Find the Inverse f(x)=1/(4x)-7

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

Step 1

Write $f (x) = \frac{1}{4 x} - 7$ as an equation.

$y = \frac{1}{4 x} - 7$

Step 2

Interchange the variables.

$x = \frac{1}{4 y} - 7$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{1}{4 y} - 7 = x$ .

$\frac{1}{4 y} - 7 = x$

Add $7$ to both sides of the equation.

$\frac{1}{4 y} = x + 7$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$4 y, 1, 1$

The LCM of one and any expression is the expression.

$4 y$

Multiply each term in $\frac{1}{4 y} = x + 7$ by $4 y$ to eliminate the fractions.

Multiply each term in $\frac{1}{4 y} = x + 7$ by $4 y$ .

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

Simplify the left side.

Rewrite using the commutative property of multiplication.

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

Cancel the common factor of $4$ .

Factor $4$ out of $4 y$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $y$ .

Cancel the common factor.

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

Rewrite the expression.

$1 = x (4 y) + 7 (4 y)$

Simplify the right side.

Simplify each term.

Rewrite using the commutative property of multiplication.

$1 = 4 x y + 7 (4 y)$

Multiply $4$ by $7$ .

$1 = 4 x y + 28 y$

Solve the equation.

Rewrite the equation as $4 x y + 28 y = 1$ .

$4 x y + 28 y = 1$

Factor $4 y$ out of $4 x y + 28 y$ .

Factor $4 y$ out of $4 x y$ .

$4 y (x) + 28 y = 1$

Factor $4 y$ out of $28 y$ .

$4 y (x) + 4 y (7) = 1$

Factor $4 y$ out of $4 y (x) + 4 y (7)$ .

$4 y (x + 7) = 1$

Divide each term in $4 y (x + 7) = 1$ by $4 (x + 7)$ and simplify.

Divide each term in $4 y (x + 7) = 1$ by $4 (x + 7)$ .

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Rewrite the expression.

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

Cancel the common factor of $x + 7$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Step 4

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

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

Step 5

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

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

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

Simplify the denominator.

Add $- 7$ and $7$ .

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

Add $\frac{1}{4 x}$ and $0$ .

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

Simplify terms.

Combine $4$ and $\frac{1}{4 x}$ .

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

Reduce the expression $\frac{4}{4 x}$ by cancelling the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Simplify each term.

Combine $4$ and $\frac{1}{4 (x + 7)}$ .

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

Reduce the expression $\frac{4}{4 (x + 7)}$ by cancelling the common factors.

Cancel the common factor.

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

Rewrite the expression.

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

Multiply the numerator by the reciprocal of the denominator.

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

Multiply $x + 7$ by $1$ .

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

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

Subtract $7$ from $7$ .

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

Add $x$ and $0$ .

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

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

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

Find the Inverse f(x)=1/(4x)-7

Related Questions

Support LCMGCF.com by disabling your adblocker.