Section 3.3 Functions 183 3c. Let c(x) be given by Y1. Find c(4) and find x such that c(x) = -6. 3d. Let f (x) be given by the graph. Find f (2) and find x such that f (x) = 1. 2 2 4 –2 –4 –2 –4 –6 x y Solutions 3a. g(x) = x2 + 1 g(-3) = (-3)2 + 1 Replace x with -3. g(-3) = 9 + 1 Simplify the exponent. g(-3) = 10 Add. So, g(-3) = 10. 3b. f (x) = 3x - 12 f (0) = 3(0) - 12 Replace x with 0. f (0) = 0 - 12 Multiply. f (0) = -12 Subtract. So, f (0) = -12. Now to solve f (x) = 0, replace f (x) with 0. f (x) = 3x - 12 0 = 3x - 12 Replace f (x) with 0. 0 + 12 = 3x - 12 + 12 Add 12 to each side. 12 = 3x Simplify. 12 3 = 3x 3 Divide each side by 3. 4 = x Simplify. So, f (x) = 0 when x = 4. 3c. To find c(4), we must find the point in the table whose x-value is 4. The point (4, -8) is one of the ordered pairs in the table. So, c(4) = -8. To solve c(x) = -6, we must find the point(s) in the table whose Y1-value is -6. The point (2, -6) is in the table. So, c(x) = -6 when x = 2. 3d. To find f (2), we need to find the point on the graph whose x-value is 2. The point (2, -4) lies on the graph of the function. So, f (2) = -4. To solve f (x) = 1, we need to find the point on the graph whose y-value is 1. The point (-3, 1) lies on the graph. So, f (x) = 1 when x = -3. 2 4 –2 –4 –2 –4 –6 x y (2 –4) 2 4 –4 –2 2 –2 –4 –6 x y (–3 1)
hendricks_intermediate_algebra_1e_ch1_3
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