Answer:
Part 1) x = 0 and x = two thirds
Part 2) x = −3; x = 0 is an extraneous solution
Part 3) (−∞, −3) ∪ (−3, ∞)
Step-by-step explanation:
Part 1) we have
![\frac{x}{4}=\frac{x^{2}}{x+2}](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B4%7D%3D%5Cfrac%7Bx%5E%7B2%7D%7D%7Bx%2B2%7D)
Solve for x
Multiply by 4(x+2) both sides to remove the fractions
![x(x+2)=4x^2](https://tex.z-dn.net/?f=x%28x%2B2%29%3D4x%5E2)
![x^2+2x=4x^2\\3x^2-2x=0](https://tex.z-dn.net/?f=x%5E2%2B2x%3D4x%5E2%5C%5C3x%5E2-2x%3D0)
Solve the quadratic equation
Factor 3
![3(x^2-(2/3)x)=0](https://tex.z-dn.net/?f=3%28x%5E2-%282%2F3%29x%29%3D0)
Complete the square
![3(x^2-(2/3)x+4/36)=0+1/3](https://tex.z-dn.net/?f=3%28x%5E2-%282%2F3%29x%2B4%2F36%29%3D0%2B1%2F3)
Rewrite as perfect squares
![3(x-(1/3))^2=1/3](https://tex.z-dn.net/?f=3%28x-%281%2F3%29%29%5E2%3D1%2F3)
![(x-(1/3))^2=1/9](https://tex.z-dn.net/?f=%28x-%281%2F3%29%29%5E2%3D1%2F9)
square root both sides
![(x-(1/3))=(+/-)1/3](https://tex.z-dn.net/?f=%28x-%281%2F3%29%29%3D%28%2B%2F-%291%2F3)
![x=(1/3)(+/-)1/3](https://tex.z-dn.net/?f=x%3D%281%2F3%29%28%2B%2F-%291%2F3)
![x=(1/3)(+)1/3=2/3](https://tex.z-dn.net/?f=x%3D%281%2F3%29%28%2B%291%2F3%3D2%2F3)
![x=(1/3)(-)1/3=0](https://tex.z-dn.net/?f=x%3D%281%2F3%29%28-%291%2F3%3D0)
Part 2) we have
![\frac{3}{x}=\frac{4x+3}{x^{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7Bx%7D%3D%5Cfrac%7B4x%2B3%7D%7Bx%5E%7B2%7D%7D)
Multiply by x^2 both sides to remove fractions
![3x=4x+3](https://tex.z-dn.net/?f=3x%3D4x%2B3)
![x=-3](https://tex.z-dn.net/?f=x%3D-3)
Part 3) we have
![f(x)=\frac{x-1}{x+3}](https://tex.z-dn.net/?f=f%28x%29%3D%5Cfrac%7Bx-1%7D%7Bx%2B3%7D)
we know that
The denominator cannot be equal to zero
so
The value of x cannot be equal to -3
so
The domain of f(x) is all real numbers except the value of x=-3
therefore
(−∞, −3) ∪ (−3, ∞)