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

$f (x) = 4 x + 3$

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

$y = 4 x + 3$

Interchange the variables.

$x = 4 y + 3$

Solve for $y$ .

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

$4 y + 3 = x$

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.

Move the negative in front of the fraction.

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

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

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

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

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

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

Combine the numerators over the common denominator.

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

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

Subtract $3$ from $3$ .

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

Add $4 x$ and $0$ .

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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 $f$ .

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{3}{4}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Add $- 3$ and $3$ .

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

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

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

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

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