Question

which statement best describes the domain and range of the function $f(x) = \frac{2x^2 - 2}{3x^2 - 6x + 3}$?

the domain consists of all real numbers such that $x \
eq 1$, while the range consists of all real numbers such that $y \
eq \frac{2}{3}$.

the domain consists of all real numbers such that $x \
eq 1$, while the range consists of all real numbers such that $y \
eq -\frac{2}{3}$.

the domain consists of all real numbers such that $x \
eq -1$ or $1$, while the range consists of all real numbers such that $y \
eq \frac{2}{3}$.

the domain consists of all real numbers such that $x \
eq -1$ or $1$, while the range consists of all real numbers such that $y \
eq -\frac{2}{3}$.

Explanation:

Step1: Find domain (denominator ≠0)

Solve $3x^2 -6x +3
eq 0$
Factor: $3(x^2-2x+1)=3(x-1)^2
eq 0$
Thus $x
eq 1$

Step2: Find range (solve for x in y=f(x))

Set $y = \frac{2x^2 -2}{3x^2 -6x +3}$
Multiply denominator: $y(3x^2-6x+3)=2x^2-2$
Rearrange: $(3y-2)x^2 -6yx + (3y+2)=0$

Step3: Real x requires discriminant ≥0

Discriminant $\Delta = (-6y)^2 -4(3y-2)(3y+2) \geq 0$
Calculate: $36y^2 -4(9y^2-4) = 36y^2-36y^2+16 = 16 \geq 0$

Step4: Exclude y where coefficient=0

When $3y-2=0 \implies y=\frac{2}{3}$, the equation becomes $-4x + 4=0$, which has no solution matching original function (since $x
eq 1$). Thus $y
eq \frac{2}{3}$.

Answer:

The domain consists of all real numbers such that $x
eq 1$, while the range consists of all real numbers such that $y
eq \frac{2}{3}$.