Find the Inverse 4-5x

$4 - 5 x$

Step 1

Interchange the variables.

$x = 4 - 5 y$

Step 2

Solve for $y$ .

Rewrite the equation as $4 - 5 y = x$ .

$4 - 5 y = x$

Subtract $4$ from both sides of the equation.

$- 5 y = x - 4$

Divide each term in $- 5 y = x - 4$ by $- 5$ and simplify.

Divide each term in $- 5 y = x - 4$ by $- 5$ .

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

Simplify the left side.

Cancel the common factor of $- 5$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

Dividing two negative values results in a positive value.

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

Step 3

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

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

Step 4

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

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

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the distributive property.

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

Multiply $- 1$ by $4$ .

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

Multiply $- 5$ by $- 1$ .

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

Simplify terms.

Combine the opposite terms in $- 4 + 5 x + 4$ .

Add $- 4$ and $4$ .

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

Add $5 x$ and $0$ .

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

Cancel the common factor of $5$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

$f^{- 1} (4 - 5 x) = x$

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{x}{5}$ into the numerator.

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

Factor $5$ out of $- 5$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Cancel the common factor of $5$ .

Factor $5$ out of $- 5$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $4$ .

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

Combine the opposite terms in $4 + x - 4$ .

Subtract $4$ from $4$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse 4-5x

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