Find the Inverse f^-1(x)=-3/4x+2

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

Step 1

Write $f^{- 1} (x) = - \frac{3}{4} x + 2$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

Rewrite the equation as $- \frac{3}{4} y + 2 = x$ .

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

Simplify each term.

Combine $y$ and $\frac{3}{4}$ .

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

Move $3$ to the left of $y$ .

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

Subtract $2$ from both sides of the equation.

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

Multiply both sides of the equation by $- \frac{4}{3}$ .

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

Simplify both sides of the equation.

Simplify the left side.

Simplify $- \frac{4}{3} (- \frac{3 y}{4})$ .

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{4}{3}$ into the numerator.

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

Move the leading negative in $- \frac{3 y}{4}$ into the numerator.

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

Factor $4$ out of $- 4$ .

$\frac{4 (- 1)}{3} \cdot \frac{- 3 y}{4} = - \frac{4}{3} (x - 2)$

Cancel the common factor.

$\frac{4 \cdot - 1}{3} \cdot \frac{- 3 y}{4} = - \frac{4}{3} (x - 2)$

Rewrite the expression.

$\frac{- 1}{3} (- 3 y) = - \frac{4}{3} (x - 2)$

Cancel the common factor of $3$ .

Factor $3$ out of $- 3 y$ .

$\frac{- 1}{3} (3 (- y)) = - \frac{4}{3} (x - 2)$

Cancel the common factor.

$\frac{- 1}{3} (3 (- y)) = - \frac{4}{3} (x - 2)$

Rewrite the expression.

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

Multiply.

Multiply $- 1$ by $- 1$ .

$1 y = - \frac{4}{3} (x - 2)$

Multiply $y$ by $1$ .

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

Simplify the right side.

Simplify $- \frac{4}{3} (x - 2)$ .

Apply the distributive property.

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

Combine $x$ and $\frac{4}{3}$ .

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

Multiply $- \frac{4}{3} \cdot - 2$ .

Multiply $- 2$ by $- 1$ .

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

Combine $2$ and $\frac{4}{3}$ .

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

Multiply $2$ by $4$ .

$y = - \frac{x \cdot 4}{3} + \frac{8}{3}$

Move $4$ to the left of $x$ .

$y = - \frac{4 x}{3} + \frac{8}{3}$

Step 4

Replace $y$ with $f (x)$ to show the final answer.

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

Step 5

Verify if $f (x) = - \frac{4 x}{3} + \frac{8}{3}$ is the inverse of $f^{- 1} (x) = - \frac{3}{4} x + 2$ .

To verify the inverse, check if $f (f^{- 1} (x)) = x$ and $f^{- 1} (f (x)) = x$ .

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

Set up the composite result function.

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

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

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

Combine the numerators over the common denominator.

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

Simplify each term.

Combine $x$ and $\frac{3}{4}$ .

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

Move $3$ to the left of $x$ .

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

Apply the distributive property.

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

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{3 x}{4}$ into the numerator.

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

Factor $4$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 3$ by $- 1$ .

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

Multiply $- 4$ by $2$ .

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

Simplify terms.

Combine the opposite terms in $3 x - 8 + 8$ .

Add $- 8$ and $8$ .

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

Add $3 x$ and $0$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

Evaluate $f^{- 1} (- \frac{4 x}{3} + \frac{8}{3})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{3}{4}$ into the numerator.

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

Move the leading negative in $- \frac{4 x}{3}$ into the numerator.

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

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $4$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{3}{4}$ into the numerator.

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

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $4$ .

Factor $4$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $2$ .

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

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

Add $- 2$ and $2$ .

$f^{- 1} (- \frac{4 x}{3} + \frac{8}{3}) = x + 0$

Add $x$ and $0$ .

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

Since $f (f^{- 1} (x)) = x$ and $f^{- 1} (f (x)) = x$ , then $f (x) = - \frac{4 x}{3} + \frac{8}{3}$ is the inverse of $f^{- 1} (x) = - \frac{3}{4} x + 2$ .

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

Find the Inverse f^-1(x)=-3/4x+2

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