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

$f (x) = - x + 3$

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

$y = - x + 3$

Interchange the variables.

$x = - y + 3$

Solve for $y$ .

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

$- y + 3 = x$

Subtract $3$ from both sides of the equation.

$- y = x - 3$

Divide each term in $- y = x - 3$ by $- 1$ and simplify.

Divide each term in $- y = x - 3$ by $- 1$ .

$\frac{- y}{- 1} = \frac{x}{- 1} + \frac{- 3}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y}{1} = \frac{x}{- 1} + \frac{- 3}{- 1}$

Divide $y$ by $1$ .

$y = \frac{x}{- 1} + \frac{- 3}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{x}{- 1}$ .

$y = - 1 \cdot x + \frac{- 3}{- 1}$

Rewrite $- 1 \cdot x$ as $- x$ .

$y = - x + \frac{- 3}{- 1}$

Divide $- 3$ by $- 1$ .

$y = - x + 3$

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

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

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

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

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

Simplify each term.

Apply the distributive property.

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

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 1$ by $3$ .

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

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

Add $- 3$ and $3$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

$f (- x + 3) = x - 1 \cdot 3 + 3$

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

$f (- x + 3) = 1 x - 1 \cdot 3 + 3$

Multiply $x$ by $1$ .

$f (- x + 3) = x - 1 \cdot 3 + 3$

Multiply $- 1$ by $3$ .

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

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

Add $- 3$ and $3$ .

$f (- x + 3) = x + 0$

Add $x$ and $0$ .

$f (- x + 3) = x$

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

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

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

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