Find the Inverse f=9/5c+32

$f = \frac{9}{5} c + 32$

Interchange the variables.

$c = \frac{9}{5} y + 32$

Solve for $y$ .

Rewrite the equation as $\frac{9}{5} y + 32 = c$ .

$\frac{9}{5} y + 32 = c$

Combine $\frac{9}{5}$ and $y$ .

$\frac{9 y}{5} + 32 = c$

Subtract $32$ from both sides of the equation.

$\frac{9 y}{5} = c - 32$

Multiply both sides of the equation by $\frac{5}{9}$ .

$\frac{5}{9} \cdot \frac{9 y}{5} = \frac{5}{9} (c - 32)$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{5}{9} \cdot \frac{9 y}{5}$ .

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5}{9} \cdot \frac{9 y}{5} = \frac{5}{9} (c - 32)$

Rewrite the expression.

$\frac{1}{9} (9 y) = \frac{5}{9} (c - 32)$

Cancel the common factor of $9$ .

Factor $9$ out of $9 y$ .

$\frac{1}{9} (9 (y)) = \frac{5}{9} (c - 32)$

Cancel the common factor.

$\frac{1}{9} (9 y) = \frac{5}{9} (c - 32)$

Rewrite the expression.

$y = \frac{5}{9} (c - 32)$

Simplify the right side.

Simplify $\frac{5}{9} (c - 32)$ .

Apply the distributive property.

$y = \frac{5}{9} c + \frac{5}{9} \cdot - 32$

Combine $\frac{5}{9}$ and $c$ .

$y = \frac{5 c}{9} + \frac{5}{9} \cdot - 32$

Multiply $\frac{5}{9} \cdot - 32$ .

Combine $\frac{5}{9}$ and $- 32$ .

$y = \frac{5 c}{9} + \frac{5 \cdot - 32}{9}$

Multiply $5$ by $- 32$ .

$y = \frac{5 c}{9} + \frac{- 160}{9}$

Move the negative in front of the fraction.

$y = \frac{5 c}{9} - \frac{160}{9}$

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

$f^{- 1} (c) = \frac{5 c}{9} - \frac{160}{9}$

Verify if $f^{- 1} (c) = \frac{5 c}{9} - \frac{160}{9}$ is the inverse of $f (c) = \frac{9}{5} c + 32$ .

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

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

Set up the composite result function.

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

Evaluate $f^{- 1} (\frac{9}{5} c + 32)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{9}{5} c + 32) = \frac{5 (\frac{9}{5} c + 32)}{9} - \frac{160}{9}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{9}{5} c + 32) = \frac{5 (\frac{9}{5} c + 32) - 160}{9}$

Simplify each term.

Combine $\frac{9}{5}$ and $c$ .

$f^{- 1} (\frac{9}{5} c + 32) = \frac{5 (\frac{9 c}{5} + 32) - 160}{9}$

Apply the distributive property.

$f^{- 1} (\frac{9}{5} c + 32) = \frac{5 (\frac{9 c}{5}) + 5 \cdot 32 - 160}{9}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f^{- 1} (\frac{9}{5} c + 32) = \frac{5 (\frac{9 c}{5}) + 5 \cdot 32 - 160}{9}$

Rewrite the expression.

$f^{- 1} (\frac{9}{5} c + 32) = \frac{9 c + 5 \cdot 32 - 160}{9}$

Multiply $5$ by $32$ .

$f^{- 1} (\frac{9}{5} c + 32) = \frac{9 c + 160 - 160}{9}$

Simplify terms.

Combine the opposite terms in $9 c + 160 - 160$ .

Subtract $160$ from $160$ .

$f^{- 1} (\frac{9}{5} c + 32) = \frac{9 c + 0}{9}$

Add $9 c$ and $0$ .

$f^{- 1} (\frac{9}{5} c + 32) = \frac{9 c}{9}$

Cancel the common factor of $9$ .

Cancel the common factor.

$f^{- 1} (\frac{9}{5} c + 32) = \frac{9 c}{9}$

Divide $c$ by $1$ .

$f^{- 1} (\frac{9}{5} c + 32) = c$

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

Set up the composite result function.

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

Evaluate $f (\frac{5 c}{9} - \frac{160}{9})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{5 c}{9} - \frac{160}{9}) = \frac{9}{5} \cdot (\frac{5 c}{9} - \frac{160}{9}) + 32$

Simplify each term.

Apply the distributive property.

$f (\frac{5 c}{9} - \frac{160}{9}) = \frac{9}{5} \cdot \frac{5 c}{9} + \frac{9}{5} \cdot (- \frac{160}{9}) + 32$

Cancel the common factor of $9$ .

Cancel the common factor.

$f (\frac{5 c}{9} - \frac{160}{9}) = \frac{9}{5} \cdot \frac{5 c}{9} + \frac{9}{5} \cdot (- \frac{160}{9}) + 32$

Rewrite the expression.

$f (\frac{5 c}{9} - \frac{160}{9}) = \frac{1}{5} \cdot (5 c) + \frac{9}{5} \cdot (- \frac{160}{9}) + 32$

Cancel the common factor of $5$ .

Factor $5$ out of $5 c$ .

$f (\frac{5 c}{9} - \frac{160}{9}) = \frac{1}{5} \cdot (5 (c)) + \frac{9}{5} \cdot (- \frac{160}{9}) + 32$

Cancel the common factor.

$f (\frac{5 c}{9} - \frac{160}{9}) = \frac{1}{5} \cdot (5 c) + \frac{9}{5} \cdot (- \frac{160}{9}) + 32$

Rewrite the expression.

$f (\frac{5 c}{9} - \frac{160}{9}) = c + \frac{9}{5} \cdot (- \frac{160}{9}) + 32$

Cancel the common factor of $9$ .

Move the leading negative in $- \frac{160}{9}$ into the numerator.

$f (\frac{5 c}{9} - \frac{160}{9}) = c + \frac{9}{5} \cdot \frac{- 160}{9} + 32$

Cancel the common factor.

$f (\frac{5 c}{9} - \frac{160}{9}) = c + \frac{9}{5} \cdot \frac{- 160}{9} + 32$

Rewrite the expression.

$f (\frac{5 c}{9} - \frac{160}{9}) = c + \frac{1}{5} \cdot - 160 + 32$

Cancel the common factor of $5$ .

Factor $5$ out of $- 160$ .

$f (\frac{5 c}{9} - \frac{160}{9}) = c + \frac{1}{5} \cdot (5 (- 32)) + 32$

Cancel the common factor.

$f (\frac{5 c}{9} - \frac{160}{9}) = c + \frac{1}{5} \cdot (5 \cdot - 32) + 32$

Rewrite the expression.

$f (\frac{5 c}{9} - \frac{160}{9}) = c - 32 + 32$

Combine the opposite terms in $c - 32 + 32$ .

Add $- 32$ and $32$ .

$f (\frac{5 c}{9} - \frac{160}{9}) = c + 0$

Add $c$ and $0$ .

$f (\frac{5 c}{9} - \frac{160}{9}) = c$

Since $f^{- 1} (f (c)) = c$ and $f (f^{- 1} (c)) = c$ , then $f^{- 1} (c) = \frac{5 c}{9} - \frac{160}{9}$ is the inverse of $f (c) = \frac{9}{5} c + 32$ .

$f^{- 1} (c) = \frac{5 c}{9} - \frac{160}{9}$

Find the Inverse f=9/5c+32

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