Find the Inverse c(n)=-6+5(n-1)

$c (n) = - 6 + 5 (n - 1)$

Step 1

Write $c (n) = - 6 + 5 (n - 1)$ as an equation.

$y = - 6 + 5 (n - 1)$

Step 2

Interchange the variables.

$n = - 6 + 5 (y - 1)$

Step 3

Solve for $y$ .

Rewrite the equation as $- 6 + 5 (y - 1) = n$ .

$- 6 + 5 (y - 1) = n$

Simplify $- 6 + 5 (y - 1)$ .

Simplify each term.

Apply the distributive property.

$- 6 + 5 y + 5 \cdot - 1 = n$

Multiply $5$ by $- 1$ .

$- 6 + 5 y - 5 = n$

Subtract $5$ from $- 6$ .

$5 y - 11 = n$

Add $11$ to both sides of the equation.

$5 y = n + 11$

Divide each term in $5 y = n + 11$ by $5$ and simplify.

Divide each term in $5 y = n + 11$ by $5$ .

$\frac{5 y}{5} = \frac{n}{5} + \frac{11}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 y}{5} = \frac{n}{5} + \frac{11}{5}$

Divide $y$ by $1$ .

$y = \frac{n}{5} + \frac{11}{5}$

Step 4

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

$c^{- 1} (n) = \frac{n}{5} + \frac{11}{5}$

Step 5

Verify if $c^{- 1} (n) = \frac{n}{5} + \frac{11}{5}$ is the inverse of $c (n) = - 6 + 5 (n - 1)$ .

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

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

Set up the composite result function.

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

Evaluate $c^{- 1} (- 6 + 5 (n - 1))$ by substituting in the value of $c$ into $c^{- 1}$ .

$c^{- 1} (- 6 + 5 (n - 1)) = \frac{- 6 + 5 (n - 1)}{5} + \frac{11}{5}$

Combine the numerators over the common denominator.

$c^{- 1} (- 6 + 5 (n - 1)) = \frac{- 6 + 5 (n - 1) + 11}{5}$

Simplify each term.

Apply the distributive property.

$c^{- 1} (- 6 + 5 (n - 1)) = \frac{- 6 + 5 n + 5 \cdot - 1 + 11}{5}$

Multiply $5$ by $- 1$ .

$c^{- 1} (- 6 + 5 (n - 1)) = \frac{- 6 + 5 n - 5 + 11}{5}$

Simplify terms.

Subtract $5$ from $- 6$ .

$c^{- 1} (- 6 + 5 (n - 1)) = \frac{5 n - 11 + 11}{5}$

Combine the opposite terms in $5 n - 11 + 11$ .

Add $- 11$ and $11$ .

$c^{- 1} (- 6 + 5 (n - 1)) = \frac{5 n + 0}{5}$

Add $5 n$ and $0$ .

$c^{- 1} (- 6 + 5 (n - 1)) = \frac{5 n}{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

$c^{- 1} (- 6 + 5 (n - 1)) = \frac{5 n}{5}$

Divide $n$ by $1$ .

$c^{- 1} (- 6 + 5 (n - 1)) = n$

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

Set up the composite result function.

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

Evaluate $c (\frac{n}{5} + \frac{11}{5})$ by substituting in the value of $c^{- 1}$ into $c$ .

$c (\frac{n}{5} + \frac{11}{5}) = - 6 + 5 ((\frac{n}{5} + \frac{11}{5}) - 1)$

Simplify each term.

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{5}{5}$ .

$c (\frac{n}{5} + \frac{11}{5}) = - 6 + 5 (\frac{n}{5} + \frac{11}{5} - 1 \cdot \frac{5}{5})$

Combine $- 1$ and $\frac{5}{5}$ .

$c (\frac{n}{5} + \frac{11}{5}) = - 6 + 5 (\frac{n}{5} + \frac{11}{5} + \frac{- 1 \cdot 5}{5})$

Combine the numerators over the common denominator.

$c (\frac{n}{5} + \frac{11}{5}) = - 6 + 5 (\frac{n}{5} + \frac{11 - 1 \cdot 5}{5})$

Simplify the numerator.

Multiply $- 1$ by $5$ .

$c (\frac{n}{5} + \frac{11}{5}) = - 6 + 5 (\frac{n}{5} + \frac{11 - 5}{5})$

Subtract $5$ from $11$ .

$c (\frac{n}{5} + \frac{11}{5}) = - 6 + 5 (\frac{n}{5} + \frac{6}{5})$

Apply the distributive property.

$c (\frac{n}{5} + \frac{11}{5}) = - 6 + 5 (\frac{n}{5}) + 5 (\frac{6}{5})$

Cancel the common factor of $5$ .

Cancel the common factor.

$c (\frac{n}{5} + \frac{11}{5}) = - 6 + 5 (\frac{n}{5}) + 5 (\frac{6}{5})$

Rewrite the expression.

$c (\frac{n}{5} + \frac{11}{5}) = - 6 + n + 5 (\frac{6}{5})$

Cancel the common factor of $5$ .

Cancel the common factor.

$c (\frac{n}{5} + \frac{11}{5}) = - 6 + n + 5 (\frac{6}{5})$

Rewrite the expression.

$c (\frac{n}{5} + \frac{11}{5}) = - 6 + n + 6$

Combine the opposite terms in $- 6 + n + 6$ .

Add $- 6$ and $6$ .

$c (\frac{n}{5} + \frac{11}{5}) = n + 0$

Add $n$ and $0$ .

$c (\frac{n}{5} + \frac{11}{5}) = n$

Since $c^{- 1} (c (n)) = n$ and $c (c^{- 1} (n)) = n$ , then $c^{- 1} (n) = \frac{n}{5} + \frac{11}{5}$ is the inverse of $c (n) = - 6 + 5 (n - 1)$ .

$c^{- 1} (n) = \frac{n}{5} + \frac{11}{5}$

Find the Inverse c(n)=-6+5(n-1)

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