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1 We want to find the inverse of the relation f (x) = 2x + 4 If we call the dependent variable y, then y = 2x + 4 Swapping the names of the variables, we get x = 2y + 4
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which can be manipulated with algebra to obtain y = x/2 2 If we replace the new variable y by the relation notation f 1 (x), we get f 1 (x) = x / 2 2 2 We want to find the inverse of the relation g (x) = x 2 4x + 4 Calling the dependent variable y, we can write this as y = x 2 4x + 4 which factors to y = (x 2)2 When we swap the names of the variables, we get x = (y 2)2 Taking the complete square root of both sides produces x1/2 = y 2 which can be manipulated with algebra to obtain y = 2 x1/2 Replacing y by g 1 (x), we get g 1(x) = 2 x1/2 3 We want to find the inverse of the relation h (x) = x 3 5 Calling the dependent variable y, we have y = x3 5 Swapping the names of the variables yields x = y3 5
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522 Worked-Out Solutions to Exercises: 11-19
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which can be morphed algebraically into y = (x + 5)1/3 Replacing y by h 1 (x), we get h 1 (x) = (x + 5)1/3 4 Let s examine each situation by imagining what happens when we try to input real numbers to the original relations, and working from there: Problem 1 and its solution: The relation and its inverse are f (x) = 2x + 4 and f 1(x) = x / 2 2 We can input any real number into either of these relations, and we always get a meaningful result Therefore, the domains of f and f 1 both span the entire set of real numbers The range of f 1 is the same as the domain of f, and the range of f is the same as the domain of f 1 Therefore, the ranges of f and f 1 both include all real numbers Problem 2 and its solution: The relation and its inverse are g (x) = x 2 4x + 4 and g 1(x) = 2 x1/2 We can input any real number we choose into g, and we always get a meaningful result However, that result is always nonnegative, because it s the square of the real-number quantity (x 2) It s possible for the output of g to equal 0; that happens when the input is 2 The range of g is therefore the set of all nonnegative reals The domain of g 1 is identical to the range of g, so the domain of g 1 spans the set of all nonnegative reals The range of g 1 is the same as the domain of g, which is the set of all reals Problem 3 and its solution: The relation and its inverse are h (x) = x 3 5 and h 1(x) = (x + 5)1/3 We can input any real number into h, and we always get a real-number output value Therefore, the domain of h spans the entire set of reals The same is true of the domain
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of h 1; the relation is defined for all possible real-number input values The range of h 1 is the same as the domain of h, and the range of h is the same as the domain of h 1 Therefore, the ranges of h and h 1 both span the set of all reals 5 The equation of our relation, as given in the problem, is x 2 /4 y 2 /9 = 1 Let s use algebra to morph this equation so it s stated with y all by itself on the left-hand side and an expression containing only the variable x on the right-hand side If we multiply through by 36, we get 9x 2 4y 2 = 36 which can be rewritten as 4y 2 = 36 9x 2 Multiplying through by 1, we get 4y 2 = 9x 2 36 When we divide through by 4, we obtain y 2 = 9x 2 /4 9 Taking the complete square root of both sides yields y = (9x 2 /4 9)1/2 Using relation notation to express this equation and name the relation f, we get f (x) = (9x 2 /4 9)1/2 That s the relation! Now let s find its inverse We begin by restating the relation with y on the left instead of f (x), so we have y = (9x 2 /4 9)1/2 Swapping the names of the variables gives us x = (9y 2 /4 9)1/2 When we square both sides, we obtain x 2 = 9y 2 /4 9
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524 Worked-Out Solutions to Exercises: 11-19
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Adding 9 to each side, we get x 2 + 9 = 9y 2 /4 Multiplying through by 4, and then transposing the left- and right-hand sides of the equation, we come up with 9y 2 = 4x 2 + 36 Dividing through by 9 yields y 2 = 4x 2 /9 + 4 After we take the complete square root of both sides, we have y = (4x 2 /9 + 4)1/2 Replacing the variable y by the relation notation f 1 (x), we conclude that f 1 (x) = (4x 2 /9 + 4)1/2 6 With the information stated in Problem 5, along with the graph of the relation shown in Fig 12-2, we can see that the real-number domain of f is the set of all reals greater than or equal to 2, or smaller than or equal to 2 Another way of stating this is to say that the domain of f is the set of all reals except those in the open interval ( 2,2) The range of f is clearly the set of all reals 7 Figure B-3 shows the graph of the original relation f (x) = (9x 2 /4 9)1/2 as a pair of gray curves, and the graph of the inverse relation f 1 (x) = (4x 2 /9 + 4)1/2 as a pair of black curves The point reflector is the dashed line The real-number domain of f 1 is the same as the real-number range of f ; that s the set of all reals The real-number range of f 1 is identical to the real-number domain of f ; that s the set of all reals except those in the open interval ( 2,2) 8 Figure B-4 shows the graph of the original relation with its domain restricted to the reals greater than or equal to 2 ( gray curve) and its inverse (black curve) The curve for the inverse relation is a half-hyperbola that opens upward and intersects the y axis at (0,2) A movable vertical line never intersects the black curve at more than one point Therefore, the inverse relation f 1 (x) = (4x 2 /9 + 4)1/2 is a true function of x
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