If $f(x) = \dfrac{x^3+x-2}{x^3-1}$ and $g(x) = \dfrac{x^2+x+2}{x^2+x+1}$, then ($D_f$ represents domain and $R_f$ represents range of $f(x)$)
$D_f = \mathbb{R}-\{1\},\; R_f = \left(1,\frac{7}{3}\right]-\left\{\frac{4}{3}\right\},\; D_g = \mathbb{R},\; R_g = \left(1,\frac{7}{3}\right]$
$D_f = \mathbb{R}-\{1\},\; R_f = \left(1,\frac{7}{3}\right],\; D_g = \mathbb{R},\; R_g = \left(1,\frac{7}{3}\right]$
$D_f = \mathbb{R}-\{1\},\; R_f = \left[1,\frac{7}{3}\right],\; D_g = \mathbb{R},\; R_g = \left[1,\frac{7}{3}\right]$
$D_f = \mathbb{R}-\{1\},\; R_f = \left(1,\frac{7}{3}\right],\; D_g = \mathbb{R},\; R_g = \left(1,\frac{7}{3}\right]$
Step-by-Step Solution
Key Concept: Factor: $f(x) = \frac{(x-1)(x^2+x+2)}{(x-1)(x^2+x+1)} = \frac{x^2+x+2}{x^2+x+1}$ for $x \neq 1$. Write as $1 + \frac{1}{x^2+x+1}$ and find range using the minimum of $x^2+x+1$.
For $x\neq 1$: $f(x)=g(x)=1+\frac{1}{x^2+x+1}$. Since $x^2+x+1\in[\frac{3}{4},\infty)$, we have $\frac{1}{x^2+x+1}\in(0,\frac{4}{3}]$, so range $=(1,\frac{7}{3}]$.
Correct Answer: 4