find $f \circ g$ and $g \circ f$.
$ f(x) = \dfrac{7}{x^2 - 1},\ g(x) = x + 1 $
(a) $ f \circ g $
$ \dfrac{7}{x(x + 2)} $
(b) $ g \circ f $
$ \dfrac{7}{x^2 - 1} + 1 $
find the domain of each function and each composite function. (enter your answers using interval notation.)
domain of $f$
domain of $g$ $ (-\infty, \infty) $
domain of $f \circ g$
domain of $g \circ f$
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Accepted Answer

### Part (b): Find $ g \circ f $
# Explanation:
## Step 1: Recall the definition of function composition
The composition $ g \circ f $ means $ g(f(x)) $. So we substitute $ f(x) $ into $ g(x) $. Given $ g(x)=x + 1 $ and $ f(x)=\frac{7}{x^{2}-1} $, we replace $ x $ in $ g(x) $ with $ f(x) $.

## Step 2: Substitute $ f(x) $ into $ g(x) $
$ g(f(x))=f(x)+1 $

Substitute $ f(x)=\frac{7}{x^{2}-1} $ into the above equation:

$ g(f(x))=\frac{7}{x^{2}-1}+1 $

Now, we simplify this expression. First, find a common denominator, which is $ x^{2}-1 $:

$ g(f(x))=\frac{7+(x^{2}-1)}{x^{2}-1} $

Simplify the numerator: $ 7+(x^{2}-1)=x^{2}+6 $

Wait, no, wait. Wait, $ g(x)=x + 1 $, so when we substitute $ f(x) $ into $ g(x) $, it's $ \frac{7}{x^{2}-1}+1 $, but let's check again. Wait, maybe I made a mistake. Wait, $ f(x)=\frac{7}{x^{2}-1} $, so $ g(f(x))=f(x)+1=\frac{7}{x^{2}-1}+1 $. To combine these terms, we can write $ 1=\frac{x^{2}-1}{x^{2}-1} $, so:

$ g(f(x))=\frac{7+(x^{2}-1)}{x^{2}-1}=\frac{x^{2}+6}{x^{2}-1} $? Wait, no, that's not right. Wait, no, $ x^{2}-1+7=x^{2}+6 $? Wait, no, $ 7+(x^{2}-1)=x^{2}+6 $? Wait, 7 - 1 is 6, so yes. But wait, maybe the original problem's (b) part was marked wrong because of simplification. Wait, let's re-express:

$ g \circ f(x)=g(f(x))=f(x)+1=\frac{7}{x^{2}-1}+1 $

To combine the fractions:

$ \frac{7}{x^{2}-1}+\frac{x^{2}-1}{x^{2}-1}=\frac{7 + x^{2}-1}{x^{2}-1}=\frac{x^{2}+6}{x^{2}-1} $? Wait, but maybe the problem expects it in a different form? Wait, no, maybe I messed up the composition. Wait, $ g(x)=x + 1 $, so $ g(f(x))=f(x)+1 $, which is $ \frac{7}{x^{2}-1}+1 $, but let's check the domain first, but the question here is to find $ g \circ f $. Wait, maybe the initial answer was wrong. Wait, let's do it again.

Wait, $ f(x)=\frac{7}{x^{2}-1} $, $ g(x)=x + 1 $. So $ g \circ f(x)=g(f(x))=f(x)+1=\frac{7}{x^{2}-1}+1 $. To simplify:

$ \frac{7 + (x^{2}-1)}{x^{2}-1}=\frac{x^{2}+6}{x^{2}-1} $? Wait, 7 - 1 is 6, so numerator is $ x^{2}+6 $, denominator is $ x^{2}-1=(x - 1)(x + 1) $. So $ g \circ f(x)=\frac{x^{2}+6}{(x - 1)(x + 1)} $ or $ \frac{7}{x^{2}-1}+1 $. But maybe the problem's (b) part was marked wrong because of a miscalculation. Wait, let's check the original problem's (b) part: the user had $ \frac{7}{x^{2}-1}+1 $ marked wrong. So maybe we need to simplify it correctly.

Wait, $ g(x)=x + 1 $, so $ g(f(x))=f(x)+1=\frac{7}{x^{2}-1}+1 $. Let's combine the terms:

$ \frac{7 + (x^{2}-1)}{x^{2}-1}=\frac{x^{2}+6}{x^{2}-1} $. Wait, but maybe the problem expects a different form? Wait, no, maybe I made a mistake in the composition. Wait, no, $ g \circ f $ is $ g(f(x)) $, so substituting $ f(x) $ into $ g(x) $, which is $ x + 1 $, so replacing $ x $ with $ f(x) $ gives $ f(x)+1 $, which is $ \frac{7}{x^{2}-1}+1 $. So that's correct, but maybe the system expects it simplified. Let's check:

$ \frac{7}{x^{2}-1}+1=\frac{7 + x^{2}-1}{x^{2}-1}=\frac{x^{2}+6}{x^{2}-1} $. So that's the simplified form.

### Domain of $ f $
# Explanation:
## Step 1: Recall the domain of a rational function
The function $ f(x)=\frac{7}{x^{2}-1} $ is a rational function, so its domain is all real numbers except where the denominator is zero.

## Step 2: Find where the denominator is zero
Set $ x^{2}-1 = 0 $

Solve for $ x $: $ x^{2}=1 $ so $ x = 1 $ or $ x=-1 $

Thus, the domain of $ f $ is all real numbers except $ x = 1 $ and $ x=-1 $. In interval notation, this is $ (-\infty, -1) \cup (-1, 1) \cup (1, \infty) $

### Domain of $ f \circ g $
# Explanation:
## Step 1: Recall the domain of a composite function $ f \circ g $
The domain of $ f \circ g $ is the set of all $ x $ in the domain of $ g $ such that $ g(x) $ is in the domain of $ f $.

## Step 2: Find the domain of $ g $
The function $ g(x)=x + 1 $ is a linear function, so its domain is $ (-\infty, \infty) $ (which we already know is correct).

## Step 3: Find $ x $ such that $ g(x) $ is in the domain of $ f $
The domain of $ f $ is all real numbers except $ x = 1 $ and $ x=-1 $. So we need $ g(x)
eq1 $ and $ g(x)
eq -1 $

Set $ g(x)=1 $: $ x + 1 = 1 $ ⇒ $ x = 0 $

Set $ g(x)=-1 $: $ x + 1 = -1 $ ⇒ $ x=-2 $

Thus, $ x $ cannot be 0 or -2 (because $ g(0)=1 $ and $ g(-2)=-1 $, which are not in the domain of $ f $). Also, we need to ensure that $ g(x) $ is defined, but $ g(x) $ is defined for all $ x $, so the domain of $ f \circ g $ is all real numbers except $ x=-2 $ and $ x = 0 $. In interval notation, this is $ (-\infty, -2) \cup (-2, 0) \cup (0, \infty) $

### Domain of $ g \circ f $
# Explanation:
## Step 1: Recall the domain of a composite function $ g \circ f $
The domain of $ g \circ f $ is the set of all $ x $ in the domain of $ f $ such that $ f(x) $ is in the domain of $ g $.

## Step 2: Domain of $ f $
We already found that the domain of $ f $ is $ (-\infty, -1) \cup (-1, 1) \cup (1, \infty) $

## Step 3: Domain of $ g $
The function $ g(x)=x + 1 $ is defined for all real numbers, so $ f(x) $ (which is a real number except when the denominator is zero) is in the domain of $ g $ as long as $ f(x) $ is defined. Wait, no: $ g(x) $ is defined for all real numbers, so as long as $ f(x) $ is defined (i.e., $ x
eq\pm1 $), then $ g(f(x)) $ is defined. Wait, but we also need to check if $ g(f(x)) $ has any additional restrictions. Wait, $ g(f(x))=\frac{7}{x^{2}-1}+1=\frac{x^{2}+6}{x^{2}-1} $ (from earlier simplification). The denominator here is $ x^{2}-1=(x - 1)(x + 1) $, so we need to exclude $ x = 1 $ and $ x=-1 $ (since those make the denominator zero). So the domain of $ g \circ f $ is the same as the domain of $ f $ excluding any additional points? Wait, no: the domain of $ g \circ f $ is the set of $ x $ such that $ x $ is in the domain of $ f $ (so $ x
eq\pm1 $) and $ f(x) $ is in the domain of $ g $. But $ g $ is defined for all real numbers, so $ f(x) $ is a real number (since $ x
eq\pm1 $, $ f(x) $ is defined and real), so the domain of $ g \circ f $ is the domain of $ f $, which is $ (-\infty, -1) \cup (-1, 1) \cup (1, \infty) $. Wait, but let's check the simplified form of $ g \circ f(x)=\frac{x^{2}+6}{(x - 1)(x + 1)} $. The denominator is zero when $ x = 1 $ or $ x=-1 $, so the domain is $ (-\infty, -1) \cup (-1, 1) \cup (1, \infty) $

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### Final Answers:

#### (b) $ g \circ f $
# Explanation:
To find $ g \circ f $, we substitute $ f(x) $ into $ g(x) $. Given $ g(x) = x + 1 $ and $ f(x) = \frac{7}{x^2 - 1} $, we have:
$ g \circ f(x) = g(f(x)) = f(x) + 1 = \frac{7}{x^2 - 1} + 1 $.
Combining the fractions over a common denominator ($ x^2 - 1 $):
$ \frac{7 + (x^2 - 1)}{x^2 - 1} = \frac{x^2 + 6}{x^2 - 1} $ (or equivalently, $ \frac{7}{x^2 - 1} + 1 $ is also correct, but simplifying gives $ \frac{x^2 + 6}{(x - 1)(x + 1)} $).

# Answer:
$ \frac{x^2 + 6}{x^2 - 1} $ (or $ \frac{7}{x^2 - 1} + 1 $)

#### Domain of $ f $
# Explanation:
The function $ f(x) = \frac{7}{x^2 - 1} $ is a rational function, so its domain excludes values where the denominator is zero. Solving $ x^2 - 1 = 0 $ gives $ x = \pm 1 $. Thus, the domain is all real numbers except $ x = 1 $ and $ x = -1 $.

# Answer:
$ (-\infty, -1) \cup (-1, 1) \cup (1, \infty) $

#### Domain of $ f \circ g $
# Explanation:
The domain of $ f \circ g $ requires $ x $ to be in the domain of $ g $ (all real numbers) and $ g(x) $ to be in the domain of $ f $ (i.e., $ g(x)
eq \pm 1 $). Solving $ g(x) = x + 1 = 1 $ gives $ x = 0 $, and $ g(x) = x + 1 = -1 $ gives $ x = -2 $. Thus, $ x
eq -2, 0 $, and the domain is all real numbers except $ -2 $ and $ 0 $.

# Answer:
$ (-\infty, -2) \cup (-2, 0) \cup (0, \infty) $

#### Domain of $ g \circ f $
# Explanation:
The domain of $ g \circ f $ requires $ x $ to be in the domain of $ f $ (i.e., $ x
eq \pm 1 $) and $ f(x) $ to be in the domain of $ g $ (all real numbers, which is satisfied since $ f(x) $ is real for $ x
eq \pm 1 $). Thus, the domain is the same as the domain of $ f $, excluding $ x = \pm 1 $.

# Answer:
$ (-\infty, -1) \cup (-1, 1) \cup (1, \infty) $

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QUESTION IMAGE

Question

Explanation:

Part (b): Find \( g \circ f \)

Step 1: Recall the definition of function composition

Step 2: Substitute \( f(x) \) into \( g(x) \)

Domain of \( f \)

Step 1: Recall the domain of a rational function

Step 2: Find where the denominator is zero

Domain of \( f \circ g \)

Step 1: Recall the domain of a composite function \( f \circ g \)

Step 2: Find the domain of \( g \)

Step 3: Find \( x \) such that \( g(x) \) is in the domain of \( f \)

Domain of \( g \circ f \)

Answer:

Domain of \( f \)