Math / Algebra

QuestionFor the functions $f(x)=\frac{3}{x+4}$ and $g(x)=\frac{7}{x+1}$ , find the composition $f \circ g$ and simplify your answer as much as possible. Write the domain using interval notation. $(f \circ g)(x)=$ $\square$
Domain of $f \circ g$ : $\square$

Studdy Solution

STEP 1

What is this asking? We're asked to find a new function that first applies $g(x)$ and then applies $f(x)$ to the result, and then to figure out which $x$ values are allowed. Watch out! Remember that the domain of the composition $f \circ g$ not only depends on the domain of $f$ , but also on the range of $g$ that can be used as input for $f$ .

STEP 2

1. Compute the composition
2. Simplify the composition
3. Find the domain

STEP 3

Let's start by writing down the definition of function composition.
The composition $f \circ g$ means we apply $g$ first, and then apply $f$ to the result.
So, $(f \circ g)(x) = f(g(x))$ .

STEP 4

We know that $g(x) = \frac{7}{x+1}$ .
So, we substitute that into $f(x)$ .
This gives us $f(g(x)) = f\left(\frac{7}{x+1}\right)$ .

STEP 5

Now, we know $f(x) = \frac{3}{x+4}$ .
So, to find $f\left(\frac{7}{x+1}\right)$ , we replace every $x$ in the formula for $f(x)$ with $\frac{7}{x+1}$ .
This gives us: $f\left(\frac{7}{x+1}\right) = \frac{3}{\frac{7}{x+1} + 4}.$

STEP 6

To simplify the expression, we want to get rid of the fraction in the denominator.
We can do this by multiplying the numerator and denominator by $(x+1)$ .
Remember, multiplying by $\frac{x+1}{x+1}$ is like multiplying by $1$ , so it doesn't change the value of the expression! $\frac{3}{\frac{7}{x+1} + 4} \cdot \frac{x+1}{x+1} = \frac{3(x+1)}{7 + 4(x+1)}.$

STEP 7

Now, we distribute the $4$ in the denominator to get: $\frac{3(x+1)}{7 + 4x + 4} = \frac{3(x+1)}{4x + 11}.$

STEP 8

The domain of $f \circ g$ is the set of all $x$ values for which $(f \circ g)(x)$ is defined.

STEP 9

First, we look at the original function $g(x)$ . $g(x)$ is undefined when $x+1=0$ , which means $x = -1$ .
So, $x$ cannot be $-1$ .

STEP 10

Next, we look at $f(g(x))$ .
We found that $(f \circ g)(x) = \frac{3(x+1)}{4x+11}$ .
This expression is undefined when the denominator is zero.
So, we set $4x+11=0$ and solve for $x$ .
This gives us $4x = -11$ , so $x = -\frac{11}{4}$ .

STEP 11

Therefore, the domain of $f \circ g$ is all real numbers except $-1$ and $-\frac{11}{4}$ .
In interval notation, this is $(-\infty, -11/4) \cup (-11/4, -1) \cup (-1, \infty)$ .
We use round brackets because the endpoints are not included.

STEP 12

$(f \circ g)(x) = \frac{3(x+1)}{4x+11}$ .
Domain of $f \circ g$ : $(-\infty, -11/4) \cup (-11/4, -1) \cup (-1, \infty)$ .

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QuestionFor the functions $f(x)=\frac{3}{x+4}$ and $g(x)=\frac{7}{x+1}$ , find the composition $f \circ g$ and simplify your answer as much as possible. Write the domain using interval notation. $(f \circ g)(x)=$ $\square$
Domain of $f \circ g$ : $\square$

Studdy Solution

STEP 1

STEP 2

1. Compute the composition
2. Simplify the composition
3. Find the domain

STEP 3

Let's start by writing down the definition of function composition.
The composition $f \circ g$ means we apply $g$ first, and then apply $f$ to the result.
So, $(f \circ g)(x) = f(g(x))$ .

STEP 4

We know that $g(x) = \frac{7}{x+1}$ .
So, we substitute that into $f(x)$ .
This gives us $f(g(x)) = f\left(\frac{7}{x+1}\right)$ .

STEP 5

Now, we know $f(x) = \frac{3}{x+4}$ .
So, to find $f\left(\frac{7}{x+1}\right)$ , we replace every $x$ in the formula for $f(x)$ with $\frac{7}{x+1}$ .
This gives us: $f\left(\frac{7}{x+1}\right) = \frac{3}{\frac{7}{x+1} + 4}.$

STEP 6

STEP 7

Now, we distribute the $4$ in the denominator to get: $\frac{3(x+1)}{7 + 4x + 4} = \frac{3(x+1)}{4x + 11}.$

STEP 8

The domain of $f \circ g$ is the set of all $x$ values for which $(f \circ g)(x)$ is defined.

STEP 9

First, we look at the original function $g(x)$ . $g(x)$ is undefined when $x+1=0$ , which means $x = -1$ .
So, $x$ cannot be $-1$ .

STEP 10

Next, we look at $f(g(x))$ .
We found that $(f \circ g)(x) = \frac{3(x+1)}{4x+11}$ .
This expression is undefined when the denominator is zero.
So, we set $4x+11=0$ and solve for $x$ .
This gives us $4x = -11$ , so $x = -\frac{11}{4}$ .

STEP 11

Therefore, the domain of $f \circ g$ is all real numbers except $-1$ and $-\frac{11}{4}$ .
In interval notation, this is $(-\infty, -11/4) \cup (-11/4, -1) \cup (-1, \infty)$ .
We use round brackets because the endpoints are not included.

STEP 12

$(f \circ g)(x) = \frac{3(x+1)}{4x+11}$ .
Domain of $f \circ g$ : $(-\infty, -11/4) \cup (-11/4, -1) \cup (-1, \infty)$ .

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Studdy Solution

STEP 1

STEP 2

1. Compute the composition2. Simplify the composition3. Find the domain

STEP 3

Let's **start** by writing down the definition of function composition.The composition f∘gf \circ gf∘g means we apply ggg first, and *then* apply fff to the result.So, (f∘g)(x)=f(g(x))(f \circ g)(x) = f(g(x))(f∘g)(x)=f(g(x)).

STEP 4

We know that g(x)=7x+1g(x) = \frac{7}{x+1}g(x)=x+17​.So, we **substitute** that into f(x)f(x)f(x).This gives us f(g(x))=f(7x+1)f(g(x)) = f\left(\frac{7}{x+1}\right)f(g(x))=f(x+17​).

STEP 5

STEP 6

STEP 7

Now, we **distribute** the 444 in the denominator to get: 3(x+1)7+4x+4=3(x+1)4x+11. \frac{3(x+1)}{7 + 4x + 4} = \frac{3(x+1)}{4x + 11}. 7+4x+43(x+1)​=4x+113(x+1)​.

STEP 8

The domain of f∘gf \circ gf∘g is the set of all xxx values for which (f∘g)(x)(f \circ g)(x)(f∘g)(x) is defined.

STEP 9

First, we look at the **original** function g(x)g(x)g(x). g(x)g(x)g(x) is undefined when x+1=0x+1=0x+1=0, which means x=−1x = -1x=−1.So, xxx cannot be −1-1−1.

STEP 10

STEP 11

STEP 12

(f∘g)(x)=3(x+1)4x+11(f \circ g)(x) = \frac{3(x+1)}{4x+11}(f∘g)(x)=4x+113(x+1)​.Domain of f∘gf \circ gf∘g: (−∞,−11/4)∪(−11/4,−1)∪(−1,∞)(-\infty, -11/4) \cup (-11/4, -1) \cup (-1, \infty)(−∞,−11/4)∪(−11/4,−1)∪(−1,∞).

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1. Compute the composition
2. Simplify the composition
3. Find the domain

Let's start by writing down the definition of function composition.
The composition $f \circ g$ means we apply $g$ first, and then apply $f$ to the result.
So, $(f \circ g)(x) = f(g(x))$ .

We know that $g(x) = \frac{7}{x+1}$ .
So, we substitute that into $f(x)$ .
This gives us $f(g(x)) = f\left(\frac{7}{x+1}\right)$ .

Now, we distribute the $4$ in the denominator to get: $\frac{3(x+1)}{7 + 4x + 4} = \frac{3(x+1)}{4x + 11}.$

The domain of $f \circ g$ is the set of all $x$ values for which $(f \circ g)(x)$ is defined.

First, we look at the original function $g(x)$ . $g(x)$ is undefined when $x+1=0$ , which means $x = -1$ .
So, $x$ cannot be $-1$ .

$(f \circ g)(x) = \frac{3(x+1)}{4x+11}$ .
Domain of $f \circ g$ : $(-\infty, -11/4) \cup (-11/4, -1) \cup (-1, \infty)$ .