Find the Inverse g(x)=4-3/2x

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

Step 1

Write $g (x) = 4 - \frac{3}{2} x$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

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

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

Simplify each term.

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

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

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

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

Subtract $4$ from both sides of the equation.

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

Multiply both sides of the equation by $- \frac{2}{3}$ .

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

Simplify both sides of the equation.

Simplify the left side.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{2}{3}$ into the numerator.

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

Move the leading negative in $- \frac{3 y}{2}$ into the numerator.

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

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $3$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply.

Multiply $- 1$ by $- 1$ .

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

Multiply $y$ by $1$ .

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

Simplify the right side.

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

Apply the distributive property.

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

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

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

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

Multiply $- 4$ by $- 1$ .

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

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

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

Multiply $4$ by $2$ .

$y = - \frac{x \cdot 2}{3} + \frac{8}{3}$

Move $2$ to the left of $x$ .

$y = - \frac{2 x}{3} + \frac{8}{3}$

Step 4

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

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

Step 5

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

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

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

Set up the composite result function.

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

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

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

Combine the numerators over the common denominator.

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

Simplify each term.

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

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

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

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

Apply the distributive property.

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

Multiply $- 2$ by $4$ .

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{3 x}{2}$ into the numerator.

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

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 3$ by $- 1$ .

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

Simplify terms.

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

Add $- 8$ and $8$ .

$g^{- 1} (4 - \frac{3}{2} x) = \frac{3 x + 0}{3}$

Add $3 x$ and $0$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

Evaluate $g (- \frac{2 x}{3} + \frac{8}{3})$ by substituting in the value of $g^{- 1}$ into $g$ .

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{3}{2}$ into the numerator.

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

Move the leading negative in $- \frac{2 x}{3}$ into the numerator.

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

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{3}{2}$ into the numerator.

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

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Factor $2$ out of $8$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $4$ .

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

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

Subtract $4$ from $4$ .

$g (- \frac{2 x}{3} + \frac{8}{3}) = x + 0$

Add $x$ and $0$ .

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

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

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

Find the Inverse g(x)=4-3/2x

Related Questions

Support LCMGCF.com by disabling your adblocker.