Question

1. if f is a real - valued function, which of the following values are not in the domain of f, where $f(x)=\frac{sqrt{x + 2}}{2x^{3}+x^{2}-2x - 1}$? select all that apply. a - 3 b - 1 c $-\frac{1}{2}$ d 1

Explanation:

Step1: Analyze the square root

For the square root $\sqrt{x + 2}$, the expression inside must be non - negative, so $x+2\geq0$, which gives $x\geq - 2$.

Step2: Analyze the denominator

Factor the denominator $2x^{3}+x^{2}-2x - 1$. Group terms: $(2x^{3}+x^{2})+(-2x - 1)=x^{2}(2x + 1)-1(2x + 1)=(2x + 1)(x^{2}-1)=(2x + 1)(x - 1)(x + 1)$. The denominator cannot be zero, so $2x+1
eq0\Rightarrow x
eq-\frac{1}{2}$, $x - 1
eq0\Rightarrow x
eq1$, $x + 1
eq0\Rightarrow x
eq - 1$.

Step3: Check each option

- Option A: For $x=-3$, $-3<-2$, so it does not satisfy the square root condition. But we also need to check other conditions. Wait, first, from the square root, $x\geq - 2$, so $x = - 3$ is not in the domain because of the square root. But let's check the other options too.
- Option B: For $x=-1$, the denominator is zero (since $x + 1=0$), so $x=-1$ is not in the domain.
- Option C: For $x =-\frac{1}{2}$, the denominator is zero (since $2x+1 = 0$), so $x=-\frac{1}{2}$ is not in the domain.
- Option D: For $x = 1$, the denominator is zero (since $x - 1=0$), so $x = 1$ is not in the domain. Wait, but let's re - check the square root for $x=-3$: $x=-3$, $x + 2=-1<0$, so the square root is not defined. But the question is which values are NOT in the domain. Let's list the reasons:
- $x=-3$: Fails the square root condition ($x+2=-1<0$).
- $x=-1$: Makes denominator zero ($(x + 1)=0$).
- $x=-\frac{1}{2}$: Makes denominator zero ($2x + 1=0$).
- $x = 1$: Makes denominator zero ($x - 1=0$). But let's check the options again. Wait, the options are A: - 3, B: - 1, C: $-\frac{1}{2}$, D:1.

Wait, let's re - evaluate:

From the square root: $x\geq - 2$. So $x=-3$ is out because $x<-2$.

From the denominator: $x
eq - 1$, $x
eq-\frac{1}{2}$, $x
eq1$.

So:

- $x=-3$: Not in domain (square root).
- $x=-1$: Not in domain (denominator zero).
- $x=-\frac{1}{2}$: Not in domain (denominator zero).
- $x = 1$: Not in domain (denominator zero).

But let's check each option:

- Option A: $x=-3$: $x+2=-1<0$, so $\sqrt{x + 2}$ is not real. So $x=-3$ is not in the domain.
- Option B: $x=-1$: Denominator $(2x + 1)(x - 1)(x + 1)=( - 2 + 1)(-1 - 1)(-1 + 1)=(-1)(-2)(0)=0$. So $x=-1$ is not in the domain.
- Option C: $x=-\frac{1}{2}$: Denominator $(2\times(-\frac{1}{2})+1)(-\frac{1}{2}-1)(-\frac{1}{2}+1)=( - 1 + 1)(-\frac{3}{2})(\frac{1}{2})=0\times(-\frac{3}{2})\times\frac{1}{2}=0$. So $x=-\frac{1}{2}$ is not in the domain.
- Option D: $x = 1$: Denominator $(2\times1+1)(1 - 1)(1 + 1)=(3)(0)(2)=0$. So $x = 1$ is not in the domain.

But the question says "Select all that apply". Let's check the original problem's options. The options are A: - 3, B: - 1, C: $-\frac{1}{2}$, D:1.

Wait, let's check $x=-3$: The function has a square root $\sqrt{x + 2}$, so $x+2\geq0\Rightarrow x\geq - 2$. So $x=-3$ is not in the domain because of the square root. $x=-1$: denominator zero. $x=-\frac{1}{2}$: denominator zero. $x = 1$: denominator zero.

So all of A, B, C, D? But that can't be. Wait, maybe I made a mistake. Wait, let's re - factor the denominator:

$2x^{3}+x^{2}-2x - 1$. Let's use rational root theorem. Possible roots are $\pm1,\pm\frac{1}{2}$.

For $x = 1$: $2(1)^{3}+1^{2}-2(1)-1=2 + 1-2 - 1=0$. So $(x - 1)$ is a factor. Then we divide $2x^{3}+x^{2}-2x - 1$ by $(x - 1)$? Wait, no, when $x = 1$, the value is zero, so $(x - 1)$ is a factor. Wait, earlier factoring: $2x^{3}+x^{2}-2x - 1=x^{2}(2x + 1)-1(2x + 1)=(2x + 1)(x^{2}-1)=(2x + 1)(x - 1)(x + 1)$. Let's check for $x = 1$: $(2(1)+1)(1 - 1)(1 + 1)=3\times0\times2 = 0$. Correct. For $x=-1$: $(2(-1)+1)(-1 - 1)(-1 + 1)=(-2 + 1)(-2)(0)=(-1)\times(-2)\times0 = 0$. Correct. For $x=-\frac{1}{2}$: $(2(-\frac{1}{2})+1)(-\frac{1}{2}-1)(-\frac{1}{2}+1)=( - 1+1)(-\frac{3}{2})(\frac{1}{2})=0\times(-\frac{3}{2})\times\frac{1}{2}=0$. Correct.

For the square root: $x+2\geq0\Rightarrow x\geq - 2$. So $x=-3$: $x=-3<-2$, so $\sqrt{x + 2}$ is not real. So $x=-3$ is not in the domain. $x=-1$: denominator zero. $x=-\frac{1}{2}$: denominator zero. $x = 1$: denominator zero.

So the values not in the domain are A: - 3, B: - 1, C: $-\frac{1}{2}$, D:1. But let's check the options again. The problem says "Select all that apply". So we need to check each:

- A: $x=-3$: Not in domain (square root).
- B: $x=-1$: Not in domain (denominator zero).
- C: $x=-\frac{1}{2}$: Not in domain (denominator zero).
- D: $x = 1$: Not in domain (denominator zero).

So all of A, B, C, D are not in the domain? But that seems odd. Wait, maybe the question is which of the following values are NOT in the domain, so we have to select all that are not. So the answer is A, B, C, D? But let's check $x=-3$: the function has $\sqrt{x + 2}$, so $x+2$ must be non - negative. So $x\geq - 2$. So $x=-3$ is out. $x=-1$: denominator zero. $x=-\frac{1}{2}$: denominator zero. $x = 1$: denominator zero. So all four options are not in the domain.

Answer:

A. - 3, B. - 1, C. $-\frac{1}{2}$, D. 1