Find the Inverse f(x)=-4x+7

$f (x) = - 4 x + 7$

Step 1

Write $f (x) = - 4 x + 7$ as an equation.

$y = - 4 x + 7$

Step 2

Interchange the variables.

$x = - 4 y + 7$

Step 3

Solve for $y$ .

Rewrite the equation as $- 4 y + 7 = x$ .

$- 4 y + 7 = x$

Subtract $7$ from both sides of the equation.

$- 4 y = x - 7$

Divide each term in $- 4 y = x - 7$ by $- 4$ and simplify.

Divide each term in $- 4 y = x - 7$ by $- 4$ .

$\frac{- 4 y}{- 4} = \frac{x}{- 4} + \frac{- 7}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 y}{- 4} = \frac{x}{- 4} + \frac{- 7}{- 4}$

Divide $y$ by $1$ .

$y = \frac{x}{- 4} + \frac{- 7}{- 4}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y = - \frac{x}{4} + \frac{- 7}{- 4}$

Dividing two negative values results in a positive value.

$y = - \frac{x}{4} + \frac{7}{4}$

Step 4

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

$f^{- 1} (x) = - \frac{x}{4} + \frac{7}{4}$

Step 5

Verify if $f^{- 1} (x) = - \frac{x}{4} + \frac{7}{4}$ is the inverse of $f (x) = - 4 x + 7$ .

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 $f^{- 1} (- 4 x + 7)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (- 4 x + 7) = - \frac{- 4 x + 7}{4} + \frac{7}{4}$

Combine the numerators over the common denominator.

$f^{- 1} (- 4 x + 7) = \frac{- (- 4 x + 7) + 7}{4}$

Simplify each term.

Apply the distributive property.

$f^{- 1} (- 4 x + 7) = \frac{- (- 4 x) - 1 \cdot 7 + 7}{4}$

Multiply $- 4$ by $- 1$ .

$f^{- 1} (- 4 x + 7) = \frac{4 x - 1 \cdot 7 + 7}{4}$

Multiply $- 1$ by $7$ .

$f^{- 1} (- 4 x + 7) = \frac{4 x - 7 + 7}{4}$

Simplify terms.

Combine the opposite terms in $4 x - 7 + 7$ .

Add $- 7$ and $7$ .

$f^{- 1} (- 4 x + 7) = \frac{4 x + 0}{4}$

Add $4 x$ and $0$ .

$f^{- 1} (- 4 x + 7) = \frac{4 x}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$f^{- 1} (- 4 x + 7) = \frac{4 x}{4}$

Divide $x$ by $1$ .

$f^{- 1} (- 4 x + 7) = x$

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

Set up the composite result function.

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

Evaluate $f (- \frac{x}{4} + \frac{7}{4})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (- \frac{x}{4} + \frac{7}{4}) = - 4 (- \frac{x}{4} + \frac{7}{4}) + 7$

Simplify each term.

Apply the distributive property.

$f (- \frac{x}{4} + \frac{7}{4}) = - 4 (- \frac{x}{4}) - 4 (\frac{7}{4}) + 7$

Cancel the common factor of $4$ .

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

$f (- \frac{x}{4} + \frac{7}{4}) = - 4 \frac{- x}{4} - 4 (\frac{7}{4}) + 7$

Factor $4$ out of $- 4$ .

$f (- \frac{x}{4} + \frac{7}{4}) = 4 (- 1) (\frac{- x}{4}) - 4 (\frac{7}{4}) + 7$

Cancel the common factor.

$f (- \frac{x}{4} + \frac{7}{4}) = 4 \cdot (- 1 \frac{- x}{4}) - 4 (\frac{7}{4}) + 7$

Rewrite the expression.

$f (- \frac{x}{4} + \frac{7}{4}) = - 1 (- x) - 4 (\frac{7}{4}) + 7$

Multiply $- 1$ by $- 1$ .

$f (- \frac{x}{4} + \frac{7}{4}) = 1 x - 4 (\frac{7}{4}) + 7$

Multiply $x$ by $1$ .

$f (- \frac{x}{4} + \frac{7}{4}) = x - 4 (\frac{7}{4}) + 7$

Cancel the common factor of $4$ .

Factor $4$ out of $- 4$ .

$f (- \frac{x}{4} + \frac{7}{4}) = x + 4 (- 1) (\frac{7}{4}) + 7$

Cancel the common factor.

$f (- \frac{x}{4} + \frac{7}{4}) = x + 4 \cdot (- 1 (\frac{7}{4})) + 7$

Rewrite the expression.

$f (- \frac{x}{4} + \frac{7}{4}) = x - 1 \cdot 7 + 7$

Multiply $- 1$ by $7$ .

$f (- \frac{x}{4} + \frac{7}{4}) = x - 7 + 7$

Combine the opposite terms in $x - 7 + 7$ .

Add $- 7$ and $7$ .

$f (- \frac{x}{4} + \frac{7}{4}) = x + 0$

Add $x$ and $0$ .

$f (- \frac{x}{4} + \frac{7}{4}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = - \frac{x}{4} + \frac{7}{4}$ is the inverse of $f (x) = - 4 x + 7$ .

$f^{- 1} (x) = - \frac{x}{4} + \frac{7}{4}$

Find the Inverse f(x)=-4x+7

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