Find the Inverse f(n)=(-16+n)/4

$f (n) = \frac{- 16 + n}{4}$

Write $f (n) = \frac{- 16 + n}{4}$ as an equation.

$y = \frac{- 16 + n}{4}$

Interchange the variables.

$n = \frac{- 16 + y}{4}$

Solve for $y$ .

Rewrite the equation as $\frac{- 16 + y}{4} = n$ .

$\frac{- 16 + y}{4} = n$

Multiply both sides of the equation by $4$ .

$4 \frac{- 16 + y}{4} = 4 n$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$4 \frac{- 16 + y}{4} = 4 n$

Rewrite the expression.

$- 16 + y = 4 n$

Add $16$ to both sides of the equation.

$y = 4 n + 16$

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

$f^{- 1} (n) = 4 n + 16$

Verify if $f^{- 1} (n) = 4 n + 16$ is the inverse of $f (n) = \frac{- 16 + n}{4}$ .

To verify the inverse, check if $f^{- 1} (f (n)) = n$ and $f (f^{- 1} (n)) = n$ .

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

Set up the composite result function.

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

Evaluate $f^{- 1} (\frac{- 16 + n}{4})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{- 16 + n}{4}) = 4 (\frac{- 16 + n}{4}) + 16$

Cancel the common factor of $4$ .

Cancel the common factor.

$f^{- 1} (\frac{- 16 + n}{4}) = 4 (\frac{- 16 + n}{4}) + 16$

Rewrite the expression.

$f^{- 1} (\frac{- 16 + n}{4}) = - 16 + n + 16$

Combine the opposite terms in $- 16 + n + 16$ .

Add $- 16$ and $16$ .

$f^{- 1} (\frac{- 16 + n}{4}) = n + 0$

Add $n$ and $0$ .

$f^{- 1} (\frac{- 16 + n}{4}) = n$

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

Set up the composite result function.

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

Evaluate $f (4 n + 16)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (4 n + 16) = \frac{- 16 + 4 n + 16}{4}$

Remove parentheses.

$f (4 n + 16) = \frac{- 16 + 4 n + 16}{4}$

Cancel the common factor of $- 16 + 4 n + 16$ and $4$ .

Factor $4$ out of $- 16$ .

$f (4 n + 16) = \frac{4 \cdot - 4 + 4 n + 16}{4}$

Factor $4$ out of $4 n$ .

$f (4 n + 16) = \frac{4 \cdot - 4 + 4 (n) + 16}{4}$

Factor $4$ out of $16$ .

$f (4 n + 16) = \frac{4 \cdot - 4 + 4 (n) + 4 \cdot 4}{4}$

Factor $4$ out of $4 (n) + 4 (4)$ .

$f (4 n + 16) = \frac{4 \cdot - 4 + 4 (n + 4)}{4}$

Factor $4$ out of $4 (- 4) + 4 (n + 4)$ .

$f (4 n + 16) = \frac{4 (- 4 + n + 4)}{4}$

Cancel the common factors.

Factor $4$ out of $4$ .

$f (4 n + 16) = \frac{4 (- 4 + n + 4)}{4 (1)}$

Cancel the common factor.

$f (4 n + 16) = \frac{4 (- 4 + n + 4)}{4 \cdot 1}$

Rewrite the expression.

$f (4 n + 16) = \frac{- 4 + n + 4}{1}$

Divide $- 4 + n + 4$ by $1$ .

$f (4 n + 16) = - 4 + n + 4$

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

Add $- 4$ and $4$ .

$f (4 n + 16) = n + 0$

Add $n$ and $0$ .

$f (4 n + 16) = n$

Since $f^{- 1} (f (n)) = n$ and $f (f^{- 1} (n)) = n$ , then $f^{- 1} (n) = 4 n + 16$ is the inverse of $f (n) = \frac{- 16 + n}{4}$ .

$f^{- 1} (n) = 4 n + 16$

Find the Inverse f(n)=(-16+n)/4

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