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

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

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

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

Interchange the variables.

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

Solve for $y$ .

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

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

Add $3$ to both sides of the equation.

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

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.

$2 y, 1, 1$

The LCM of one and any expression is the expression.

$2 y$

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

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

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

Simplify the left side.

Rewrite using the commutative property of multiplication.

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

Cancel the common factor of $2$ .

Factor $2$ out of $2 y$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $y$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Simplify each term.

Rewrite using the commutative property of multiplication.

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

Multiply $2$ by $3$ .

$1 = 2 x y + 6 y$

Solve the equation.

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

$2 x y + 6 y = 1$

Factor $2 y$ out of $2 x y + 6 y$ .

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

$2 y (x) + 6 y = 1$

Factor $2 y$ out of $6 y$ .

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

Factor $2 y$ out of $2 y (x) + 2 y (3)$ .

$2 y (x + 3) = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x + 3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

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

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

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

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

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

Simplify the denominator.

Add $- 3$ and $3$ .

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

Add $\frac{1}{2 x}$ and $0$ .

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

Simplify terms.

Combine $2$ and $\frac{1}{2 x}$ .

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

Reduce the expression $\frac{2}{2 x}$ by cancelling the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Combine $2$ and $\frac{1}{2 (x + 3)}$ .

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

Reduce the expression $\frac{2}{2 (x + 3)}$ by cancelling the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x + 3$ by $1$ .

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

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

Subtract $3$ from $3$ .

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

Add $x$ and $0$ .

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

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

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

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

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