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Chapter 1 | Functions and Graphs
a. f ( x ) = ( x −4) 2 +5 b. f ( x ) = 3 x +2−1 c. f ( x ) = 3 x −2
Solution
a. Consider f ( x ) = ( x −4) 2 +5. i. Since f ( x ) = ( x −4) 2 +5 is a real number for any real number x , the domain of f is the interval (−∞, ∞). ii. Since ( x −4) 2 ≥0, we know f ( x ) = ( x −4) 2 +5≥5. Therefore, the range must be a subset of ⎧ ⎩ ⎨ y | y ≥5 ⎫ ⎭ ⎬ . To show that every element in this set is in the range, we need to show that for a given y in that set, there is a real number x such that f ( x ) = ( x −4) 2 +5= y . Solving this equation for x , we see that we need x such that ( x −4) 2 = y −5.
This equation is satisfied as long as there exists a real number x such that x −4= ± y −5.
Since y ≥5, the square root is well-defined. We conclude that for x =4± y −5, f ( x ) = y , and therefore the range is ⎧ ⎩ ⎨ y | y ≥5 ⎫ ⎭ ⎬ . b. Consider f ( x ) = 3 x +2−1. i. To find the domain of f , we need the expression 3 x +2≥0. Solving this inequality, we conclude that the domain is { x | x ≥−2/3}. ii. To find the range of f , we note that since 3 x +2≥0, f ( x ) = 3 x +2−1≥−1. Therefore, the range of f must be a subset of the set ⎧ ⎩ ⎨ y | y ≥−1 ⎫ ⎭ ⎬ . To show that every element in this set is in the range of f , we need to show that for all y in this set, there exists a real number x in the domain such that f ( x ) = y . Let y ≥−1. Then, f ( x ) = y if and only if 3 x +2−1= y .
Solving this equation for x , we see that x must solve the equation 3 x +2= y +1.
Since y ≥−1, such an x could exist. Squaring both sides of this equation, we have 3 x +2= ( y +1) 2 . Therefore, we need 3 x = ( y +1) 2 −2,
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