Find the Inverse 5/2x-1

$\frac{5}{2} x - 1$

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Combine $\frac{5}{2}$ and $y$ .

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

Add $1$ to both sides of the equation.

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

Multiply both sides of the equation by $\frac{2}{5}$ .

$\frac{2}{5} \cdot \frac{5 y}{2} = \frac{2}{5} (x + 1)$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{2}{5} \cdot \frac{5 y}{2}$ .

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2}{5} \cdot \frac{5 y}{2} = \frac{2}{5} (x + 1)$

Rewrite the expression.

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

Cancel the common factor of $5$ .

Factor $5$ out of $5 y$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify the right side.

Simplify $\frac{2}{5} (x + 1)$ .

Apply the distributive property.

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

Combine $\frac{2}{5}$ and $x$ .

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

Multiply $\frac{2}{5}$ by $1$ .

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

Step 3

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

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

Step 4

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

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

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

Combine the numerators over the common denominator.

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

Simplify each term.

Combine $\frac{5}{2}$ and $x$ .

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

Apply the distributive property.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $2$ by $- 1$ .

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

Simplify terms.

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

Add $- 2$ and $2$ .

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

Add $5 x$ and $0$ .

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

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

Subtract $1$ from $1$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse 5/2x-1

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