Find the Inverse f(x)=-5+1/2x

$f (x) = - 5 + \frac{1}{2} x$

Step 1

Write $f (x) = - 5 + \frac{1}{2} x$ as an equation.

$y = - 5 + \frac{1}{2} x$

Step 2

Interchange the variables.

$x = - 5 + \frac{1}{2} y$

Step 3

Solve for $y$ .

Rewrite the equation as $- 5 + \frac{1}{2} y = x$ .

$- 5 + \frac{1}{2} y = x$

Combine $\frac{1}{2}$ and $y$ .

$- 5 + \frac{y}{2} = x$

Add $5$ to both sides of the equation.

$\frac{y}{2} = x + 5$

Multiply both sides of the equation by $2$ .

$2 \frac{y}{2} = 2 (x + 5)$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$2 \frac{y}{2} = 2 (x + 5)$

Rewrite the expression.

$y = 2 (x + 5)$

Simplify the right side.

Simplify $2 (x + 5)$ .

Apply the distributive property.

$y = 2 x + 2 \cdot 5$

Multiply $2$ by $5$ .

$y = 2 x + 10$

Step 4

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

$f^{- 1} (x) = 2 x + 10$

Step 5

Verify if $f^{- 1} (x) = 2 x + 10$ is the inverse of $f (x) = - 5 + \frac{1}{2} x$ .

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 $f^{- 1} (- 5 + \frac{1}{2} x)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (- 5 + \frac{1}{2} x) = 2 (- 5 + \frac{1}{2} x) + 10$

Simplify each term.

Combine $\frac{1}{2}$ and $x$ .

$f^{- 1} (- 5 + \frac{1}{2} x) = 2 (- 5 + \frac{x}{2}) + 10$

Apply the distributive property.

$f^{- 1} (- 5 + \frac{1}{2} x) = 2 \cdot - 5 + 2 (\frac{x}{2}) + 10$

Multiply $2$ by $- 5$ .

$f^{- 1} (- 5 + \frac{1}{2} x) = - 10 + 2 (\frac{x}{2}) + 10$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (- 5 + \frac{1}{2} x) = - 10 + 2 (\frac{x}{2}) + 10$

Rewrite the expression.

$f^{- 1} (- 5 + \frac{1}{2} x) = - 10 + x + 10$

Combine the opposite terms in $- 10 + x + 10$ .

Add $- 10$ and $10$ .

$f^{- 1} (- 5 + \frac{1}{2} x) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (- 5 + \frac{1}{2} x) = x$

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

Set up the composite result function.

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

Evaluate $f (2 x + 10)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (2 x + 10) = - 5 + \frac{1}{2} \cdot (2 x + 10)$

Simplify each term.

Apply the distributive property.

$f (2 x + 10) = - 5 + \frac{1}{2} \cdot (2 x) + \frac{1}{2} \cdot 10$

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

$f (2 x + 10) = - 5 + \frac{1}{2} \cdot (2 (x)) + \frac{1}{2} \cdot 10$

Cancel the common factor.

$f (2 x + 10) = - 5 + \frac{1}{2} \cdot (2 x) + \frac{1}{2} \cdot 10$

Rewrite the expression.

$f (2 x + 10) = - 5 + x + \frac{1}{2} \cdot 10$

Cancel the common factor of $2$ .

Factor $2$ out of $10$ .

$f (2 x + 10) = - 5 + x + \frac{1}{2} \cdot (2 (5))$

Cancel the common factor.

$f (2 x + 10) = - 5 + x + \frac{1}{2} \cdot (2 \cdot 5)$

Rewrite the expression.

$f (2 x + 10) = - 5 + x + 5$

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

Add $- 5$ and $5$ .

$f (2 x + 10) = x + 0$

Add $x$ and $0$ .

$f (2 x + 10) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 2 x + 10$ is the inverse of $f (x) = - 5 + \frac{1}{2} x$ .

$f^{- 1} (x) = 2 x + 10$

Find the Inverse f(x)=-5+1/2x

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