Find the Inverse f(x)=1/3x-5

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

Step 1

Write $f (x) = \frac{1}{3} x - 5$ as an equation.

$y = \frac{1}{3} x - 5$

Step 2

Interchange the variables.

$x = \frac{1}{3} y - 5$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{1}{3} y - 5 = x$ .

$\frac{1}{3} y - 5 = x$

Combine $\frac{1}{3}$ and $y$ .

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

Add $5$ to both sides of the equation.

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

Multiply both sides of the equation by $3$ .

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

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

$y = 3 (x + 5)$

Simplify the right side.

Simplify $3 (x + 5)$ .

Apply the distributive property.

$y = 3 x + 3 \cdot 5$

Multiply $3$ by $5$ .

$y = 3 x + 15$

Step 4

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

$f^{- 1} (x) = 3 x + 15$

Step 5

Verify if $f^{- 1} (x) = 3 x + 15$ is the inverse of $f (x) = \frac{1}{3} x - 5$ .

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

$f^{- 1} (\frac{1}{3} x - 5) = 3 (\frac{1}{3} x - 5) + 15$

Simplify each term.

Combine $\frac{1}{3}$ and $x$ .

$f^{- 1} (\frac{1}{3} x - 5) = 3 (\frac{x}{3} - 5) + 15$

Apply the distributive property.

$f^{- 1} (\frac{1}{3} x - 5) = 3 (\frac{x}{3}) + 3 \cdot - 5 + 15$

Cancel the common factor of $3$ .

Cancel the common factor.

$f^{- 1} (\frac{1}{3} x - 5) = 3 (\frac{x}{3}) + 3 \cdot - 5 + 15$

Rewrite the expression.

$f^{- 1} (\frac{1}{3} x - 5) = x + 3 \cdot - 5 + 15$

Multiply $3$ by $- 5$ .

$f^{- 1} (\frac{1}{3} x - 5) = x - 15 + 15$

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

Add $- 15$ and $15$ .

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

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

Evaluate $f (3 x + 15)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (3 x + 15) = \frac{1}{3} \cdot (3 x + 15) - 5$

Simplify each term.

Apply the distributive property.

$f (3 x + 15) = \frac{1}{3} \cdot (3 x) + \frac{1}{3} \cdot 15 - 5$

Cancel the common factor of $3$ .

Factor $3$ out of $3 x$ .

$f (3 x + 15) = \frac{1}{3} \cdot (3 (x)) + \frac{1}{3} \cdot 15 - 5$

Cancel the common factor.

$f (3 x + 15) = \frac{1}{3} \cdot (3 x) + \frac{1}{3} \cdot 15 - 5$

Rewrite the expression.

$f (3 x + 15) = x + \frac{1}{3} \cdot 15 - 5$

Cancel the common factor of $3$ .

Factor $3$ out of $15$ .

$f (3 x + 15) = x + \frac{1}{3} \cdot (3 (5)) - 5$

Cancel the common factor.

$f (3 x + 15) = x + \frac{1}{3} \cdot (3 \cdot 5) - 5$

Rewrite the expression.

$f (3 x + 15) = x + 5 - 5$

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

Subtract $5$ from $5$ .

$f (3 x + 15) = x + 0$

Add $x$ and $0$ .

$f (3 x + 15) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 3 x + 15$ is the inverse of $f (x) = \frac{1}{3} x - 5$ .

$f^{- 1} (x) = 3 x + 15$

Find the Inverse f(x)=1/3x-5

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