Find the Inverse g(x)=-4x+1

$g (x) = - 4 x + 1$

Step 1

Write $g (x) = - 4 x + 1$ as an equation.

$y = - 4 x + 1$

Step 2

Interchange the variables.

$x = - 4 y + 1$

Step 3

Solve for $y$ .

Rewrite the equation as $- 4 y + 1 = x$ .

$- 4 y + 1 = x$

Subtract $1$ from both sides of the equation.

$- 4 y = x - 1$

Divide each term in $- 4 y = x - 1$ by $- 4$ and simplify.

Divide each term in $- 4 y = x - 1$ by $- 4$ .

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

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

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

Dividing two negative values results in a positive value.

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

Step 4

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

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

Step 5

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

To verify the inverse, check if $g^{- 1} (g (x)) = x$ and $g (g^{- 1} (x)) = x$ .

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

Set up the composite result function.

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

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

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

Combine the numerators over the common denominator.

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

Simplify each term.

Apply the distributive property.

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

Multiply $- 4$ by $- 1$ .

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

Multiply $- 1$ by $1$ .

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

Simplify terms.

Combine the opposite terms in $4 x - 1 + 1$ .

Add $- 1$ and $1$ .

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

Add $4 x$ and $0$ .

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

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Apply the distributive property.

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

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{x}{4}$ into the numerator.

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

Factor $4$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Multiply $x$ by $1$ .

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

Cancel the common factor of $4$ .

Factor $4$ out of $- 4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

Add $- 1$ and $1$ .

$g (- \frac{x}{4} + \frac{1}{4}) = x + 0$

Add $x$ and $0$ .

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

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

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

Find the Inverse g(x)=-4x+1

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