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

$f (x) = 3 - 4 x$

Step 1

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

$y = 3 - 4 x$

Step 2

Interchange the variables.

$x = 3 - 4 y$

Step 3

Solve for $y$ .

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

$3 - 4 y = x$

Subtract $3$ from both sides of the equation.

$- 4 y = x - 3$

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

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

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

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

Dividing two negative values results in a positive value.

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

Step 4

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

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

Step 5

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

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

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the distributive property.

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

Multiply $- 1$ by $3$ .

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

Multiply $- 4$ by $- 1$ .

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

Simplify terms.

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

Add $- 3$ and $3$ .

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

Add $4 x$ and $0$ .

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $4$ .

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

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

Factor $4$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Cancel the common factor of $4$ .

Factor $4$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $3$ .

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

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

Subtract $3$ from $3$ .

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

Add $x$ and $0$ .

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

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

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

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

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