Find the Inverse 6-4x

$6 - 4 x$

Step 1

Interchange the variables.

$x = 6 - 4 y$

Step 2

Solve for $y$ .

Rewrite the equation as $6 - 4 y = x$ .

$6 - 4 y = x$

Subtract $6$ from both sides of the equation.

$- 4 y = x - 6$

Divide each term in $- 4 y = x - 6$ by $- 4$ and simplify.

Divide each term in $- 4 y = x - 6$ by $- 4$ .

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

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

Cancel the common factor of $- 6$ and $- 4$ .

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

$y = - \frac{x}{4} + \frac{- 2 (3)}{- 4}$

Cancel the common factors.

Factor $- 2$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = - \frac{x}{4} + \frac{3}{2}$ is the inverse of $f (x) = 6 - 4 x$ .

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

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

Simplify terms.

Cancel the common factor of $6 - 4 x$ and $4$ .

Factor $2$ out of $6$ .

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

Factor $2$ out of $- 4 x$ .

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

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

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

Cancel the common factors.

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the distributive property.

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

Multiply $- 1$ by $3$ .

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

Multiply $- 2$ by $- 1$ .

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

Simplify terms.

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

Add $- 3$ and $3$ .

$f^{- 1} (6 - 4 x) = \frac{2 x + 0}{2}$

Add $2 x$ and $0$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

$f^{- 1} (6 - 4 x) = x$

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

Set up the composite result function.

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

Evaluate $f (- \frac{x}{4} + \frac{3}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{x}{4}$ into the numerator.

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

Factor $4$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Cancel the common factor of $2$ .

Factor $2$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 2$ by $3$ .

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

Combine the opposite terms in $6 + x - 6$ .

Subtract $6$ from $6$ .

$f (- \frac{x}{4} + \frac{3}{2}) = x + 0$

Add $x$ and $0$ .

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

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{x}{4} + \frac{3}{2}$ is the inverse of $f (x) = 6 - 4 x$ .

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

Find the Inverse 6-4x

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