Find the Inverse h(x)=-3x+6

$h (x) = - 3 x + 6$

Step 1

Write $h (x) = - 3 x + 6$ as an equation.

$y = - 3 x + 6$

Step 2

Interchange the variables.

$x = - 3 y + 6$

Step 3

Solve for $y$ .

Rewrite the equation as $- 3 y + 6 = x$ .

$- 3 y + 6 = x$

Subtract $6$ from both sides of the equation.

$- 3 y = x - 6$

Divide each term in $- 3 y = x - 6$ by $- 3$ and simplify.

Divide each term in $- 3 y = x - 6$ by $- 3$ .

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

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

Divide $- 6$ by $- 3$ .

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

Step 4

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

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

Step 5

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

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Cancel the common factor of $- 3 x + 6$ and $3$ .

Factor $3$ out of $- 3 x$ .

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

Factor $3$ out of $6$ .

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

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

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

Cancel the common factors.

Factor $3$ out of $3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- x + 2$ by $1$ .

$h^{- 1} (- 3 x + 6) = - (- x + 2) + 2$

Apply the distributive property.

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

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 1$ by $2$ .

$h^{- 1} (- 3 x + 6) = x - 2 + 2$

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

Add $- 2$ and $2$ .

$h^{- 1} (- 3 x + 6) = x + 0$

Add $x$ and $0$ .

$h^{- 1} (- 3 x + 6) = x$

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $3$ .

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

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

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 3$ by $2$ .

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

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

Add $- 6$ and $6$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse h(x)=-3x+6

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