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 $2 y - 10 = x$ .

$2 y - 10 = x$

Add $10$ to 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.

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 $f^{- 1}$ .

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

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 $1$ .

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

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 $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $2$ by $5$ .

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

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

Subtract $10$ from $10$ .

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

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

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