Find the Inverse f(x)=2+1/(x+3)

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

Write $f (x) = 2 + \frac{1}{x + 3}$ as an equation.

$y = 2 + \frac{1}{x + 3}$

Interchange the variables.

$x = 2 + \frac{1}{y + 3}$

Solve for $y$ .

Rewrite the equation as $2 + \frac{1}{y + 3} = x$ .

$2 + \frac{1}{y + 3} = x$

Subtract $2$ from both sides of the equation.

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

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.

$y + 3, 1, 1$

Remove parentheses.

$y + 3, 1, 1$

The LCM of one and any expression is the expression.

$y + 3$

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

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

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

Simplify the left side.

Cancel the common factor of $y + 3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Simplify each term.

Apply the distributive property.

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

Move $3$ to the left of $x$ .

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

Apply the distributive property.

$1 = x y + 3 x - 2 y - 2 \cdot 3$

Multiply $- 2$ by $3$ .

$1 = x y + 3 x - 2 y - 6$

Solve the equation.

Rewrite the equation as $x y + 3 x - 2 y - 6 = 1$ .

$x y + 3 x - 2 y - 6 = 1$

Move all terms not containing $y$ to the right side of the equation.

Subtract $3 x$ from both sides of the equation.

$x y - 2 y - 6 = 1 - 3 x$

Add $6$ to both sides of the equation.

$x y - 2 y = 1 - 3 x + 6$

Add $1$ and $6$ .

$x y - 2 y = - 3 x + 7$

Factor $y$ out of $x y - 2 y$ .

Factor $y$ out of $x y$ .

$y x - 2 y = - 3 x + 7$

Factor $y$ out of $- 2 y$ .

$y x + y \cdot - 2 = - 3 x + 7$

Factor $y$ out of $y x + y \cdot - 2$ .

$y (x - 2) = - 3 x + 7$

Divide each term in $y (x - 2) = - 3 x + 7$ by $x - 2$ and simplify.

Divide each term in $y (x - 2) = - 3 x + 7$ by $x - 2$ .

$\frac{y (x - 2)}{x - 2} = \frac{- 3 x}{x - 2} + \frac{7}{x - 2}$

Simplify the left side.

Cancel the common factor of $x - 2$ .

Cancel the common factor.

$\frac{y (x - 2)}{x - 2} = \frac{- 3 x}{x - 2} + \frac{7}{x - 2}$

Divide $y$ by $1$ .

$y = \frac{- 3 x}{x - 2} + \frac{7}{x - 2}$

Simplify the right side.

Combine the numerators over the common denominator.

$y = \frac{- 3 x + 7}{x - 2}$

Factor $- 1$ out of $- 3 x$ .

$y = \frac{- (3 x) + 7}{x - 2}$

Rewrite $7$ as $- 1 (- 7)$ .

$y = \frac{- (3 x) - 1 (- 7)}{x - 2}$

Factor $- 1$ out of $- (3 x) - 1 (- 7)$ .

$y = \frac{- (3 x - 7)}{x - 2}$

Simplify the expression.

Rewrite $- (3 x - 7)$ as $- 1 (3 x - 7)$ .

$y = \frac{- 1 (3 x - 7)}{x - 2}$

Move the negative in front of the fraction.

$y = - \frac{3 x - 7}{x - 2}$

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = - \frac{3 x - 7}{x - 2}$

Verify if $f^{- 1} (x) = - \frac{3 x - 7}{x - 2}$ is the inverse of $f (x) = 2 + \frac{1}{x + 3}$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (2 + \frac{1}{x + 3})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{3 (2 + \frac{1}{x + 3}) - 7}{(2 + \frac{1}{x + 3}) - 2}$

Multiply the numerator and denominator of the complex fraction by $x + 3$ .

Multiply $\frac{3 (2 + \frac{1}{x + 3}) - 7}{2 + \frac{1}{x + 3} - 2}$ by $\frac{x + 3}{x + 3}$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - (\frac{x + 3}{x + 3} \cdot \frac{3 (2 + \frac{1}{x + 3}) - 7}{2 + \frac{1}{x + 3} - 2})$

Combine.

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (3 (2 + \frac{1}{x + 3}) - 7)}{(x + 3) (2 + \frac{1}{x + 3} - 2)}$

Apply the distributive property.

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (3 (2 + \frac{1}{x + 3})) + (x + 3) \cdot - 7}{(x + 3) \cdot 2 + (x + 3) (\frac{1}{x + 3}) + (x + 3) \cdot - 2}$

Cancel the common factor of $x + 3$ .

Cancel the common factor.

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (3 (2 + \frac{1}{x + 3})) + (x + 3) \cdot - 7}{(x + 3) \cdot 2 + (x + 3) (\frac{1}{x + 3}) + (x + 3) \cdot - 2}$

Rewrite the expression.

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (3 (2 + \frac{1}{x + 3})) + (x + 3) \cdot - 7}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Simplify the numerator.

Factor $x + 3$ out of $(x + 3) \cdot 3 (2 + \frac{1}{x + 3}) + (x + 3) \cdot - 7$ .

Factor $x + 3$ out of $(x + 3) \cdot 3 (2 + \frac{1}{x + 3})$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (3 (2 + \frac{1}{x + 3})) + (x + 3) \cdot - 7}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Factor $x + 3$ out of $(x + 3) \cdot - 7$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (3 (2 + \frac{1}{x + 3})) + (x + 3) (- 7)}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Factor $x + 3$ out of $(x + 3) (3 (2 + \frac{1}{x + 3})) + (x + 3) (- 7)$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (3 (2 + \frac{1}{x + 3}) - 7)}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

To write $2$ as a fraction with a common denominator, multiply by $\frac{x + 3}{x + 3}$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (3 (\frac{2 (x + 3)}{x + 3} + \frac{1}{x + 3}) - 7)}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Combine the numerators over the common denominator.

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (3 (\frac{2 (x + 3) + 1}{x + 3}) - 7)}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Simplify the numerator.

Apply the distributive property.

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (3 (\frac{2 x + 2 \cdot 3 + 1}{x + 3}) - 7)}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Multiply $2$ by $3$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (3 (\frac{2 x + 6 + 1}{x + 3}) - 7)}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Add $6$ and $1$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (3 (\frac{2 x + 7}{x + 3}) - 7)}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Combine $3$ and $\frac{2 x + 7}{x + 3}$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (\frac{3 (2 x + 7)}{x + 3} - 7)}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

To write $- 7$ as a fraction with a common denominator, multiply by $\frac{x + 3}{x + 3}$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (\frac{3 (2 x + 7)}{x + 3} - 7 \cdot \frac{x + 3}{x + 3})}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Combine $- 7$ and $\frac{x + 3}{x + 3}$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (\frac{3 (2 x + 7)}{x + 3} + \frac{- 7 (x + 3)}{x + 3})}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Combine the numerators over the common denominator.

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (\frac{3 (2 x + 7) - 7 (x + 3)}{x + 3})}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Rewrite $\frac{3 (2 x + 7) - 7 (x + 3)}{x + 3}$ in a factored form.

Apply the distributive property.

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (\frac{3 (2 x) + 3 \cdot 7 - 7 (x + 3)}{x + 3})}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Multiply $2$ by $3$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (\frac{6 x + 3 \cdot 7 - 7 (x + 3)}{x + 3})}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Multiply $3$ by $7$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (\frac{6 x + 21 - 7 (x + 3)}{x + 3})}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Apply the distributive property.

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (\frac{6 x + 21 - 7 x - 7 \cdot 3}{x + 3})}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Multiply $- 7$ by $3$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (\frac{6 x + 21 - 7 x - 21}{x + 3})}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Subtract $7 x$ from $6 x$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{(x + 3) (\frac{- x + 21 - 21}{x + 3})}{(x + 3) \cdot 2 + 1 + (x + 3) \cdot - 2}$

Subtract $21$ from $21$ .

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

Add $- x$ and $0$ .

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

Move the negative in front of the fraction.

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

Factor out negative.

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

Simplify the denominator.

Apply the distributive property.

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

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

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

Multiply $3$ by $2$ .

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

Apply the distributive property.

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

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

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

Multiply $3$ by $- 2$ .

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

Subtract $2 x$ from $2 x$ .

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

Add $0$ and $6$ .

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

Add $6$ and $1$ .

$f^{- 1} (2 + \frac{1}{x + 3}) = - \frac{- (x + 3) \frac{x}{x + 3}}{7 - 6}$

Subtract $6$ from $7$ .

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

Reduce the expression by cancelling the common factors.

Divide $- (x + 3) \frac{x}{x + 3}$ by $1$ .

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

Cancel the common factor of $x + 3$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Rewrite $- 1 x$ as $- x$ .

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

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (- \frac{3 x - 7}{x - 2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \frac{3 x - 7}{x - 2}) = 2 + \frac{1}{(- \frac{3 x - 7}{x - 2}) + 3}$

Simplify each term.

Simplify the denominator.

To write $3$ as a fraction with a common denominator, multiply by $\frac{x - 2}{x - 2}$ .

$f (- \frac{3 x - 7}{x - 2}) = 2 + \frac{1}{- \frac{3 x - 7}{x - 2} + \frac{3 (x - 2)}{x - 2}}$

Combine the numerators over the common denominator.

$f (- \frac{3 x - 7}{x - 2}) = 2 + \frac{1}{\frac{- (3 x - 7) + 3 (x - 2)}{x - 2}}$

Reorder terms.

$f (- \frac{3 x - 7}{x - 2}) = 2 + \frac{1}{\frac{3 (x - 2) - (3 x - 7)}{x - 2}}$

Rewrite $\frac{3 (x - 2) - (3 x - 7)}{x - 2}$ in a factored form.

Apply the distributive property.

$f (- \frac{3 x - 7}{x - 2}) = 2 + \frac{1}{\frac{3 x + 3 \cdot - 2 - (3 x - 7)}{x - 2}}$

Multiply $3$ by $- 2$ .

$f (- \frac{3 x - 7}{x - 2}) = 2 + \frac{1}{\frac{3 x - 6 - (3 x - 7)}{x - 2}}$

Apply the distributive property.

$f (- \frac{3 x - 7}{x - 2}) = 2 + \frac{1}{\frac{3 x - 6 - (3 x) + 7}{x - 2}}$

Multiply $3$ by $- 1$ .

$f (- \frac{3 x - 7}{x - 2}) = 2 + \frac{1}{\frac{3 x - 6 - 3 x + 7}{x - 2}}$

Multiply $- 1$ by $- 7$ .

$f (- \frac{3 x - 7}{x - 2}) = 2 + \frac{1}{\frac{3 x - 6 - 3 x + 7}{x - 2}}$

Subtract $3 x$ from $3 x$ .

$f (- \frac{3 x - 7}{x - 2}) = 2 + \frac{1}{\frac{0 - 6 + 7}{x - 2}}$

Subtract $6$ from $0$ .

$f (- \frac{3 x - 7}{x - 2}) = 2 + \frac{1}{\frac{- 6 + 7}{x - 2}}$

Add $- 6$ and $7$ .

$f (- \frac{3 x - 7}{x - 2}) = 2 + \frac{1}{\frac{1}{x - 2}}$

Multiply the numerator by the reciprocal of the denominator.

$f (- \frac{3 x - 7}{x - 2}) = 2 + 1 (x - 2)$

Multiply $x - 2$ by $1$ .

$f (- \frac{3 x - 7}{x - 2}) = 2 + x - 2$

Combine the opposite terms in $2 + x - 2$ .

Subtract $2$ from $2$ .

$f (- \frac{3 x - 7}{x - 2}) = x + 0$

Add $x$ and $0$ .

$f (- \frac{3 x - 7}{x - 2}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{3 x - 7}{x - 2}$ is the inverse of $f (x) = 2 + \frac{1}{x + 3}$ .

$f^{- 1} (x) = - \frac{3 x - 7}{x - 2}$

Find the Inverse f(x)=2+1/(x+3)

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