Find the Inverse f(x)=-8x+4

$f (x) = - 8 x + 4$

Step 1

Write $f (x) = - 8 x + 4$ as an equation.

$y = - 8 x + 4$

Step 2

Interchange the variables.

$x = - 8 y + 4$

Step 3

Solve for $y$ .

Rewrite the equation as $- 8 y + 4 = x$ .

$- 8 y + 4 = x$

Subtract $4$ from both sides of the equation.

$- 8 y = x - 4$

Divide each term in $- 8 y = x - 4$ by $- 8$ and simplify.

Divide each term in $- 8 y = x - 4$ by $- 8$ .

$\frac{- 8 y}{- 8} = \frac{x}{- 8} + \frac{- 4}{- 8}$

Simplify the left side.

Cancel the common factor of $- 8$ .

Cancel the common factor.

$\frac{- 8 y}{- 8} = \frac{x}{- 8} + \frac{- 4}{- 8}$

Divide $y$ by $1$ .

$y = \frac{x}{- 8} + \frac{- 4}{- 8}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y = - \frac{x}{8} + \frac{- 4}{- 8}$

Cancel the common factor of $- 4$ and $- 8$ .

Factor $- 4$ out of $- 4$ .

$y = - \frac{x}{8} + \frac{- 4 (1)}{- 8}$

Cancel the common factors.

Factor $- 4$ out of $- 8$ .

$y = - \frac{x}{8} + \frac{- 4 \cdot 1}{- 4 \cdot 2}$

Cancel the common factor.

$y = - \frac{x}{8} + \frac{- 4 \cdot 1}{- 4 \cdot 2}$

Rewrite the expression.

$y = - \frac{x}{8} + \frac{1}{2}$

Step 4

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

$f^{- 1} (x) = - \frac{x}{8} + \frac{1}{2}$

Step 5

Verify if $f^{- 1} (x) = - \frac{x}{8} + \frac{1}{2}$ is the inverse of $f (x) = - 8 x + 4$ .

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 $f^{- 1} (- 8 x + 4)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (- 8 x + 4) = - \frac{- 8 x + 4}{8} + \frac{1}{2}$

Simplify terms.

Cancel the common factor of $- 8 x + 4$ and $8$ .

Factor $4$ out of $- 8 x$ .

$f^{- 1} (- 8 x + 4) = - \frac{4 (- 2 x) + 4}{8} + \frac{1}{2}$

Factor $4$ out of $4$ .

$f^{- 1} (- 8 x + 4) = - \frac{4 (- 2 x) + 4 (1)}{8} + \frac{1}{2}$

Factor $4$ out of $4 (- 2 x) + 4 (1)$ .

$f^{- 1} (- 8 x + 4) = - \frac{4 (- 2 x + 1)}{8} + \frac{1}{2}$

Cancel the common factors.

Factor $4$ out of $8$ .

$f^{- 1} (- 8 x + 4) = - \frac{4 (- 2 x + 1)}{4 (2)} + \frac{1}{2}$

Cancel the common factor.

$f^{- 1} (- 8 x + 4) = - \frac{4 (- 2 x + 1)}{4 \cdot 2} + \frac{1}{2}$

Rewrite the expression.

$f^{- 1} (- 8 x + 4) = - \frac{- 2 x + 1}{2} + \frac{1}{2}$

Combine the numerators over the common denominator.

$f^{- 1} (- 8 x + 4) = \frac{- (- 2 x + 1) + 1}{2}$

Simplify each term.

Apply the distributive property.

$f^{- 1} (- 8 x + 4) = \frac{- (- 2 x) - 1 \cdot 1 + 1}{2}$

Multiply $- 2$ by $- 1$ .

$f^{- 1} (- 8 x + 4) = \frac{2 x - 1 \cdot 1 + 1}{2}$

Multiply $- 1$ by $1$ .

$f^{- 1} (- 8 x + 4) = \frac{2 x - 1 + 1}{2}$

Simplify terms.

Combine the opposite terms in $2 x - 1 + 1$ .

Add $- 1$ and $1$ .

$f^{- 1} (- 8 x + 4) = \frac{2 x + 0}{2}$

Add $2 x$ and $0$ .

$f^{- 1} (- 8 x + 4) = \frac{2 x}{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (- 8 x + 4) = \frac{2 x}{2}$

Divide $x$ by $1$ .

$f^{- 1} (- 8 x + 4) = x$

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

Set up the composite result function.

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

Evaluate $f (- \frac{x}{8} + \frac{1}{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \frac{x}{8} + \frac{1}{2}) = - 8 (- \frac{x}{8} + \frac{1}{2}) + 4$

Simplify each term.

Apply the distributive property.

$f (- \frac{x}{8} + \frac{1}{2}) = - 8 (- \frac{x}{8}) - 8 (\frac{1}{2}) + 4$

Cancel the common factor of $8$ .

Move the leading negative in $- \frac{x}{8}$ into the numerator.

$f (- \frac{x}{8} + \frac{1}{2}) = - 8 \frac{- x}{8} - 8 (\frac{1}{2}) + 4$

Factor $8$ out of $- 8$ .

$f (- \frac{x}{8} + \frac{1}{2}) = 8 (- 1) (\frac{- x}{8}) - 8 (\frac{1}{2}) + 4$

Cancel the common factor.

$f (- \frac{x}{8} + \frac{1}{2}) = 8 \cdot (- 1 \frac{- x}{8}) - 8 (\frac{1}{2}) + 4$

Rewrite the expression.

$f (- \frac{x}{8} + \frac{1}{2}) = - 1 (- x) - 8 (\frac{1}{2}) + 4$

Multiply $- 1$ by $- 1$ .

$f (- \frac{x}{8} + \frac{1}{2}) = 1 x - 8 (\frac{1}{2}) + 4$

Multiply $x$ by $1$ .

$f (- \frac{x}{8} + \frac{1}{2}) = x - 8 (\frac{1}{2}) + 4$

Cancel the common factor of $2$ .

Factor $2$ out of $- 8$ .

$f (- \frac{x}{8} + \frac{1}{2}) = x + 2 (- 4) (\frac{1}{2}) + 4$

Cancel the common factor.

$f (- \frac{x}{8} + \frac{1}{2}) = x + 2 \cdot (- 4 (\frac{1}{2})) + 4$

Rewrite the expression.

$f (- \frac{x}{8} + \frac{1}{2}) = x - 4 + 4$

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

Add $- 4$ and $4$ .

$f (- \frac{x}{8} + \frac{1}{2}) = x + 0$

Add $x$ and $0$ .

$f (- \frac{x}{8} + \frac{1}{2}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{x}{8} + \frac{1}{2}$ is the inverse of $f (x) = - 8 x + 4$ .

$f^{- 1} (x) = - \frac{x}{8} + \frac{1}{2}$

Find the Inverse f(x)=-8x+4

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