Find the Inverse F(x)=2x+5

$F (x) = 2 x + 5$

Step 1

Write $F (x) = 2 x + 5$ as an equation.

$y = 2 x + 5$

Step 2

Interchange the variables.

$x = 2 y + 5$

Step 3

Solve for $y$ .

Rewrite the equation as $2 y + 5 = x$ .

$2 y + 5 = x$

Subtract $5$ from both sides of the equation.

$2 y = x - 5$

Divide each term in $2 y = x - 5$ by $2$ and simplify.

Divide each term in $2 y = x - 5$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 4

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

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

Step 5

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

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

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

Combine the numerators over the common denominator.

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

Combine the opposite terms in $2 x + 5 - 5$ .

Subtract $5$ from $5$ .

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

Add $2 x$ and $0$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{5}{2}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Add $- 5$ and $5$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse F(x)=2x+5

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