Evaluate f^-1((3+x)/(x-2))=x+1

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

Step 1

Combine $f^{- 1}$ and $\frac{3 + x}{x - 2}$ .

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

Step 2

Multiply each term in $\frac{f^{- 1} (3 + x)}{x - 2} = x + 1$ by $\frac{x - 2}{3 + x}$ to eliminate the fractions.

Multiply each term in $\frac{f^{- 1} (3 + x)}{x - 2} = x + 1$ by $\frac{x - 2}{3 + x}$ .

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

Simplify the left side.

Cancel the common factor of $3 + x$ .

Factor $3 + x$ out of $f^{- 1} (3 + x)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x - 2$ .

Cancel the common factor.

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

Rewrite the expression.

$f^{- 1} = x \frac{x - 2}{3 + x} + 1 \frac{x - 2}{3 + x}$

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

$\frac{1}{f} = x \frac{x - 2}{3 + x} + 1 \frac{x - 2}{3 + x}$

Simplify the right side.

Simplify terms.

Simplify each term.

Combine $x$ and $\frac{x - 2}{3 + x}$ .

$\frac{1}{f} = \frac{x (x - 2)}{3 + x} + 1 \frac{x - 2}{3 + x}$

Multiply $\frac{x - 2}{3 + x}$ by $1$ .

$\frac{1}{f} = \frac{x (x - 2)}{3 + x} + \frac{x - 2}{3 + x}$

Combine the numerators over the common denominator.

$\frac{1}{f} = \frac{x (x - 2) + x - 2}{x + 3}$

Simplify each term.

Apply the distributive property.

$\frac{1}{f} = \frac{x \cdot x + x \cdot - 2 + x - 2}{x + 3}$

Multiply $x$ by $x$ .

$\frac{1}{f} = \frac{x^{2} + x \cdot - 2 + x - 2}{x + 3}$

Move $- 2$ to the left of $x$ .

$\frac{1}{f} = \frac{x^{2} - 2 x + x - 2}{x + 3}$

Add $- 2 x$ and $x$ .

$\frac{1}{f} = \frac{x^{2} - x - 2}{x + 3}$

Factor $x^{2} - x - 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Write the factored form using these integers.

$\frac{1}{f} = \frac{(x - 2) (x + 1)}{x + 3}$

Step 3

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$f, x + 3$

Since $f, x + 3$ contain both numbers and variables, there are four steps to find the LCM. Find LCM for the numeric, variable, and compound variable parts. Then, multiply them all together.

Steps to find the LCM for $f, x + 3$ are:

1. Find the LCM for the numeric part $1, 1$ .

2. Find the LCM for the variable part $f^{1}$ .

3. Find the LCM for the compound variable part $x + 3$ .

4. Multiply each LCM together.

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

The number $1$ is not a prime number because it only has one positive factor, which is itself.

Not prime

The LCM of $1, 1$ is the result of multiplying all prime factors the greatest number of times they occur in either number.

$1$

The factor for $f^{1}$ is $f$ itself.

$f^{1} = f$

$f$ occurs $1$ time.

The LCM of $f^{1}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$f$

The factor for $x + 3$ is $x + 3$ itself.

$(x + 3) = x + 3$

$(x + 3)$ occurs $1$ time.

The LCM of $x + 3$ is the result of multiplying all factors the greatest number of times they occur in either term.

$x + 3$

The Least Common Multiple $LCM$ of some numbers is the smallest number that the numbers are factors of.

$f (x + 3)$

Step 4

Multiply each term in $\frac{1}{f} = \frac{(x - 2) (x + 1)}{x + 3}$ by $f (x + 3)$ to eliminate the fractions.

Multiply each term in $\frac{1}{f} = \frac{(x - 2) (x + 1)}{x + 3}$ by $f (x + 3)$ .

$\frac{1}{f} (f (x + 3)) = \frac{(x - 2) (x + 1)}{x + 3} (f (x + 3))$

Simplify the left side.

Cancel the common factor of $f$ .

Cancel the common factor.

$\frac{1}{f} (f (x + 3)) = \frac{(x - 2) (x + 1)}{x + 3} (f (x + 3))$

Rewrite the expression.

$x + 3 = \frac{(x - 2) (x + 1)}{x + 3} (f (x + 3))$

Simplify the right side.

Cancel the common factor of $x + 3$ .

Factor $x + 3$ out of $f (x + 3)$ .

$x + 3 = \frac{(x - 2) (x + 1)}{x + 3} ((x + 3) f)$

Cancel the common factor.

$x + 3 = \frac{(x - 2) (x + 1)}{x + 3} ((x + 3) f)$

Rewrite the expression.

$x + 3 = (x - 2) (x + 1) f$

Expand $(x - 2) (x + 1)$ using the FOIL Method.

Apply the distributive property.

$x + 3 = (x (x + 1) - 2 (x + 1)) f$

Apply the distributive property.

$x + 3 = (x \cdot x + x \cdot 1 - 2 (x + 1)) f$

Apply the distributive property.

$x + 3 = (x \cdot x + x \cdot 1 - 2 x - 2 \cdot 1) f$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 2$ by $1$ .

$x + 3 = (x^{2} + x - 2 x - 2) f$

Subtract $2 x$ from $x$ .

$x + 3 = (x^{2} - x - 2) f$

Apply the distributive property.

$x + 3 = x^{2} f - x f - 2 f$

Step 5

Solve the equation.

Rewrite the equation as $x^{2} f - x f - 2 f = x + 3$ .

$x^{2} f - x f - 2 f = x + 3$

Factor $f$ out of $x^{2} f - x f - 2 f$ .

Factor $f$ out of $x^{2} f$ .

$f x^{2} - x f - 2 f = x + 3$

Factor $f$ out of $- x f$ .

$f x^{2} + f (- x) - 2 f = x + 3$

Factor $f$ out of $- 2 f$ .

$f x^{2} + f (- x) + f \cdot - 2 = x + 3$

Factor $f$ out of $f x^{2} + f (- x)$ .

$f (x^{2} - x) + f \cdot - 2 = x + 3$

Factor $f$ out of $f (x^{2} - x) + f \cdot - 2$ .

$f (x^{2} - x - 2) = x + 3$

Factor.

Factor $x^{2} - x - 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Write the factored form using these integers.

$f ((x - 2) (x + 1)) = x + 3$

Remove unnecessary parentheses.

$f (x - 2) (x + 1) = x + 3$

Divide each term in $f (x - 2) (x + 1) = x + 3$ by $(x - 2) (x + 1)$ and simplify.

Divide each term in $f (x - 2) (x + 1) = x + 3$ by $(x - 2) (x + 1)$ .

$\frac{f (x - 2) (x + 1)}{(x - 2) (x + 1)} = \frac{x}{(x - 2) (x + 1)} + \frac{3}{(x - 2) (x + 1)}$

Simplify the left side.

Cancel the common factor of $x - 2$ .

Cancel the common factor.

$\frac{f (x - 2) (x + 1)}{(x - 2) (x + 1)} = \frac{x}{(x - 2) (x + 1)} + \frac{3}{(x - 2) (x + 1)}$

Rewrite the expression.

$\frac{f (x + 1)}{x + 1} = \frac{x}{(x - 2) (x + 1)} + \frac{3}{(x - 2) (x + 1)}$

Cancel the common factor of $x + 1$ .

Cancel the common factor.

$\frac{f (x + 1)}{x + 1} = \frac{x}{(x - 2) (x + 1)} + \frac{3}{(x - 2) (x + 1)}$

Divide $f$ by $1$ .

$f = \frac{x}{(x - 2) (x + 1)} + \frac{3}{(x - 2) (x + 1)}$

Evaluate f^-1((3+x)/(x-2))=x+1

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