Find the Inverse f(x)=4/5*(x-15)

$f (x) = \frac{4}{5} \cdot (x - 15)$

Step 1

Write $f (x) = \frac{4}{5} \cdot (x - 15)$ as an equation.

$y = \frac{4}{5} \cdot (x - 15)$

Step 2

Interchange the variables.

$x = \frac{4}{5} \cdot (y - 15)$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{4}{5} \cdot (y - 15) = x$ .

$\frac{4}{5} \cdot (y - 15) = x$

Multiply both sides of the equation by $\frac{5}{4}$ .

$\frac{5}{4} (\frac{4}{5} \cdot (y - 15)) = \frac{5}{4} x$

Simplify both sides of the equation.

Simplify the left side.

Simplify $\frac{5}{4} (\frac{4}{5} \cdot (y - 15))$ .

Apply the distributive property.

$\frac{5}{4} (\frac{4}{5} y + \frac{4}{5} \cdot - 15) = \frac{5}{4} x$

Combine $\frac{4}{5}$ and $y$ .

$\frac{5}{4} (\frac{4 y}{5} + \frac{4}{5} \cdot - 15) = \frac{5}{4} x$

Cancel the common factor of $5$ .

Factor $5$ out of $- 15$ .

$\frac{5}{4} (\frac{4 y}{5} + \frac{4}{5} \cdot (5 (- 3))) = \frac{5}{4} x$

Cancel the common factor.

$\frac{5}{4} (\frac{4 y}{5} + \frac{4}{5} \cdot (5 \cdot - 3)) = \frac{5}{4} x$

Rewrite the expression.

$\frac{5}{4} (\frac{4 y}{5} + 4 \cdot - 3) = \frac{5}{4} x$

Multiply $4$ by $- 3$ .

$\frac{5}{4} (\frac{4 y}{5} - 12) = \frac{5}{4} x$

Apply the distributive property.

$\frac{5}{4} \cdot \frac{4 y}{5} + \frac{5}{4} \cdot - 12 = \frac{5}{4} x$

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5}{4} \cdot \frac{4 y}{5} + \frac{5}{4} \cdot - 12 = \frac{5}{4} x$

Rewrite the expression.

$\frac{1}{4} (4 y) + \frac{5}{4} \cdot - 12 = \frac{5}{4} x$

Cancel the common factor of $4$ .

Factor $4$ out of $4 y$ .

$\frac{1}{4} (4 (y)) + \frac{5}{4} \cdot - 12 = \frac{5}{4} x$

Cancel the common factor.

$\frac{1}{4} (4 y) + \frac{5}{4} \cdot - 12 = \frac{5}{4} x$

Rewrite the expression.

$y + \frac{5}{4} \cdot - 12 = \frac{5}{4} x$

Cancel the common factor of $4$ .

Factor $4$ out of $- 12$ .

$y + \frac{5}{4} \cdot (4 (- 3)) = \frac{5}{4} x$

Cancel the common factor.

$y + \frac{5}{4} \cdot (4 \cdot - 3) = \frac{5}{4} x$

Rewrite the expression.

$y + 5 \cdot - 3 = \frac{5}{4} x$

Multiply $5$ by $- 3$ .

$y - 15 = \frac{5}{4} x$

Simplify the right side.

Combine $\frac{5}{4}$ and $x$ .

$y - 15 = \frac{5 x}{4}$

Add $15$ to both sides of the equation.

$y = \frac{5 x}{4} + 15$

Step 4

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

$f^{- 1} (x) = \frac{5 x}{4} + 15$

Step 5

Verify if $f^{- 1} (x) = \frac{5 x}{4} + 15$ is the inverse of $f (x) = \frac{4}{5} \cdot (x - 15)$ .

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 $f^{- 1} (\frac{4}{5} \cdot (x - 15))$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{4}{5} \cdot (x - 15)) = \frac{5 (\frac{4}{5} \cdot (x - 15))}{4} + 15$

Simplify each term.

Combine $5$ and $\frac{4}{5}$ .

$f^{- 1} (\frac{4}{5} \cdot (x - 15)) = \frac{\frac{5 \cdot 4}{5} \cdot (x - 15)}{4} + 15$

Multiply $5$ by $4$ .

$f^{- 1} (\frac{4}{5} \cdot (x - 15)) = \frac{\frac{20}{5} \cdot (x - 15)}{4} + 15$

Divide $20$ by $5$ .

$f^{- 1} (\frac{4}{5} \cdot (x - 15)) = \frac{4 (x - 15)}{4} + 15$

Cancel the common factor of $4$ .

Cancel the common factor.

$f^{- 1} (\frac{4}{5} \cdot (x - 15)) = \frac{4 (x - 15)}{4} + 15$

Divide $x - 15$ by $1$ .

$f^{- 1} (\frac{4}{5} \cdot (x - 15)) = x - 15 + 15$

Combine the opposite terms in $x - 15 + 15$ .

Add $- 15$ and $15$ .

$f^{- 1} (\frac{4}{5} \cdot (x - 15)) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\frac{4}{5} \cdot (x - 15)) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{5 x}{4} + 15)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{5 x}{4} + 15) = \frac{4}{5} \cdot ((\frac{5 x}{4} + 15) - 15)$

Combine the opposite terms in $(\frac{5 x}{4} + 15) - 15$ .

Subtract $15$ from $15$ .

$f (\frac{5 x}{4} + 15) = \frac{4}{5} \cdot (\frac{5 x}{4} + 0)$

Add $\frac{5 x}{4}$ and $0$ .

$f (\frac{5 x}{4} + 15) = \frac{4}{5} \cdot \frac{5 x}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$f (\frac{5 x}{4} + 15) = \frac{4}{5} \cdot \frac{5 x}{4}$

Rewrite the expression.

$f (\frac{5 x}{4} + 15) = \frac{1}{5} \cdot (5 x)$

Cancel the common factor of $5$ .

Factor $5$ out of $5 x$ .

$f (\frac{5 x}{4} + 15) = \frac{1}{5} \cdot (5 (x))$

Cancel the common factor.

$f (\frac{5 x}{4} + 15) = \frac{1}{5} \cdot (5 x)$

Rewrite the expression.

$f (\frac{5 x}{4} + 15) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{5 x}{4} + 15$ is the inverse of $f (x) = \frac{4}{5} \cdot (x - 15)$ .

$f^{- 1} (x) = \frac{5 x}{4} + 15$

Find the Inverse f(x)=4/5*(x-15)

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