Math 115 Spring 2014 Written Homework 8-SOLUTIONS Due Friday, April 11

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1 Math 115 Spring 2014 Written Homework 8-SOLUTIONS Due Friday, April 11 Instructions: Write complete solutions on separate paper (not spiral bound). If multiple pieces of paper are used, THEY MUST BE STAPLED with your name and lecture written on each page. Please review the Course Information document for more complete instructions. 1. (a) If f 1 (x) = x 3 + 2x 2, for what value of x will f(x) = 1 (do not determine the formula for f(x)). Solution: Determining the value of x for which f(x) = 1 is equivalent to determining f 1 ( 1): f 1 ( 1) = ( 1) 3 + 2( 1) 2 = = 1 Thus, f(1) = 1. (Note: While the process shown in this solution is correct in general, we unfortunately mistakenly chose a function that is not one-to-one. Due to this error on our part, this problem is not one of the ones that was graded.) (b) Suppose g(x) = x 3 x 4. Evaluate g 1 (2) without finding the formula for g 1 (x). Solution: Determining the value of g 1 (2) is equivalent to determining the x value when g(x) = 2. Thus we need to solve the equation x 3 x 4 = 2: Thus, g 1 (2) = 5. x 3 x 4 = 2 x 3 = 2(x 4) x 3 = 2x 8 3 = x 8 5 = x

2 (c) Determine f 1 (x) for f(x) = and f 1 (x). 4. Determine the domain and range for both f(x) 5x 3 Solution: To find f 1, we reverse the roles of x and y in the equation y = f(x) and solve for y. y = 4 5x 3 x = 4 5y 3 x(5y 3) = 4 5y 3 = 4 x 5y = 4 x + 3 y = x 5 or y = 4 + 3x 5x 4 Therefore, we define the inverse of f to be f 1 x (x) = 3. 5 The domain of the function f(x) is all real numbers except x = 3 5 and the domain of f 1 (x) is all real numbers except x = 0. Since the domain of f(x) is the range of f 1 (x) and vice versa: Domain of f(x) is (, 3 5 ) (3 5, ). Range of f(x) is (, 0) (0, ). Domain of f 1 (x) is (, 0) (0, ). Range of f 1 (x) is (, 3 5 ) (3 5, ).

3 (d) Determine f 1 (x) for f(x) = 2x. Determine the domain and range for both f(x) x + 3 and f 1 (x). Solution: To find f 1, we reverse the roles of x and y in the equation y = f(x) and solve for y. y = 2y x + 3 x = 2y y + 3 x(y + 3) = 2y xy + 3x = 2y 3x = 2y xy 3x = y(2 x) 3x 2 x = y Therefore, we define the inverse of f to be f 1 (x) = 3x 2 x. The domain of the function f(x) is all real numbers except x = 3 and the domain of f 1 (x) is all real numbers except x = 2. Since the domain of f(x) is the range of f 1 (x) and vice versa: Domain of f(x) is (, 3) ( 3, ). Range of f(x) is (, 2) (2, ). Domain of f 1 (x) is (, 2) (2, ). Range of f 1 (x) is (, 3) ( 3, ).

4 2. Use the graph of y = g(x) to answer the questions below. You should assume the function is not defined beyond what is shown in this picture. In other words, assume the domain of the function is [ 2, 4]. (a) Estimate the value of g(2). Solution: It appears that the graph of y = g(x) goes through the point (2, 3). Thus, we know that g(2) = 3. (b) Is g a one-to-one function? How do you know? Solution: Yes because the graph of y = g(x) passes the horizontal line test. Each output corresponds to only one input. (c) Estimate the value of g 1 (2). Solution: By the definition of inverse functions, g 1 (2) = a if and only if g(a) = 2. Thus, we need to determine the input/x-value on this graph of y = g(x) that corresponds to an output/y-value of 2. It appears that the graph goes through the point, (0.25, 2) so we have g(0.25) = 2 and thus g 1 (2) = (d) Estimate the domain of g 1 (x). Solution: By the definition of inverse functions, the domain of g 1 is the same as the range of g(x). According to this graph, the range of g(x) is R : [ 1, 3.5]. Thus, the domain of g 1 is D : [ 1, 3.5]. (e) Sketch the graph of g 1 (x). Solution: The graph of an inverse function is the reflection of the graph of y = g(x) over the line y = x. Another way to think about this is to switch the order of all ordered pairs (i.e. if the graph of y = g(x) goes through the point (a, b), then the graph of y = g 1 (x) will go through the point (b, a).) Below is a sketch of y = g 1 (x).

5 3. For the following functions f and g, verify that they are inverses by showing f(g(x)) = x AND g(f(x)) = x. (a) f(x) := x and g(x) := (x 3) 1/3 Solution: f(g(x)) := ((x 3) 1/3 ) = (x 3) + 3 = x = x g(f(x)) := ((x 3 + 3) 3) 1/3 = (x ) 1/3 = (x 3 ) 1/3 = x Since f(g(x)) := x and g(f(x)) := x, the functions f and g are inverses.

6 (b) f(x) := x + 1 x 1 and g(x) := x + 1 x 1 Solution: Note that here f and g are the same function. Hence f(g(x)) = g(f(x)). In other words, we are showing that the inverse of the function f is f itself. ( x+1 f(g(x) := x 1) + 1 ( x+1 x 1) 1 = = = x+1 + x 1 x 1 x 1 x+1 x 1 x 1 x 1 (x+1)+(x 1) x 1 (x+1) (x 1) x 1 (x + 1) + (x 1) (x + 1) (x 1) = x x 1 x + 1 x + 1 = 2x 2 = x Hence, the functions f and g are inverses.

7 4. Sketch the graph of each function and its inverse. State the domain and range of each. (a) g(t) := t 3 Solution: The domain of g is (, ) since g is a polynomial. We know what the graph of the function looks like. Figure 1: y = t 3 Since the graph of g passes the horizontal line test, we know that g is a one-to-one function on its domain. Hence, it is invertible. To find g 1, we reverse the roles of t and y in the equation y = g(t) and solve for y. t = y 3 3 t = y y = 3 t Therefore, we define the inverse of g to be g 1 (t) = 3 t. Again, to graph the inverse of g, I am going to demonstrate that the graphs y = g(t) (the blue curve) and y = g 1 (t) (the green curve) are symmetric about the line y = t (the red line).

8 (b) f(x) := 1 x + 1 Solution: The domain of f is (, 1) ( 1, ). We know what the graph of the function looks like. It is the graph of 1 x translated to the left by 1. Figure 2: y = f(x) Since the graph of f passes the horizontal line test, we know that f is a one-to-one function on its domain. Hence, it is invertible.

9 To find f 1, we reverse the roles of x and y in the equation y = f(x) and solve for y. x = 1 y + 1 x(y + 1) = 1 y + 1 = 1 x y = 1 x 1 Therefore, we define the inverse of f to be f 1 (x) = 1 x 1. There are multiple ways to graph this function. The easiest being to realize that this is the graph y = 1/x shifted down 1. Here, I am going to demonstrate that the graphs y = f(x) (the blue curve) and y = f 1 (x) (the green curve) are symmetric about the line y = x (the red line).

10 5. Consider f(x) := (x 2) 2 (x + 2) 2. (a) What is the domain of f(x)? What is the range of f(x)? Justify your answers. Solution: Since f is a polynomial, we know that the domain of f is all of R. For the range, note that f is the product of the two terms (x 2) 2 and (x + 2) 2 and both of these terms are greater than or equal to 0 for all values of x, the range of f is [0, ). (b) Sketch a rough graph of f(x) using long-run behavior and intercepts. Solution: Long-Run Behavior: Since f is a polynomial, the long-run behavior is dictated by the highest order term of the polynomial. For f, the highest order term is (x 2 )(x 2 ) = x 4. Since the highest degree of f is even, we also know that Intercepts lim f(x) = x lim f(x) = x + lim x + x4 =. x-intercepts: Since f is already in factored form, we see the x-intercepts to be ( 2, 0) and (2, 0). The y-intercept occurs at (0, f(0)) = (0, 16). Combining these conditions, we get a frame for our graph:

11 Using the domain, range and the knowledge that f is a continuous function, we get something that looks like: Figure 3: y = (x 2) 2 (x + 2) 2 (c) Explain how you know that f(x) is not a one-to-one function. Describe a domain on which f(x) is one-to-one. Solution: Note that the graph spectacularly fails the horizontal line test. Given any line y = k that intersects the graph y = f(x), the horizontal line will intersect the curve in at least 2 points (if not three or four). Hence, f(x) is not a one-to-one function. However, the graph can be broken into pieces on which f is one-to-one. The biggest such intervals are (, 2], [ 2, 0], [0, 2] and [2, ). (d) Does your domain in (c) cover the entire range stated in (a)? If not, find another domain on which f(x) is one-to-one such that the image of the restricted domain is the entire range. Solution: Recall that the range of f being [0, ) is represented graphically by the y values of the points on the graph. The intervals x (, 2] and x [2, ) both cover the entire range of f.

12 6. Determine the domain of each of the following functions. (a) f(x) = ln(x 2) Solution: x 2 > 0 x > 2 The domain is D : (2, ). (b) g(x) = 4 + log 3 ( x) Solution: x > 0 x < 0 The domain is D : (, 0). (c) h(x) = log(x + 3) Solution: x + 3 > 0 x > 3 The domain is D : ( 3, ).

13 7. Using the domains you found in Problem 6, determine the limits of each function at the ends of their domain. What do these limits tell you about the asymptote(s) of each function? (a) f(x) = ln(x 2) Solution: Using the domain from 6a and graph from 8a: lim f(x) = x 2 + (b) g(x) = 4 + log 3 ( x) lim f(x) = x Solution: Using the domain from 6b and graph from 8b: lim g(x) = x (c) h(x) = log(x + 3) lim g(x) = x 0 Solution: Using the domain from 6c and graph from 8c: lim f(x) = x 3 + lim f(x) = x

14 8. Using a proper sequence of transformations and the information from Problems 6 and 7, sketch a rough graph of the following functions. For each graph, clearly label any asymptotes and at least one point on the graph. (a) f(x) = ln(x 2) Solution: We will begin with the graph of y = ln x. Then reflect that graph of the x-axis and shift it horizontally to the right 2. Figure 4: y = ln x Figure 5: y = ln x Figure 6: y = ln(x 2) On this final graph, you should sketch in and label a vertical asymptote at x = 2 and the most obvious point to label is (3, 0) (shift the original x-intercept right 2.)

15 (b) g(x) = 4 + log 3 ( x) Solution: We will begin with the graph of y = log 3 x. Then reflect that graph over the y-axis and shift it vertically up 4. Figure 7: y = log 3 x Figure 8: y = log 3 ( x) Figure 9: y = 4 + log 3 ( x) On this final graph, you should sketch in and label a vertical asymptote at x = 0 and the most obvious point to label is ( 1, 4) (reflect the original x-intercept over y-axis and shift up 4.)

16 (c) h(x) = log(x + 3) Solution: We will begin with the graph of y = log x. Then shift it horizontally to the left 3. Figure 10: y = log x Figure 11: y = log(x + 3) On this final graph, you should sketch in and label a vertical asymptote at x = 3 and the most obvious point to label is ( 2, 0) (shift the original x-intercept left 3.) 9. Explain why it does not make sense to ask for the value of lim x (ln x). Solution: The domain of y = ln x is all real numbers greater than 0: x > 0. Thus, negative values are not in the domain so it makes no sense to ask about the behavior as x goes to.