APPM 1340 Midterm: appm1340fall2014exam1_sol

• School: CU-Boulder
• Department: Applied Math
• Course: APPM 1340
• Professor: All

Document Summary

Fall 2014: (20 points) let f (x) = 1 (a) simplify f . and g(x) = 5 2x (5 2x) (3 x) 5 2x (b) from the de nition of f we observe that 3 x 6= 0 x 6= 3. 2 , 3(cid:1) (3, ) . (c) f (x) = 4x2 20x + 25 = x2 4x + 4. 3x2 16x + 21 = 0 (3x 7)(x 3) = 0. Since x = 3 is not in the domain of f , the only solution is x = 7/3 . If f (x) = 1/f (x), then [f (x)]2 = 1 and f (x) = 1. First solve f (x) = 1. f (x) = x + 2. = 1 x + 2 = 5 2x x = 3. Note, however, that x = 3 is not in the domain of f . Next solve f (x) = 1. f (x) = x + 2.