Find the Inverse f(x)=-2x+10

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

Step 1

Write $f (x) = - 2 x + 10$ as an equation.

$y = - 2 x + 10$

Step 2

Interchange the variables.

$x = - 2 y + 10$

Step 3

Solve for $y$ .

Rewrite the equation as $- 2 y + 10 = x$ .

$- 2 y + 10 = x$

Subtract $10$ from both sides of the equation.

$- 2 y = x - 10$

Divide each term in $- 2 y = x - 10$ by $- 2$ and simplify.

Divide each term in $- 2 y = x - 10$ by $- 2$ .

$\frac{- 2 y}{- 2} = \frac{x}{- 2} + \frac{- 10}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 y}{- 2} = \frac{x}{- 2} + \frac{- 10}{- 2}$

Divide $y$ by $1$ .

$y = \frac{x}{- 2} + \frac{- 10}{- 2}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y = - \frac{x}{2} + \frac{- 10}{- 2}$

Divide $- 10$ by $- 2$ .

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

Step 4

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

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

Step 5

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

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

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

Simplify each term.

Cancel the common factor of $- 2 x + 10$ and $2$ .

Factor $2$ out of $- 2 x$ .

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

Factor $2$ out of $10$ .

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

Factor $2$ out of $2 (- x) + 2 (5)$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $- x + 5$ by $1$ .

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

Apply the distributive property.

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

Multiply $- - x$ .

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 1$ by $5$ .

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

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

Add $- 5$ and $5$ .

$f^{- 1} (- 2 x + 10) = x + 0$

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{x}{2}$ into the numerator.

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

Factor $2$ out of $- 2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Multiply $- 2$ by $5$ .

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

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

Add $- 10$ and $10$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse f(x)=-2x+10

Related Questions

Support LCMGCF.com by disabling your adblocker.