Find the Inverse f(x)=(3x-5)/2

$f (x) = \frac{3 x - 5}{2}$

Step 1

Write $f (x) = \frac{3 x - 5}{2}$ as an equation.

$y = \frac{3 x - 5}{2}$

Step 2

Interchange the variables.

$x = \frac{3 y - 5}{2}$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{3 y - 5}{2} = x$ .

$\frac{3 y - 5}{2} = x$

Multiply both sides by $2$ .

$\frac{3 y - 5}{2} \cdot 2 = x \cdot 2$

Simplify.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{3 y - 5}{2} \cdot 2 = x \cdot 2$

Rewrite the expression.

$3 y - 5 = x \cdot 2$

Simplify the right side.

Move $2$ to the left of $x$ .

$3 y - 5 = 2 x$

Solve for $y$ .

Add $5$ to both sides of the equation.

$3 y = 2 x + 5$

Divide each term in $3 y = 2 x + 5$ by $3$ and simplify.

Divide each term in $3 y = 2 x + 5$ by $3$ .

$\frac{3 y}{3} = \frac{2 x}{3} + \frac{5}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{2 x}{3} + \frac{5}{3}$

Divide $y$ by $1$ .

$y = \frac{2 x}{3} + \frac{5}{3}$

Step 4

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

$f^{- 1} (x) = \frac{2 x}{3} + \frac{5}{3}$

Step 5

Verify if $f^{- 1} (x) = \frac{2 x}{3} + \frac{5}{3}$ is the inverse of $f (x) = \frac{3 x - 5}{2}$ .

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 $f^{- 1} (\frac{3 x - 5}{2})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{3 x - 5}{2}) = \frac{2 (\frac{3 x - 5}{2})}{3} + \frac{5}{3}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{3 x - 5}{2}) = \frac{2 (\frac{3 x - 5}{2}) + 5}{3}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\frac{3 x - 5}{2}) = \frac{2 (\frac{3 x - 5}{2}) + 5}{3}$

Rewrite the expression.

$f^{- 1} (\frac{3 x - 5}{2}) = \frac{3 x - 5 + 5}{3}$

Combine the opposite terms in $3 x - 5 + 5$ .

Add $- 5$ and $5$ .

$f^{- 1} (\frac{3 x - 5}{2}) = \frac{3 x + 0}{3}$

Add $3 x$ and $0$ .

$f^{- 1} (\frac{3 x - 5}{2}) = \frac{3 x}{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (\frac{3 x - 5}{2}) = \frac{3 x}{3}$

Divide $x$ by $1$ .

$f^{- 1} (\frac{3 x - 5}{2}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{2 x}{3} + \frac{5}{3})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{2 x}{3} + \frac{5}{3}) = \frac{3 (\frac{2 x}{3} + \frac{5}{3}) - 5}{2}$

Simplify the numerator.

Apply the distributive property.

$f (\frac{2 x}{3} + \frac{5}{3}) = \frac{3 (\frac{2 x}{3}) + 3 (\frac{5}{3}) - 5}{2}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{2 x}{3} + \frac{5}{3}) = \frac{3 (\frac{2 x}{3}) + 3 (\frac{5}{3}) - 5}{2}$

Rewrite the expression.

$f (\frac{2 x}{3} + \frac{5}{3}) = \frac{2 x + 3 (\frac{5}{3}) - 5}{2}$

Cancel the common factor of $3$ .

Cancel the common factor.

$f (\frac{2 x}{3} + \frac{5}{3}) = \frac{2 x + 3 (\frac{5}{3}) - 5}{2}$

Rewrite the expression.

$f (\frac{2 x}{3} + \frac{5}{3}) = \frac{2 x + 5 - 5}{2}$

Subtract $5$ from $5$ .

$f (\frac{2 x}{3} + \frac{5}{3}) = \frac{2 x + 0}{2}$

Add $2 x$ and $0$ .

$f (\frac{2 x}{3} + \frac{5}{3}) = \frac{2 x}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f (\frac{2 x}{3} + \frac{5}{3}) = \frac{2 x}{2}$

Divide $x$ by $1$ .

$f (\frac{2 x}{3} + \frac{5}{3}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{2 x}{3} + \frac{5}{3}$ is the inverse of $f (x) = \frac{3 x - 5}{2}$ .

$f^{- 1} (x) = \frac{2 x}{3} + \frac{5}{3}$

Find the Inverse f(x)=(3x-5)/2

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