Find the Inverse f(x)=3(x-7)+5

$f (x) = 3 (x - 7) + 5$

Write $f (x) = 3 (x - 7) + 5$ as an equation.

$y = 3 (x - 7) + 5$

Interchange the variables.

$x = 3 (y - 7) + 5$

Solve for $y$ .

Rewrite the equation as $3 (y - 7) + 5 = x$ .

$3 (y - 7) + 5 = x$

Simplify $3 (y - 7) + 5$ .

Simplify each term.

Apply the distributive property.

$3 y + 3 \cdot - 7 + 5 = x$

Multiply $3$ by $- 7$ .

$3 y - 21 + 5 = x$

Add $- 21$ and $5$ .

$3 y - 16 = x$

Add $16$ to both sides of the equation.

$3 y = x + 16$

Divide each term in $3 y = x + 16$ by $3$ and simplify.

Divide each term in $3 y = x + 16$ by $3$ .

$\frac{3 y}{3} = \frac{x}{3} + \frac{16}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{x}{3} + \frac{16}{3}$

Divide $y$ by $1$ .

$y = \frac{x}{3} + \frac{16}{3}$

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

$f^{- 1} (x) = \frac{x}{3} + \frac{16}{3}$

Verify if $f^{- 1} (x) = \frac{x}{3} + \frac{16}{3}$ is the inverse of $f (x) = 3 (x - 7) + 5$ .

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 $f^{- 1} (3 (x - 7) + 5)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (3 (x - 7) + 5) = \frac{3 (x - 7) + 5}{3} + \frac{16}{3}$

Combine the numerators over the common denominator.

$f^{- 1} (3 (x - 7) + 5) = \frac{3 (x - 7) + 5 + 16}{3}$

Simplify each term.

Apply the distributive property.

$f^{- 1} (3 (x - 7) + 5) = \frac{3 x + 3 \cdot - 7 + 5 + 16}{3}$

Multiply $3$ by $- 7$ .

$f^{- 1} (3 (x - 7) + 5) = \frac{3 x - 21 + 5 + 16}{3}$

Simplify terms.

Add $- 21$ and $5$ .

$f^{- 1} (3 (x - 7) + 5) = \frac{3 x - 16 + 16}{3}$

Combine the opposite terms in $3 x - 16 + 16$ .

Add $- 16$ and $16$ .

$f^{- 1} (3 (x - 7) + 5) = \frac{3 x + 0}{3}$

Add $3 x$ and $0$ .

$f^{- 1} (3 (x - 7) + 5) = \frac{3 x}{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (3 (x - 7) + 5) = \frac{3 x}{3}$

Divide $x$ by $1$ .

$f^{- 1} (3 (x - 7) + 5) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{x}{3} + \frac{16}{3})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x}{3} + \frac{16}{3}) = 3 ((\frac{x}{3} + \frac{16}{3}) - 7) + 5$

Simplify each term.

To write $- 7$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$f (\frac{x}{3} + \frac{16}{3}) = 3 (\frac{x}{3} + \frac{16}{3} - 7 \cdot \frac{3}{3}) + 5$

Combine $- 7$ and $\frac{3}{3}$ .

$f (\frac{x}{3} + \frac{16}{3}) = 3 (\frac{x}{3} + \frac{16}{3} + \frac{- 7 \cdot 3}{3}) + 5$

Combine the numerators over the common denominator.

$f (\frac{x}{3} + \frac{16}{3}) = 3 (\frac{x}{3} + \frac{16 - 7 \cdot 3}{3}) + 5$

Simplify the numerator.

Multiply $- 7$ by $3$ .

$f (\frac{x}{3} + \frac{16}{3}) = 3 (\frac{x}{3} + \frac{16 - 21}{3}) + 5$

Subtract $21$ from $16$ .

$f (\frac{x}{3} + \frac{16}{3}) = 3 (\frac{x}{3} + \frac{- 5}{3}) + 5$

Move the negative in front of the fraction.

$f (\frac{x}{3} + \frac{16}{3}) = 3 (\frac{x}{3} - \frac{5}{3}) + 5$

Apply the distributive property.

$f (\frac{x}{3} + \frac{16}{3}) = 3 (\frac{x}{3}) + 3 (- \frac{5}{3}) + 5$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{x}{3} + \frac{16}{3}) = 3 (\frac{x}{3}) + 3 (- \frac{5}{3}) + 5$

Rewrite the expression.

$f (\frac{x}{3} + \frac{16}{3}) = x + 3 (- \frac{5}{3}) + 5$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{5}{3}$ into the numerator.

$f (\frac{x}{3} + \frac{16}{3}) = x + 3 (\frac{- 5}{3}) + 5$

Cancel the common factor.

$f (\frac{x}{3} + \frac{16}{3}) = x + 3 (\frac{- 5}{3}) + 5$

Rewrite the expression.

$f (\frac{x}{3} + \frac{16}{3}) = x - 5 + 5$

Combine the opposite terms in $x - 5 + 5$ .

Add $- 5$ and $5$ .

$f (\frac{x}{3} + \frac{16}{3}) = x + 0$

Add $x$ and $0$ .

$f (\frac{x}{3} + \frac{16}{3}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x}{3} + \frac{16}{3}$ is the inverse of $f (x) = 3 (x - 7) + 5$ .

$f^{- 1} (x) = \frac{x}{3} + \frac{16}{3}$

Find the Inverse f(x)=3(x-7)+5

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