Find the Inverse f(x)=-3x+4

$f (x) = - 3 x + 4$

Step 1

Write $f (x) = - 3 x + 4$ as an equation.

$y = - 3 x + 4$

Step 2

Interchange the variables.

$x = - 3 y + 4$

Step 3

Solve for $y$ .

Rewrite the equation as $- 3 y + 4 = x$ .

$- 3 y + 4 = x$

Subtract $4$ from both sides of the equation.

$- 3 y = x - 4$

Divide each term in $- 3 y = x - 4$ by $- 3$ and simplify.

Divide each term in $- 3 y = x - 4$ by $- 3$ .

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

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

Dividing two negative values results in a positive value.

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

Step 4

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

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

Step 5

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

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

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the distributive property.

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

Multiply $- 3$ by $- 1$ .

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

Multiply $- 1$ by $4$ .

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

Simplify terms.

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

Add $- 4$ and $4$ .

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

Add $3 x$ and $0$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $3$ .

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

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

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Cancel the common factor of $3$ .

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $4$ .

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

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

Add $- 4$ and $4$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse f(x)=-3x+4

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