f(x)=sqrt4x^2-19x+12

Question

I have tried this problem multiple time and I can't find the answer. I'm not sure what I am doing wrong so please help.

thank you! 

Answer 1

I am guessing that you are looking for the domain of the function.          

f(x) = √(4x²-19x+12)

 

domain = {x l 4x² - 19x +12 ≥ 0]

 

           = {x l (4x-3)(x-4) ≥ 0}

 

(4x-3)(x-4) = 0 when x = 3/4 or x = 4

 

When x < 3/4, (4x-3)(x-4) > 0

 

When 3/4<x<4, (4x-3)(x-4) < 0

 

When x>4, (4x-3)(x-4) > 0

 

Domain:  (-∞, 3/4] ∪ [4,∞)

 

 

Answer 2

Hello Paige,

 

What is it in particular you are trying to do with the problem?

 

~Darrell