Question

what are the domain and range of the function $f(x) = \frac{x^2 + 8x + 7}{x + 1}$?\
\

• d: $\\{x \in \mathbb{r} | x \

eq 1\\}$; r: $\\{y \in \mathbb{r} | y \
eq 0\\}$\

• d: $\\{x \in \mathbb{r} | x \

eq -1\\}$; r: $\\{y \in \mathbb{r} | y \
eq -6\\}$\

• d: $\\{x \in \mathbb{r} | x \

eq -7\\}$; r: $\\{y \in \mathbb{r} | y \
eq 0\\}$\

• d: $\\{x \in \mathbb{r} | x \

eq -1\\}$; r: $\\{y \in \mathbb{r} | y \
eq 6\\}$

Explanation:

Step1: Simplify the function

First, factor the numerator: $x^{2}+8x + 7=(x + 1)(x+7)$. So the function $f(x)=\frac{(x + 1)(x + 7)}{x + 1}$. We can cancel out $x + 1$ (for $x
eq - 1$), so $f(x)=x + 7$, $x
eq - 1$.

Step2: Find the domain

The original function has a denominator $x + 1$. The denominator cannot be zero, so $x+1
eq0\implies x
eq - 1$. Thus, the domain $D=\{x\in\mathbb{R}|x
eq - 1\}$.

Step3: Find the range

The simplified function is $y=x + 7$ with the restriction that $x
eq - 1$. When $x=-1$, the original function is undefined, but for the simplified function (excluding $x = - 1$), when $x=-1$, $y=-1 + 7=6$. So the value $y = 6$ is not in the range. So the range $R=\{y\in\mathbb{R}|y
eq6\}$.

Answer:

D: $\{x \in \mathbb{R} | x
eq -1\}$; R: $\{y \in \mathbb{R} | y
eq 6\}$