Find the Inverse -6x+8

$- 6 x + 8$

Step 1

Interchange the variables.

$x = - 6 y + 8$

Step 2

Solve for $y$ .

Rewrite the equation as $- 6 y + 8 = x$ .

$- 6 y + 8 = x$

Subtract $8$ from both sides of the equation.

$- 6 y = x - 8$

Divide each term in $- 6 y = x - 8$ by $- 6$ and simplify.

Divide each term in $- 6 y = x - 8$ by $- 6$ .

$\frac{- 6 y}{- 6} = \frac{x}{- 6} + \frac{- 8}{- 6}$

Simplify the left side.

Cancel the common factor of $- 6$ .

Cancel the common factor.

$\frac{- 6 y}{- 6} = \frac{x}{- 6} + \frac{- 8}{- 6}$

Divide $y$ by $1$ .

$y = \frac{x}{- 6} + \frac{- 8}{- 6}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y = - \frac{x}{6} + \frac{- 8}{- 6}$

Cancel the common factor of $- 8$ and $- 6$ .

Factor $- 2$ out of $- 8$ .

$y = - \frac{x}{6} + \frac{- 2 (4)}{- 6}$

Cancel the common factors.

Factor $- 2$ out of $- 6$ .

$y = - \frac{x}{6} + \frac{- 2 \cdot 4}{- 2 \cdot 3}$

Cancel the common factor.

$y = - \frac{x}{6} + \frac{- 2 \cdot 4}{- 2 \cdot 3}$

Rewrite the expression.

$y = - \frac{x}{6} + \frac{4}{3}$

Step 3

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

$f^{- 1} (x) = - \frac{x}{6} + \frac{4}{3}$

Step 4

Verify if $f^{- 1} (x) = - \frac{x}{6} + \frac{4}{3}$ is the inverse of $f (x) = - 6 x + 8$ .

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 $f^{- 1} (- 6 x + 8)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (- 6 x + 8) = - \frac{- 6 x + 8}{6} + \frac{4}{3}$

Simplify terms.

Cancel the common factor of $- 6 x + 8$ and $6$ .

Factor $2$ out of $- 6 x$ .

$f^{- 1} (- 6 x + 8) = - \frac{2 (- 3 x) + 8}{6} + \frac{4}{3}$

Factor $2$ out of $8$ .

$f^{- 1} (- 6 x + 8) = - \frac{2 (- 3 x) + 2 (4)}{6} + \frac{4}{3}$

Factor $2$ out of $2 (- 3 x) + 2 (4)$ .

$f^{- 1} (- 6 x + 8) = - \frac{2 (- 3 x + 4)}{6} + \frac{4}{3}$

Cancel the common factors.

Factor $2$ out of $6$ .

$f^{- 1} (- 6 x + 8) = - \frac{2 (- 3 x + 4)}{2 (3)} + \frac{4}{3}$

Cancel the common factor.

$f^{- 1} (- 6 x + 8) = - \frac{2 (- 3 x + 4)}{2 \cdot 3} + \frac{4}{3}$

Rewrite the expression.

$f^{- 1} (- 6 x + 8) = - \frac{- 3 x + 4}{3} + \frac{4}{3}$

Combine the numerators over the common denominator.

$f^{- 1} (- 6 x + 8) = \frac{- (- 3 x + 4) + 4}{3}$

Simplify each term.

Apply the distributive property.

$f^{- 1} (- 6 x + 8) = \frac{- (- 3 x) - 1 \cdot 4 + 4}{3}$

Multiply $- 3$ by $- 1$ .

$f^{- 1} (- 6 x + 8) = \frac{3 x - 1 \cdot 4 + 4}{3}$

Multiply $- 1$ by $4$ .

$f^{- 1} (- 6 x + 8) = \frac{3 x - 4 + 4}{3}$

Simplify terms.

Combine the opposite terms in $3 x - 4 + 4$ .

Add $- 4$ and $4$ .

$f^{- 1} (- 6 x + 8) = \frac{3 x + 0}{3}$

Add $3 x$ and $0$ .

$f^{- 1} (- 6 x + 8) = \frac{3 x}{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (- 6 x + 8) = \frac{3 x}{3}$

Divide $x$ by $1$ .

$f^{- 1} (- 6 x + 8) = x$

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

Set up the composite result function.

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

Evaluate $f (- \frac{x}{6} + \frac{4}{3})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \frac{x}{6} + \frac{4}{3}) = - 6 (- \frac{x}{6} + \frac{4}{3}) + 8$

Simplify each term.

Apply the distributive property.

$f (- \frac{x}{6} + \frac{4}{3}) = - 6 (- \frac{x}{6}) - 6 (\frac{4}{3}) + 8$

Cancel the common factor of $6$ .

Move the leading negative in $- \frac{x}{6}$ into the numerator.

$f (- \frac{x}{6} + \frac{4}{3}) = - 6 \frac{- x}{6} - 6 (\frac{4}{3}) + 8$

Factor $6$ out of $- 6$ .

$f (- \frac{x}{6} + \frac{4}{3}) = 6 (- 1) (\frac{- x}{6}) - 6 (\frac{4}{3}) + 8$

Cancel the common factor.

$f (- \frac{x}{6} + \frac{4}{3}) = 6 \cdot (- 1 \frac{- x}{6}) - 6 (\frac{4}{3}) + 8$

Rewrite the expression.

$f (- \frac{x}{6} + \frac{4}{3}) = - 1 (- x) - 6 (\frac{4}{3}) + 8$

Multiply $- 1$ by $- 1$ .

$f (- \frac{x}{6} + \frac{4}{3}) = 1 x - 6 (\frac{4}{3}) + 8$

Multiply $x$ by $1$ .

$f (- \frac{x}{6} + \frac{4}{3}) = x - 6 (\frac{4}{3}) + 8$

Cancel the common factor of $3$ .

Factor $3$ out of $- 6$ .

$f (- \frac{x}{6} + \frac{4}{3}) = x + 3 (- 2) (\frac{4}{3}) + 8$

Cancel the common factor.

$f (- \frac{x}{6} + \frac{4}{3}) = x + 3 \cdot (- 2 (\frac{4}{3})) + 8$

Rewrite the expression.

$f (- \frac{x}{6} + \frac{4}{3}) = x - 2 \cdot 4 + 8$

Multiply $- 2$ by $4$ .

$f (- \frac{x}{6} + \frac{4}{3}) = x - 8 + 8$

Combine the opposite terms in $x - 8 + 8$ .

Add $- 8$ and $8$ .

$f (- \frac{x}{6} + \frac{4}{3}) = x + 0$

Add $x$ and $0$ .

$f (- \frac{x}{6} + \frac{4}{3}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{x}{6} + \frac{4}{3}$ is the inverse of $f (x) = - 6 x + 8$ .

$f^{- 1} (x) = - \frac{x}{6} + \frac{4}{3}$

Find the Inverse -6x+8

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