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

$f (x) = 1 - 3 x$

Write $f (x) = 1 - 3 x$ as an equation.

$y = 1 - 3 x$

Interchange the variables.

$x = 1 - 3 y$

Solve for $y$ .

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

$1 - 3 y = x$

Subtract $1$ from both sides of the equation.

$- 3 y = x - 1$

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

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

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

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

Dividing two negative values results in a positive value.

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the distributive property.

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

Multiply $- 1$ by $1$ .

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

Multiply $- 3$ by $- 1$ .

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

Simplify terms.

Combine the opposite terms in $- 1 + 3 x + 1$ .

Add $- 1$ and $1$ .

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

Add $3 x$ and $0$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $3$ .

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

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

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Cancel the common factor of $3$ .

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

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

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

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

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