Find f(g(x)) f(x)=1/(x^2+1) , g(x)=x^-6

$f (x) = \frac{1}{x^{2} + 1}$ , $g (x) = x^{- 6}$

Step 1

Set up the composite result function.

$f (g (x))$

Step 2

Evaluate $f (x^{- 6})$ by substituting in the value of $g$ into $f$ .

$f (x^{- 6}) = \frac{1}{{(x^{- 6})}^{2} + 1}$

Step 3

Simplify the denominator.

Multiply the exponents in ${(x^{- 6})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (x^{- 6}) = \frac{1}{x^{- 6 \cdot 2} + 1}$

Multiply $- 6$ by $2$ .

$f (x^{- 6}) = \frac{1}{x^{- 12} + 1}$

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$f (x^{- 6}) = \frac{1}{\frac{1}{x^{12}} + 1}$

Rewrite $\frac{1}{x^{12}} + 1$ in a factored form.

Rewrite $1$ as $1^{3}$ .

$f (x^{- 6}) = \frac{1}{\frac{1^{3}}{x^{12}} + 1}$

Rewrite $x^{12}$ as ${(x^{4})}^{3}$ .

$f (x^{- 6}) = \frac{1}{\frac{1^{3}}{{(x^{4})}^{3}} + 1}$

Rewrite $\frac{1^{3}}{{(x^{4})}^{3}}$ as ${(\frac{1}{x^{4}})}^{3}$ .

$f (x^{- 6}) = \frac{1}{{(\frac{1}{x^{4}})}^{3} + 1}$

Rewrite $1$ as $1^{3}$ .

$f (x^{- 6}) = \frac{1}{{(\frac{1}{x^{4}})}^{3} + 1^{3}}$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = \frac{1}{x^{4}}$ and $b = 1$ .

$f (x^{- 6}) = \frac{1}{(\frac{1}{x^{4}} + 1) ({(\frac{1}{x^{4}})}^{2} - \frac{1}{x^{4}} \cdot 1 + 1^{2})}$

Simplify.

Apply the product rule to $\frac{1}{x^{4}}$ .

$f (x^{- 6}) = \frac{1}{(\frac{1}{x^{4}} + 1) (\frac{1^{2}}{{(x^{4})}^{2}} - \frac{1}{x^{4}} \cdot 1 + 1^{2})}$

One to any power is one.

$f (x^{- 6}) = \frac{1}{(\frac{1}{x^{4}} + 1) (\frac{1}{{(x^{4})}^{2}} - \frac{1}{x^{4}} \cdot 1 + 1^{2})}$

Multiply the exponents in ${(x^{4})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f (x^{- 6}) = \frac{1}{(\frac{1}{x^{4}} + 1) (\frac{1}{x^{4 \cdot 2}} - \frac{1}{x^{4}} \cdot 1 + 1^{2})}$

Multiply $4$ by $2$ .

$f (x^{- 6}) = \frac{1}{(\frac{1}{x^{4}} + 1) (\frac{1}{x^{8}} - \frac{1}{x^{4}} \cdot 1 + 1^{2})}$

Multiply $- 1$ by $1$ .

$f (x^{- 6}) = \frac{1}{(\frac{1}{x^{4}} + 1) (\frac{1}{x^{8}} - \frac{1}{x^{4}} + 1^{2})}$

One to any power is one.

$f (x^{- 6}) = \frac{1}{(\frac{1}{x^{4}} + 1) (\frac{1}{x^{8}} - \frac{1}{x^{4}} + 1)}$

Reorder terms.

$f (x^{- 6}) = \frac{1}{(\frac{1}{x^{4}} + 1) (- \frac{1}{x^{4}} + \frac{1}{x^{8}} + 1)}$

Write $1$ as a fraction with a common denominator.

$f (x^{- 6}) = \frac{1}{(\frac{1}{x^{4}} + \frac{x^{4}}{x^{4}}) (- \frac{1}{x^{4}} + \frac{1}{x^{8}} + 1)}$

Combine the numerators over the common denominator.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot (- \frac{1}{x^{4}} + \frac{1}{x^{8}} + 1)}$

To write $- \frac{1}{x^{4}}$ as a fraction with a common denominator, multiply by $\frac{x^{4}}{x^{4}}$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot (- \frac{1}{x^{4}} \cdot \frac{x^{4}}{x^{4}} + \frac{1}{x^{8}} + 1)}$

Write each expression with a common denominator of $x^{8}$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{x^{4}}$ by $\frac{x^{4}}{x^{4}}$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot (- \frac{x^{4}}{x^{4} x^{4}} + \frac{1}{x^{8}} + 1)}$

Multiply $x^{4}$ by $x^{4}$ by adding the exponents.

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot (- \frac{x^{4}}{x^{4 + 4}} + \frac{1}{x^{8}} + 1)}$

Add $4$ and $4$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot (- \frac{x^{4}}{x^{8}} + \frac{1}{x^{8}} + 1)}$

Combine the numerators over the common denominator.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot (\frac{- x^{4} + 1}{x^{8}} + 1)}$

Simplify the numerator.

Rewrite $1$ as $1^{2}$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot (\frac{- x^{4} + 1^{2}}{x^{8}} + 1)}$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot (\frac{- {(x^{2})}^{2} + 1^{2}}{x^{8}} + 1)}$

Reorder $- {(x^{2})}^{2}$ and $1^{2}$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot (\frac{1^{2} - {(x^{2})}^{2}}{x^{8}} + 1)}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x^{2}$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot (\frac{(1 + x^{2}) (1 - x^{2})}{x^{8}} + 1)}$

Simplify.

Rewrite $1$ as $1^{2}$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot (\frac{(1 + x^{2}) (1^{2} - x^{2})}{x^{8}} + 1)}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = x$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot (\frac{(1 + x^{2}) (1 + x) (1 - x)}{x^{8}} + 1)}$

Write $1$ as a fraction with a common denominator.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot (\frac{(1 + x^{2}) (1 + x) (1 - x)}{x^{8}} + \frac{x^{8}}{x^{8}})}$

Combine the numerators over the common denominator.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{(1 + x^{2}) (1 + x) (1 - x) + x^{8}}{x^{8}}}$

Simplify the numerator.

Expand $(1 + x^{2}) (1 + x)$ using the FOIL Method.

Apply the distributive property.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{(1 (1 + x) + x^{2} (1 + x)) (1 - x) + x^{8}}{x^{8}}}$

Apply the distributive property.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{(1 \cdot 1 + 1 x + x^{2} (1 + x)) (1 - x) + x^{8}}{x^{8}}}$

Apply the distributive property.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{(1 \cdot 1 + 1 x + x^{2} \cdot 1 + x^{2} x) (1 - x) + x^{8}}{x^{8}}}$

Simplify each term.

Multiply $1$ by $1$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{(1 + 1 x + x^{2} \cdot 1 + x^{2} x) (1 - x) + x^{8}}{x^{8}}}$

Multiply $x$ by $1$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{(1 + x + x^{2} \cdot 1 + x^{2} x) (1 - x) + x^{8}}{x^{8}}}$

Multiply $x^{2}$ by $1$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{(1 + x + x^{2} + x^{2} x) (1 - x) + x^{8}}{x^{8}}}$

Multiply $x^{2}$ by $x$ by adding the exponents.

Multiply $x^{2}$ by $x$ .

Raise $x$ to the power of $1$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{(1 + x + x^{2} + x^{2} x) (1 - x) + x^{8}}{x^{8}}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{(1 + x + x^{2} + x^{2 + 1}) (1 - x) + x^{8}}{x^{8}}}$

Add $2$ and $1$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{(1 + x + x^{2} + x^{3}) (1 - x) + x^{8}}{x^{8}}}$

Expand $(1 + x + x^{2} + x^{3}) (1 - x)$ by multiplying each term in the first expression by each term in the second expression.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 \cdot 1 + 1 (- x) + x \cdot 1 + x (- x) + x^{2} \cdot 1 + x^{2} (- x) + x^{3} \cdot 1 + x^{3} (- x) + x^{8}}{x^{8}}}$

Simplify each term.

Multiply $1$ by $1$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 + 1 (- x) + x \cdot 1 + x (- x) + x^{2} \cdot 1 + x^{2} (- x) + x^{3} \cdot 1 + x^{3} (- x) + x^{8}}{x^{8}}}$

Multiply $- x$ by $1$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x \cdot 1 + x (- x) + x^{2} \cdot 1 + x^{2} (- x) + x^{3} \cdot 1 + x^{3} (- x) + x^{8}}{x^{8}}}$

Multiply $x$ by $1$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x + x (- x) + x^{2} \cdot 1 + x^{2} (- x) + x^{3} \cdot 1 + x^{3} (- x) + x^{8}}{x^{8}}}$

Rewrite using the commutative property of multiplication.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - x \cdot x + x^{2} \cdot 1 + x^{2} (- x) + x^{3} \cdot 1 + x^{3} (- x) + x^{8}}{x^{8}}}$

Multiply $x$ by $x$ by adding the exponents.

Move $x$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - (x \cdot x) + x^{2} \cdot 1 + x^{2} (- x) + x^{3} \cdot 1 + x^{3} (- x) + x^{8}}{x^{8}}}$

Multiply $x$ by $x$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - x^{2} + x^{2} \cdot 1 + x^{2} (- x) + x^{3} \cdot 1 + x^{3} (- x) + x^{8}}{x^{8}}}$

Multiply $x^{2}$ by $1$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - x^{2} + x^{2} + x^{2} (- x) + x^{3} \cdot 1 + x^{3} (- x) + x^{8}}{x^{8}}}$

Rewrite using the commutative property of multiplication.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - x^{2} + x^{2} - x^{2} x + x^{3} \cdot 1 + x^{3} (- x) + x^{8}}{x^{8}}}$

Multiply $x^{2}$ by $x$ by adding the exponents.

Move $x$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - x^{2} + x^{2} - (x \cdot x^{2}) + x^{3} \cdot 1 + x^{3} (- x) + x^{8}}{x^{8}}}$

Multiply $x$ by $x^{2}$ .

Raise $x$ to the power of $1$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - x^{2} + x^{2} - (x \cdot x^{2}) + x^{3} \cdot 1 + x^{3} (- x) + x^{8}}{x^{8}}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - x^{2} + x^{2} - x^{1 + 2} + x^{3} \cdot 1 + x^{3} (- x) + x^{8}}{x^{8}}}$

Add $1$ and $2$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - x^{2} + x^{2} - x^{3} + x^{3} \cdot 1 + x^{3} (- x) + x^{8}}{x^{8}}}$

Multiply $x^{3}$ by $1$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - x^{2} + x^{2} - x^{3} + x^{3} + x^{3} (- x) + x^{8}}{x^{8}}}$

Rewrite using the commutative property of multiplication.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - x^{2} + x^{2} - x^{3} + x^{3} - x^{3} x + x^{8}}{x^{8}}}$

Multiply $x^{3}$ by $x$ by adding the exponents.

Move $x$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - x^{2} + x^{2} - x^{3} + x^{3} - (x \cdot x^{3}) + x^{8}}{x^{8}}}$

Multiply $x$ by $x^{3}$ .

Raise $x$ to the power of $1$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - x^{2} + x^{2} - x^{3} + x^{3} - (x \cdot x^{3}) + x^{8}}{x^{8}}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - x^{2} + x^{2} - x^{3} + x^{3} - x^{1 + 3} + x^{8}}{x^{8}}}$

Add $1$ and $3$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x + x - x^{2} + x^{2} - x^{3} + x^{3} - x^{4} + x^{8}}{x^{8}}}$

Combine the opposite terms in $1 - x + x - x^{2} + x^{2} - x^{3} + x^{3} - x^{4}$ .

Add $- x$ and $x$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 + 0 - x^{2} + x^{2} - x^{3} + x^{3} - x^{4} + x^{8}}{x^{8}}}$

Add $1$ and $0$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x^{2} + x^{2} - x^{3} + x^{3} - x^{4} + x^{8}}{x^{8}}}$

Add $- x^{2}$ and $x^{2}$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 + 0 - x^{3} + x^{3} - x^{4} + x^{8}}{x^{8}}}$

Add $1$ and $0$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x^{3} + x^{3} - x^{4} + x^{8}}{x^{8}}}$

Add $- x^{3}$ and $x^{3}$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 + 0 - x^{4} + x^{8}}{x^{8}}}$

Add $1$ and $0$ .

$f (x^{- 6}) = \frac{1}{\frac{1 + x^{4}}{x^{4}} \cdot \frac{1 - x^{4} + x^{8}}{x^{8}}}$

Step 4

Multiply $\frac{1 + x^{4}}{x^{4}}$ by $\frac{1 - x^{4} + x^{8}}{x^{8}}$ .

$f (x^{- 6}) = \frac{1}{\frac{(1 + x^{4}) (1 - x^{4} + x^{8})}{x^{4} x^{8}}}$

Step 5

Multiply $x^{4}$ by $x^{8}$ by adding the exponents.

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f (x^{- 6}) = \frac{1}{\frac{(1 + x^{4}) (1 - x^{4} + x^{8})}{x^{4 + 8}}}$

Add $4$ and $8$ .

$f (x^{- 6}) = \frac{1}{\frac{(1 + x^{4}) (1 - x^{4} + x^{8})}{x^{12}}}$

Step 6

Multiply the numerator by the reciprocal of the denominator.

$f (x^{- 6}) = 1 (\frac{x^{12}}{(1 + x^{4}) (1 - x^{4} + x^{8})})$

Step 7

Multiply $\frac{x^{12}}{(1 + x^{4}) (1 - x^{4} + x^{8})}$ by $1$ .

$f (x^{- 6}) = \frac{x^{12}}{(1 + x^{4}) (1 - x^{4} + x^{8})}$

Find f(g(x)) f(x)=1/(x^2+1) , g(x)=x^-6

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