Find the Inverse f(x)=7x-1

$f (x) = 7 x - 1$

Write $f (x) = 7 x - 1$ as an equation.

$y = 7 x - 1$

Interchange the variables.

$x = 7 y - 1$

Solve for $y$ .

Rewrite the equation as $7 y - 1 = x$ .

$7 y - 1 = x$

Add $1$ to both sides of the equation.

$7 y = x + 1$

Divide each term in $7 y = x + 1$ by $7$ and simplify.

Divide each term in $7 y = x + 1$ by $7$ .

$\frac{7 y}{7} = \frac{x}{7} + \frac{1}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 y}{7} = \frac{x}{7} + \frac{1}{7}$

Divide $y$ by $1$ .

$y = \frac{x}{7} + \frac{1}{7}$

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

$f^{- 1} (x) = \frac{x}{7} + \frac{1}{7}$

Verify if $f^{- 1} (x) = \frac{x}{7} + \frac{1}{7}$ is the inverse of $f (x) = 7 x - 1$ .

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 $f^{- 1} (7 x - 1)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (7 x - 1) = \frac{7 x - 1}{7} + \frac{1}{7}$

Combine the numerators over the common denominator.

$f^{- 1} (7 x - 1) = \frac{7 x - 1 + 1}{7}$

Combine the opposite terms in $7 x - 1 + 1$ .

Add $- 1$ and $1$ .

$f^{- 1} (7 x - 1) = \frac{7 x + 0}{7}$

Add $7 x$ and $0$ .

$f^{- 1} (7 x - 1) = \frac{7 x}{7}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f^{- 1} (7 x - 1) = \frac{7 x}{7}$

Divide $x$ by $1$ .

$f^{- 1} (7 x - 1) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{x}{7} + \frac{1}{7})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x}{7} + \frac{1}{7}) = 7 (\frac{x}{7} + \frac{1}{7}) - 1$

Simplify each term.

Apply the distributive property.

$f (\frac{x}{7} + \frac{1}{7}) = 7 (\frac{x}{7}) + 7 (\frac{1}{7}) - 1$

Cancel the common factor of $7$ .

Cancel the common factor.

$f (\frac{x}{7} + \frac{1}{7}) = 7 (\frac{x}{7}) + 7 (\frac{1}{7}) - 1$

Rewrite the expression.

$f (\frac{x}{7} + \frac{1}{7}) = x + 7 (\frac{1}{7}) - 1$

Cancel the common factor of $7$ .

Cancel the common factor.

$f (\frac{x}{7} + \frac{1}{7}) = x + 7 (\frac{1}{7}) - 1$

Rewrite the expression.

$f (\frac{x}{7} + \frac{1}{7}) = x + 1 - 1$

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

Subtract $1$ from $1$ .

$f (\frac{x}{7} + \frac{1}{7}) = x + 0$

Add $x$ and $0$ .

$f (\frac{x}{7} + \frac{1}{7}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x}{7} + \frac{1}{7}$ is the inverse of $f (x) = 7 x - 1$ .

$f^{- 1} (x) = \frac{x}{7} + \frac{1}{7}$

Find the Inverse f(x)=7x-1

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