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

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

Step 1

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

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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

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

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.

$y + 3, 1$

Remove parentheses.

$y + 3, 1$

The LCM of one and any expression is the expression.

$y + 3$

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

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

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

Simplify the left side.

Cancel the common factor of $y + 3$ .

Cancel the common factor.

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

Rewrite the expression.

$1 = x (y + 3)$

Simplify the right side.

Apply the distributive property.

$1 = x y + x \cdot 3$

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

$1 = x y + 3 x$

Solve the equation.

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

$x y + 3 x = 1$

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

$x y = 1 - 3 x$

Divide each term in $x y = 1 - 3 x$ by $x$ and simplify.

Divide each term in $x y = 1 - 3 x$ by $x$ .

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

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Cancel the common factor of $x$ .

Cancel the common factor.

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

Divide $- 3$ by $1$ .

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

Step 4

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

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

Step 5

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

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

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

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x + 3$ by $1$ .

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

Combine the opposite terms in $x + 3 - 3$ .

Subtract $3$ from $3$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify the denominator.

Add $- 3$ and $3$ .

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

Add $\frac{1}{x}$ and $0$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x$ by $1$ .

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

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

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

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

Related Questions

Support LCMGCF.com by disabling your adblocker.