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

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

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

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

Interchange the variables.

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

Solve for $y$ .

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

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

Multiply both sides of the equation by $2$ .

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

Simplify the left side.

Simplify $2 (\frac{1}{2} \cdot (3 - 3 y))$ .

Apply the distributive property.

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

Combine $\frac{1}{2}$ and $3$ .

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

Multiply $\frac{1}{2} (- 3 y)$ .

Combine $- 3$ and $\frac{1}{2}$ .

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

Combine $\frac{- 3}{2}$ and $y$ .

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

Move the negative in front of the fraction.

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

Simplify terms.

Apply the distributive property.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3 y}{2}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

$3 - 3 y = 2 x$

Subtract $3$ from both sides of the equation.

$- 3 y = 2 x - 3$

Divide each term in $- 3 y = 2 x - 3$ by $- 3$ and simplify.

Divide each term in $- 3 y = 2 x - 3$ by $- 3$ .

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

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

Divide $- 3$ by $- 3$ .

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

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

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

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

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

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

Simplify each term.

Cancel the common factor of $3 - 3 x$ and $3$ .

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

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

Cancel the common factors.

Factor $3$ out of $3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $2 (\frac{1}{2} \cdot (1 - x))$ by $1$ .

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

Apply the distributive property.

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

Multiply $\frac{1}{2}$ by $1$ .

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

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

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

Apply the distributive property.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{x}{2}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Apply the distributive property.

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

Multiply $- 1$ by $1$ .

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

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Combine the opposite terms in $- 1 + x + 1$ .

Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{2 x}{3}$ into the numerator.

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

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 2$ by $- 1$ .

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

Multiply $- 3$ by $1$ .

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

Simplify terms.

Combine the opposite terms in $3 + 2 x - 3$ .

Subtract $3$ from $3$ .

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

Add $2 x$ and $0$ .

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

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

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

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