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

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

Step 1

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

$y = 2 x + 1$

Step 2

Interchange the variables.

$x = 2 y + 1$

Step 3

Solve for $y$ .

Rewrite the equation as $2 y + 1 = x$ .

$2 y + 1 = x$

Subtract $1$ from both sides of the equation.

$2 y = x - 1$

Divide each term in $2 y = x - 1$ by $2$ and simplify.

Divide each term in $2 y = x - 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 4

Replace $y$ with $f (x)$ to show the final answer.

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

Step 5

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

To verify the inverse, check if $f (f^{- 1} (x)) = x$ and $f^{- 1} (f (x)) = x$ .

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

Set up the composite result function.

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

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

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

Combine the numerators over the common denominator.

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

Combine the opposite terms in $2 x + 1 - 1$ .

Subtract $1$ from $1$ .

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

Add $2 x$ and $0$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

$f (2 x + 1) = x$

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Combine the opposite terms in $x - 1 + 1$ .

Add $- 1$ and $1$ .

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

Add $x$ and $0$ .

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

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

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

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

Related Questions

Support LCMGCF.com by disabling your adblocker.