Find the Inverse (x-1)/3

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Multiply both sides of the equation by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$y - 1 = 3 x$

Add $1$ to both sides of the equation.

$y = 3 x + 1$

Step 3

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

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

Step 4

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Subtract $1$ from $1$ .

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

Add $3 x$ and $0$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

$f (3 x + 1) = x$

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

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

Find the Inverse (x-1)/3

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