Find the Inverse 3x-1/4

$3 x - \frac{1}{4}$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

Rewrite the equation as $3 y - \frac{1}{4} = x$ .

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

Add $\frac{1}{4}$ to both sides of the equation.

$3 y = x + \frac{1}{4}$

Divide each term in $3 y = x + \frac{1}{4}$ by $3$ and simplify.

Divide each term in $3 y = x + \frac{1}{4}$ by $3$ .

$\frac{3 y}{3} = \frac{x}{3} + \frac{\frac{1}{4}}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{x}{3} + \frac{\frac{1}{4}}{3}$

Divide $y$ by $1$ .

$y = \frac{x}{3} + \frac{\frac{1}{4}}{3}$

Simplify the right side.

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$y = \frac{x}{3} + \frac{1}{4} \cdot \frac{1}{3}$

Multiply $\frac{1}{4} \cdot \frac{1}{3}$ .

Multiply $\frac{1}{4}$ by $\frac{1}{3}$ .

$y = \frac{x}{3} + \frac{1}{4 \cdot 3}$

Multiply $4$ by $3$ .

$y = \frac{x}{3} + \frac{1}{12}$

Step 3

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

$f^{- 1} (x) = \frac{x}{3} + \frac{1}{12}$

Step 4

Verify if $f^{- 1} (x) = \frac{x}{3} + \frac{1}{12}$ is the inverse of $f (x) = 3 x - \frac{1}{4}$ .

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

$f^{- 1} (3 x - \frac{1}{4}) = \frac{3 x - \frac{1}{4}}{3} + \frac{1}{12}$

Simplify each term.

Simplify the numerator.

To write $3 x$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$f^{- 1} (3 x - \frac{1}{4}) = \frac{3 x \cdot \frac{4}{4} - \frac{1}{4}}{3} + \frac{1}{12}$

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

$f^{- 1} (3 x - \frac{1}{4}) = \frac{\frac{3 x \cdot 4}{4} - \frac{1}{4}}{3} + \frac{1}{12}$

Combine the numerators over the common denominator.

$f^{- 1} (3 x - \frac{1}{4}) = \frac{\frac{3 x \cdot 4 - 1}{4}}{3} + \frac{1}{12}$

Multiply $4$ by $3$ .

$f^{- 1} (3 x - \frac{1}{4}) = \frac{\frac{12 x - 1}{4}}{3} + \frac{1}{12}$

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (3 x - \frac{1}{4}) = \frac{12 x - 1}{4} \cdot \frac{1}{3} + \frac{1}{12}$

Multiply $\frac{12 x - 1}{4} \cdot \frac{1}{3}$ .

Multiply $\frac{12 x - 1}{4}$ by $\frac{1}{3}$ .

$f^{- 1} (3 x - \frac{1}{4}) = \frac{12 x - 1}{4 \cdot 3} + \frac{1}{12}$

Multiply $4$ by $3$ .

$f^{- 1} (3 x - \frac{1}{4}) = \frac{12 x - 1}{12} + \frac{1}{12}$

Simplify terms.

Combine the numerators over the common denominator.

$f^{- 1} (3 x - \frac{1}{4}) = \frac{12 x - 1 + 1}{12}$

Combine the opposite terms in $12 x - 1 + 1$ .

Add $- 1$ and $1$ .

$f^{- 1} (3 x - \frac{1}{4}) = \frac{12 x + 0}{12}$

Add $12 x$ and $0$ .

$f^{- 1} (3 x - \frac{1}{4}) = \frac{12 x}{12}$

Cancel the common factor of $12$ .

Cancel the common factor.

$f^{- 1} (3 x - \frac{1}{4}) = \frac{12 x}{12}$

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

Evaluate $f (\frac{x}{3} + \frac{1}{12})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x}{3} + \frac{1}{12}) = 3 (\frac{x}{3} + \frac{1}{12}) - \frac{1}{4}$

Simplify each term.

Apply the distributive property.

$f (\frac{x}{3} + \frac{1}{12}) = 3 (\frac{x}{3}) + 3 (\frac{1}{12}) - \frac{1}{4}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{x}{3} + \frac{1}{12}) = 3 (\frac{x}{3}) + 3 (\frac{1}{12}) - \frac{1}{4}$

Rewrite the expression.

$f (\frac{x}{3} + \frac{1}{12}) = x + 3 (\frac{1}{12}) - \frac{1}{4}$

Cancel the common factor of $3$ .

Factor $3$ out of $12$ .

$f (\frac{x}{3} + \frac{1}{12}) = x + 3 (\frac{1}{3 (4)}) - \frac{1}{4}$

Cancel the common factor.

$f (\frac{x}{3} + \frac{1}{12}) = x + 3 (\frac{1}{3 \cdot 4}) - \frac{1}{4}$

Rewrite the expression.

$f (\frac{x}{3} + \frac{1}{12}) = x + \frac{1}{4} - \frac{1}{4}$

Combine the opposite terms in $x + \frac{1}{4} - \frac{1}{4}$ .

Combine the numerators over the common denominator.

$f (\frac{x}{3} + \frac{1}{12}) = x + \frac{1 - 1}{4}$

Subtract $1$ from $1$ .

$f (\frac{x}{3} + \frac{1}{12}) = x + \frac{0}{4}$

Divide $0$ by $4$ .

$f (\frac{x}{3} + \frac{1}{12}) = x + 0$

Add $x$ and $0$ .

$f (\frac{x}{3} + \frac{1}{12}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x}{3} + \frac{1}{12}$ is the inverse of $f (x) = 3 x - \frac{1}{4}$ .

$f^{- 1} (x) = \frac{x}{3} + \frac{1}{12}$

Find the Inverse 3x-1/4

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