Find the Inverse y=5x+1/3

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

Step 1

Interchange the variables.

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

Step 2

Solve for $y$ .

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

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

Subtract $\frac{1}{3}$ from both sides of the equation.

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

Divide each term in $5 y = x - \frac{1}{3}$ by $5$ and simplify.

Divide each term in $5 y = x - \frac{1}{3}$ by $5$ .

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

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $- \frac{1}{3} \cdot \frac{1}{5}$ .

Multiply $\frac{1}{5}$ by $\frac{1}{3}$ .

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

Multiply $5$ by $3$ .

$y = \frac{x}{5} - \frac{1}{15}$

Step 3

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

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

Step 4

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

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

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

Simplify each term.

Simplify the numerator.

To write $5 x$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

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

Combine $5 x$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

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

Multiply $3$ by $5$ .

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $\frac{15 x + 1}{3} \cdot \frac{1}{5}$ .

Multiply $\frac{15 x + 1}{3}$ by $\frac{1}{5}$ .

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

Multiply $3$ by $5$ .

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

Simplify terms.

Combine the numerators over the common denominator.

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

Combine the opposite terms in $15 x + 1 - 1$ .

Subtract $1$ from $1$ .

$f^{- 1} (5 x + \frac{1}{3}) = \frac{15 x + 0}{15}$

Add $15 x$ and $0$ .

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

Cancel the common factor of $15$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{1}{15}$ into the numerator.

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

Factor $5$ out of $15$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Move the negative in front of the fraction.

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

Combine the opposite terms in $x - \frac{1}{3} + \frac{1}{3}$ .

Add $- \frac{1}{3}$ and $\frac{1}{3}$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse y=5x+1/3

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