) x ⴝ 4y ⴚ 1 4y ⴝ x ⴙ 1 i) x ⴝ ⴚy2 ⴙ 3 y2 ⴝ 3 ⴚ x √ yⴝ— 3ⴚx xⴙ1 yⴝ 4 ii) 4 ii) y y x1 4 y 3 x 2 4 y 2 x x 4 2 0 2 4 4 2 0 2 2 y 4x 1 2 4 y x 2 3 4 iii) The inverse is a function because its graph passes the vertical line test. iii) The inverse is not a function because its graph does not pass the vertical line test. 7. Determine whether the functions in each pair are inverses of each other. Justify your answer. a) y = x2 + 7, x ≥ 0 and y = √ x - 7 An equation of the inverse of y ⴝ x2 ⴙ 7, x » 0 is: xⴝy ⴙ7 2 x ⴚ 7 ⴝ y2 √ yⴝ— xⴚ7 Since the domain of y ⴝ x2 ⴙ 7 is x » 0, then the range of the inverse is y » 0. √ So, an equation of the inverse is y ⴝ x ⴚ 7. Since this equation matches the given equation, the functions are inverses of each other. b) y = 5x - 3 and y = 3 - x 5 An equation of the inverse of y ⴝ 5x ⴚ 3 is: x ⴝ 5y ⴚ 3 x ⴙ 3 ⴝ 5y yⴝ xⴙ3 5 Since this is not equivalent to the given equation, the functions are not inverses of each other. 40 3.5 Inverse Relations—Solutions DO NOT COPY. ©P 04_ch03b_pre-calculas12_wncp_solution.qxd 5/17/12 7:49 PM Page 41 Home Quit √ c) y = (x + 4)2 - 2, x ≥ -4 and y = - x + 2 + 4 An equation of the inverse of y ⴝ (x ⴙ 4)2 ⴚ 2, x » ⴚ4 is: x ⴝ (y ⴙ 4) ⴚ 2 2 x ⴙ 2 ⴝ (y ⴙ 4)2 √ yⴙ4ⴝ— xⴙ2 √ yⴝ— xⴙ2ⴚ4 Since the domain of y ⴝ (x ⴙ 4)2 ⴚ 2 is x » ⴚ4, then the range of the inverse is y » ⴚ4. √ So, an equation of the inverse is y ⴝ x ⴙ 2 ⴚ 4. Since this equation is not equivalent to the given equation, the functions are not inverses of each other. 1 d) y = 2x - 6 and y = 2x + 12 1 2 An equation of the inverse of y ⴝ x ⴚ 6 is: 1 2 1 xⴙ6ⴝ y 2 xⴝ yⴚ6 y ⴝ 2x ⴙ 12 Since this matches the given equation, the functions are inverses of each other. 8. Determine whether the functions in each pair are inverses of each other. Justify your answer. a) b) y 4 y 2 2 x 4 2 0 x 2 2 4 6 4 2 2 4 6 Corresponding pairs of points are equidistant from the line y ⴝ x, so the functions are inverses of each other. ©P DO NOT COPY. Corresponding pairs of points are not equidistant from the line y ⴝ x, so the functions are not inverses of each other. 3.5 Inverse Relations—Solutions 41 04_ch03b_pre-calculas12_wncp_solution.qxd 5/17/12 7:49 PM Page 42 Home Quit 9. Sketch the graph of the inverse of each relation. Is the inverse a function? Justify your answer. a) b) x f (y) 4 y 4 y f (x) y f(x) y 2 x 4 x 4 2 0 2 4 0 2 2 2 4 x f(y) 4 4 The inverse is not a function because its graph does not pass the vertical line test. The inverse is not a function because its graph does not pass the vertical line test. 10. Determine two ways to restrict the domain of each function below so its inverse is a function. Write the equation of the inverse each time. Use a graph to illustrate each way. a) y = 2(x + 1)2 - 2 y 2(x 1)2 2, x 1 y 2(x 1)2 2, x 1 y 4 y 2 x2 1 2 x 4 2 4 4 y x2 1 2 Sketch the graphs of y ⴝ 2(x ⴙ 1)2 ⴚ 2 and its inverse. For an equation of the inverse, write x ⴝ 2(y ⴙ 1)2 ⴚ 2, then solve for y. x ⴙ 2 ⴝ 2(y ⴙ 1)2 xⴙ2 ⴝ (y ⴙ 1)2 2 yⴙ1ⴝ— yⴝ— √x ⴙ 2 √x ⴙ2 2 2 ⴚ1 The inverse is a function if the domain of y ⴝ 2(x ⴙ 1)2 ⴚ 2 is restricted to either x ◊ ⴚ1 or x » ⴚ1. For y ⴝ 2(x ⴙ 1)2 ⴚ 2, x yⴝⴚ √x ⴙ 2 2 ⴚ1 For y ⴝ 2(x ⴙ 1)2 ⴚ 2, x yⴝ 42 √x ⴙ 2 2 ◊ ⴚ1, the inverse is: » ⴚ1, the inverse is: ⴚ1 3.5 Inverse Relations—Solutions DO NOT COPY. ©P 04_ch03b_pre-calculas12_wncp_solution.qxd 5/17/12 7:49 PM Page 43 Home Quit b) y = -3(x - 2)2 + 4 y y x 4 2 3 4 y 2 x 4 2 3 4 x 2 2 4 2 4 y 3(x 2)2 4, x 2 y 3(x 2)2 4, x 2 Sketch the graphs of y ⴝ ⴚ3(x ⴚ 2)2 ⴙ 4 and its inverse. For an equation of the inverse, write x ⴝ ⴚ3(y ⴚ 2)2 ⴙ 4, then solve for y. ⴚx ⴙ 4 ⴝ 3(y ⴚ 2)2 ⴚx ⴙ 4 ⴝ (y ⴚ 2) 2 3 yⴚ2ⴝ— yⴝ— √ⴚx ⴙ 4 √ ⴚ x 3ⴙ 4 3 ⴙ2 The inverse is a function if the domain of y ⴝ ⴚ3(x ⴚ 2)2 ⴙ 4 is restricted to either x ◊ 2 or x » 2. For y ⴝ ⴚ3(x ⴚ 2)2 ⴙ 4, x yⴝⴚ √ⴚx ⴙ 4 3 ⴙ2 For y ⴝ ⴚ3(x ⴚ 2)2 ⴙ 4, x yⴝ √ⴚx ⴙ 4 3 ◊ 2, the inverse is: » 2, the inverse is: ⴙ2 11. The point A(5, ⫺6) lies on the graph of y = f(x). What are the coordinates of its image A¿ on the graph of y = f -1(x - 3)? On the graph of y ⴝ fⴚ1(x), the coordinates of the image of A are: (ⴚ6, 5) On the graph of y ⴝ f ⴚ1(x ⴚ 3), this point is translated 3 units right, so the coordinates of Aœ are: ( ⴚ3, 5) 12. a) Graph the function y = (x - 3)2 + 4, x ≥ 3. y (x 3)2 4, x 3 y The graph is the part of a parabola, congruent to y ⴝ x2, with vertex (3, 4), for which x » 3. 6 4 y √ x 43 2 b) What is the range of the function? x 0 From the graph, the range is: y ©P DO NOT COPY. 2 4 6 »4 3.5 Inverse Relations—Solutions 43 04_ch03b_pre-calculas12_wncp_solution.qxd 5/17/12 7:49 PM Page 44 Home Quit c) Determine the equation of its inverse. Sketch its graph. For an equation of the inverse, write x ⴝ (y ⴚ 3)2 ⴙ 4, then solve for y. x ⴚ 4 ⴝ (y ⴚ 3)2 √ yⴚ3ⴝ — xⴚ4 √ yⴝ— xⴚ4ⴙ3 Since the domain of y ⴝ (x ⴚ 3)2 ⴙ 4 is restricted to x √ equation of the inverse is: y ⴝ x ⴚ 4 ⴙ 3 » 3, then the d) What are the domain and range of the inverse? The domain of the inverse is: x » 4 The range of the inverse is: y » 3 13. Determine the coordinates of two points that lie on the graphs of both y = x2 + 2x - 6 and its inverse. Explain the strategy you used. For the points to lie on both graphs, the points must lie on the line y ⴝ x. I input the equation in a graphing calculator, then looked at the table of values to determine two points with matching coordinates. These points are: (2, 2) and ( ⴚ3, ⴚ3) 14. A graph was reflected in the line y = x. Its reflection image is shown. Determine an equation of the original graph in terms of x and y. a) 4 b) y 6 2 2 0 y x2 2 y g(x) x 2 4 4 x 2 Use the line y ⴝ x to sketch the graph of the inverse. This line has y-intercept ⴚ2, and slope 3, so its equation is: y ⴝ 3x ⴚ 2 44 3.5 Inverse Relations—Solutions y g(x) 2 2 y 3x 2 y 0 2 4 6 2 Use the line y ⴝ x to sketch the graph of the inverse. This curve is a parabola that has vertex (0, 2), and is congruent to y ⴝ x2. So, its equation is: y ⴝ x2 ⴙ 2 DO NOT COPY. ©P 04_ch03b_pre-calculas12_wncp_solution.qxd 5/17/12 7:49 PM Home Page 45 Quit C 15. Determine the equations of two functions such that the graphs of each function and its inverse coincide. The graphs of these functions must be symmetrical about the line y ⴝ x. For example, any linear function with slope –1 has a graph that coincides with its inverse, such as y ⴝ ⴚx ⴙ 4. 16. The graphs of y = value of k. Given y ⴝ xⴝ xⴙ xⴙ yⴙ 5 , interchange x and y. k 5 yⴙk xy ⴙ xk ⴝ y ⴙ 5 y(x ⴚ 1) ⴝ 5 ⴚ xk yⴝ x + 5 and its inverse coincide. Determine the x + k 5 ⴚ xk xⴚ1 Simplify. Solve for y. This equation must be the same as y ⴝ xⴙ5 . xⴙk Compare the denominators: x ⴚ 1 ⴝ x ⴙ k So, k ⴝ ⴚ1 ©P DO NOT COPY. 3.5 Inverse Relations—Solutions 45
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