QuestionFind $(f \circ g)(x)$ for $f(x)=\frac{x}{x+9}$ and $g(x)=\frac{8}{x^{2}-1}$ ; state the domain in interval notation.

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is given by $f(x)=\frac{x}{x+9}$ . The function $g(x)$ is given by $g(x)=\frac{8}{x^{}-1}$
3. We are asked to find the composition of the functions, denoted as $(f \circ g)(x)$

STEP 2

The composition of two functions, $f$ and $g$ , denoted as $(f \circ g)(x)$ , is defined as $f(g(x))$ . This means that we substitute $g(x)$ into the function $f(x)$ .

STEP 3

Substitute $g(x)$ into $f(x)$ to find $(f \circ g)(x)$ .
$(f \circ g)(x) = f(g(x)) = f\left(\frac{8}{x^{2}-1}\right)$

STEP 4

Now, replace $x$ in $f(x)$ with $g(x)$ .
$(f \circ g)(x) = \frac{\frac{8}{x^{2}-1}}{\frac{8}{x^{2}-1} +9}$

STEP 5

implify the expression by finding a common denominator for the fractions in the numerator and the denominator.
$(f \circ g)(x) = \frac{\frac{8}{x^{2}-1}}{\frac{8+9x^{2}-9}{x^{2}-1}}$

STEP 6

implify the expression further by multiplying the numerator and the denominator by $(x^{2}-1)$ to get rid of the complex fraction.
$(f \circ g)(x) = \frac{8}{8+9x^{2}-9}$

STEP 7

implify the denominator.
$(f \circ g)(x) = \frac{}{9x^{2}-1}$ This is the composition of the functions $f$ and $g$ .

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QuestionFind $(f \circ g)(x)$ for $f(x)=\frac{x}{x+9}$ and $g(x)=\frac{8}{x^{2}-1}$ ; state the domain in interval notation.

Studdy Solution

STEP 1

Assumptions1. The function $f(x)$ is given by $f(x)=\frac{x}{x+9}$ . The function $g(x)$ is given by $g(x)=\frac{8}{x^{}-1}$
3. We are asked to find the composition of the functions, denoted as $(f \circ g)(x)$

STEP 2

The composition of two functions, $f$ and $g$ , denoted as $(f \circ g)(x)$ , is defined as $f(g(x))$ . This means that we substitute $g(x)$ into the function $f(x)$ .

STEP 3

Substitute $g(x)$ into $f(x)$ to find $(f \circ g)(x)$ .
$(f \circ g)(x) = f(g(x)) = f\left(\frac{8}{x^{2}-1}\right)$

STEP 4

Now, replace $x$ in $f(x)$ with $g(x)$ .
$(f \circ g)(x) = \frac{\frac{8}{x^{2}-1}}{\frac{8}{x^{2}-1} +9}$

STEP 5

implify the expression by finding a common denominator for the fractions in the numerator and the denominator.
$(f \circ g)(x) = \frac{\frac{8}{x^{2}-1}}{\frac{8+9x^{2}-9}{x^{2}-1}}$

STEP 6

implify the expression further by multiplying the numerator and the denominator by $(x^{2}-1)$ to get rid of the complex fraction.
$(f \circ g)(x) = \frac{8}{8+9x^{2}-9}$

STEP 7

implify the denominator.
$(f \circ g)(x) = \frac{}{9x^{2}-1}$ This is the composition of the functions $f$ and $g$ .

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QuestionFind (f∘g)(x)(f \circ g)(x)(f∘g)(x) for f(x)=xx+9f(x)=\frac{x}{x+9}f(x)=x+9x​ and g(x)=8x2−1g(x)=\frac{8}{x^{2}-1}g(x)=x2−18​; state the domain in interval notation.

Studdy Solution

STEP 1

Assumptions1. The function f(x)f(x)f(x) is given by f(x)=xx+9f(x)=\frac{x}{x+9}f(x)=x+9x​ . The function g(x)g(x)g(x) is given by g(x)=8x−1g(x)=\frac{8}{x^{}-1}g(x)=x−18​3. We are asked to find the composition of the functions, denoted as (f∘g)(x)(f \circ g)(x)(f∘g)(x)

STEP 2

The composition of two functions, fff and ggg, denoted as (f∘g)(x)(f \circ g)(x)(f∘g)(x), is defined as f(g(x))f(g(x))f(g(x)). This means that we substitute g(x)g(x)g(x) into the function f(x)f(x)f(x).

STEP 3

Substitute g(x)g(x)g(x) into f(x)f(x)f(x) to find (f∘g)(x)(f \circ g)(x)(f∘g)(x).(f∘g)(x)=f(g(x))=f(8x2−1)(f \circ g)(x) = f(g(x)) = f\left(\frac{8}{x^{2}-1}\right)(f∘g)(x)=f(g(x))=f(x2−18​)

STEP 4

Now, replace xxx in f(x)f(x)f(x) with g(x)g(x)g(x).(f∘g)(x)=8x2−18x2−1+9(f \circ g)(x) = \frac{\frac{8}{x^{2}-1}}{\frac{8}{x^{2}-1} +9}(f∘g)(x)=x2−18​+9x2−18​​

STEP 5

implify the expression by finding a common denominator for the fractions in the numerator and the denominator.(f∘g)(x)=8x2−18+9x2−9x2−1(f \circ g)(x) = \frac{\frac{8}{x^{2}-1}}{\frac{8+9x^{2}-9}{x^{2}-1}}(f∘g)(x)=x2−18+9x2−9​x2−18​​

STEP 6

implify the expression further by multiplying the numerator and the denominator by (x2−1)(x^{2}-1)(x2−1) to get rid of the complex fraction.(f∘g)(x)=88+9x2−9(f \circ g)(x) = \frac{8}{8+9x^{2}-9}(f∘g)(x)=8+9x2−98​

STEP 7

implify the denominator.(f∘g)(x)=9x2−1(f \circ g)(x) = \frac{}{9x^{2}-1}(f∘g)(x)=9x2−1​This is the composition of the functions fff and ggg.

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QuestionFind $(f \circ g)(x)$ for $f(x)=\frac{x}{x+9}$ and $g(x)=\frac{8}{x^{2}-1}$ ; state the domain in interval notation.

Assumptions1. The function $f(x)$ is given by $f(x)=\frac{x}{x+9}$ . The function $g(x)$ is given by $g(x)=\frac{8}{x^{}-1}$
3. We are asked to find the composition of the functions, denoted as $(f \circ g)(x)$

The composition of two functions, $f$ and $g$ , denoted as $(f \circ g)(x)$ , is defined as $f(g(x))$ . This means that we substitute $g(x)$ into the function $f(x)$ .

Substitute $g(x)$ into $f(x)$ to find $(f \circ g)(x)$ .
$(f \circ g)(x) = f(g(x)) = f\left(\frac{8}{x^{2}-1}\right)$

Now, replace $x$ in $f(x)$ with $g(x)$ .
$(f \circ g)(x) = \frac{\frac{8}{x^{2}-1}}{\frac{8}{x^{2}-1} +9}$

implify the expression by finding a common denominator for the fractions in the numerator and the denominator.
$(f \circ g)(x) = \frac{\frac{8}{x^{2}-1}}{\frac{8+9x^{2}-9}{x^{2}-1}}$

implify the expression further by multiplying the numerator and the denominator by $(x^{2}-1)$ to get rid of the complex fraction.
$(f \circ g)(x) = \frac{8}{8+9x^{2}-9}$

implify the denominator.
$(f \circ g)(x) = \frac{}{9x^{2}-1}$ This is the composition of the functions $f$ and $g$ .