Question

6 (12) For each function in parts a through f, state a domain that, ifit was the domain of the given function, would make the function one-to-one, and explain your answer. If no such domain exists, explain why not. (Hint: graph the function and use the appropriate line test) a. f(x) x 2| b. g(x) 3D х? —1 с. q(х) 3D х2 - 3х — 4 d. k(x)x3 9 = е. f(x) %3D Зх — 1 f l(x) 2x + 1]

Answer 1

1 = 4 (a) f(x)= 18-21 21 = X-2, f x)2 01231 |-(2-2) if < since the graph is symmetric with respect to the line x=2. Hence of is injective in either (-«, 2], 08 [2, c), Ž > X 2 - na a * 1- 2+ = 4 (), * 4+(941). Hence og is injective in either (-0, 0], as to, c). 23 (c). Q(x) = x= 3x -4, Differentiating w.rita, we get 2! (0) = 2x-3 Differentiating Again q"(x) = 2(>0 Hence 2(x) has minimum value to at x = %. Hence og is injective in either (-00, 9%) or [12 x).

*9=9 (d) kla) = x 49 ik is one-to-one in eitter (-,). .). le). f(x) = 3x-1 y = 32-) 3x ny s) Profore fis injective in R. f 1/9).- 12001) S 21), 23-1 {-(2x+1), XL-) ! in either Therefore & is one-to-one (-.-) • E5,).