Find the Inverse (4x+1)/3

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Multiply both sides by $3$ .

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

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$4 y + 1 = x \cdot 3$

Simplify the right side.

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

$4 y + 1 = 3 x$

Solve for $y$ .

Subtract $1$ from both sides of the equation.

$4 y = 3 x - 1$

Divide each term in $4 y = 3 x - 1$ by $4$ and simplify.

Divide each term in $4 y = 3 x - 1$ by $4$ .

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 3

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

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

Step 4

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

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

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

Combine the numerators over the common denominator.

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Combine the opposite terms in $4 x + 1 - 1$ .

Subtract $1$ from $1$ .

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

Add $4 x$ and $0$ .

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify the numerator.

Apply the distributive property.

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{1}{4}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Add $- 1$ and $1$ .

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

Add $3 x$ and $0$ .

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

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

Find the Inverse (4x+1)/3

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