1. If $$ f(x)=\sqrt{x+2} $$ and $$ g(x)-\frac{1}{x+1} $$, then domain of $$ f=g(x)= $$ $$ \begin{array}{lll} ext { a. } \sqrt{x+\frac{1}{x+1}} & ext { b. } \sqrt{\frac{2 x+3}{x+1}} & ext { c. } \frac{1}{\sqrt{x+2}+1}\end{array} $$ d. None of these 2. The function $$ P(x)=\frac{x^{2}-1}{(x-1)(x+2)} $$ is discontinuous on $$ \begin{array}{ll} ext { a. }(-\infty,-2) \cup(1, \infty) & ext { b. }(-\infty,-2) \cup(-2,1) \cup(1, \infty) \ ext { c. }(-\infty,-2) \cup(-2,0) \cup(0,1) \cup(1, \infty) & ext { d. None of these }\end{array} $$

Question

1. If $$ f(x)=\sqrt{x+2} $$ and $$ g(x)-\frac{1}{x+1} $$, then domain of $$ f=g(x)= $$ $$ \begin{array}{lll}	ext { a. } \sqrt{x+\frac{1}{x+1}} & 	ext { b. } \sqrt{\frac{2 x+3}{x+1}} & 	ext { c. } \frac{1}{\sqrt{x+2}+1}\end{array} $$ d. None of these 2. The function $$ P(x)=\frac{x^{2}-1}{(x-1)(x+2)} $$ is discontinuous on $$ \begin{array}{ll}	ext { a. }(-\infty,-2) \cup(1, \infty) & 	ext { b. }(-\infty,-2) \cup(-2,1) \cup(1, \infty) \ 	ext { c. }(-\infty,-2) \cup(-2,0) \cup(0,1) \cup(1, \infty) & 	ext { d. None of these }\end{array} $$

UpStudy AI · Accepted Answer

Let's solve each part of the problem step by step.

### Part 1: Finding the Domain of $$ f = g(x) $$

Given:
- $$ f(x) = \sqrt{x + 2} $$
- $$ g(x) = \frac{1}{x + 1} $$

To find the domain of $$ f = g(x) $$, we need to ensure that both functions are defined.

1. **Domain of $$ f(x) $$**:
   - The expression under the square root must be non-negative:
     $$
     x + 2 \geq 0 \implies x \geq -2
     $$

2. **Domain of $$ g(x) $$**:
   - The denominator cannot be zero:
     $$
     x + 1 \neq 0 \implies x \neq -1
     $$

Combining these conditions, the domain of $$ f $$ and $$ g $$ is:
- $$ x \geq -2 $$ and $$ x \neq -1 $$

Thus, the domain in interval notation is:
$$
[-2, -1) \cup (-1, \infty)
$$

Now, we need to check which of the given options matches this domain.

### Checking the Options:
- **a.** $$ \sqrt{x + \frac{1}{x + 1}} $$: The domain will depend on the expression inside the square root being non-negative and $$ x + 1 \neq 0 $$.
- **b.** $$ \sqrt{\frac{2x + 3}{x + 1}} $$: The domain will depend on both the numerator being non-negative and the denominator not being zero.
- **c.** $$ \frac{1}{\sqrt{x + 2} + 1} $$: The denominator must be non-zero, which is always true for $$ x \geq -2 $$.
- **d.** None of these.

Let's analyze the options to find the correct one.

### Part 2: Finding Discontinuities of $$ P(x) $$

Given:
- $$ P(x) = \frac{x^2 - 1}{(x - 1)(x + 2)} $$

To find the points of discontinuity, we need to identify where the denominator is zero.

1. **Set the denominator to zero**:
   $$
   (x - 1)(x + 2) = 0
   $$
   This gives us:
   - $$ x - 1 = 0 \implies x = 1 $$
   - $$ x + 2 = 0 \implies x = -2 $$

Thus, $$ P(x) $$ is discontinuous at $$ x = 1 $$ and $$ x = -2 $$.

### Checking the Options:
- **a.** $$ (-\infty, -2) \cup (1, \infty) $$: This does not include the point of discontinuity at $$ -2 $$.
- **b.** $$ (-\infty, -2) \cup (-2, 1) \cup (1, \infty) $$: This correctly includes both points of discontinuity.
- **c.** $$ (-\infty, -2) \cup (-2, 0) \cup (0, 1) \cup (1, \infty) $$: This does not include the point of discontinuity at $$ -2 $$.
- **d.** None of these.

The correct answer for the discontinuity is option **b**.

### Summary of Answers:
1. The domain of $$ f = g(x) $$ is **d. None of these** (as none of the options match the derived domain).
2. The function $$ P(x) $$ is discontinuous on **b.** $$ (-\infty, -2) \cup (-2, 1) \cup (1, \infty) $$.
1. The domain of $$ f = g(x) $$ is **d. None of these**. 2. The function $$ P(x) $$ is discontinuous on **b.** $$ (-\infty, -2) \cup (-2, 1) \cup (1, \infty) $$.

If and , then domain of
d. None of these

The function is discontinuous on

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