Question

Given f(x)=x^2 + x - 2, find the roots of g(x)=3f (-2x). Hint: Use the mapping rule.

Answer 1

`f(x) = {x}^(2) + x - 2 \\ f( - 2x) = ( - 2x) ^(2) + ( - 2x) - 2 \\ = 4 {x}^(2) - 2x - 2 \\ \\ g(x) = 3f( - 2x) \\ g(x) = 3(4 {x}^(2) - 2x - 2) \\ = 12 {x}^(2) - 6x - 6 \\ = 6(2x + 1)(x - 1) \\ \\ g(x) = √(3f( - 2x)) \\ = \sqrt{3(4 {x}^(2) - 2x - 2)} \\ = \sqrt{12 {x}^(2) - 6x - 6} \\ = √(6(2x + 1)(x - 1)) \\ x = - (1)/(2) ,1`

From what I understood from the question I answered, I'm not sure about it , I hope this helps you ^_^