Find the Inverse f(x)=(4x-3)/7

$f (x) = \frac{4 x - 3}{7}$

Step 1

Write $f (x) = \frac{4 x - 3}{7}$ as an equation.

$y = \frac{4 x - 3}{7}$

Step 2

Interchange the variables.

$x = \frac{4 y - 3}{7}$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{4 y - 3}{7} = x$ .

$\frac{4 y - 3}{7} = x$

Multiply both sides by $7$ .

$\frac{4 y - 3}{7} \cdot 7 = x \cdot 7$

Simplify.

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{4 y - 3}{7} \cdot 7 = x \cdot 7$

Rewrite the expression.

$4 y - 3 = x \cdot 7$

Simplify the right side.

Move $7$ to the left of $x$ .

$4 y - 3 = 7 x$

Solve for $y$ .

Add $3$ to both sides of the equation.

$4 y = 7 x + 3$

Divide each term in $4 y = 7 x + 3$ by $4$ and simplify.

Divide each term in $4 y = 7 x + 3$ by $4$ .

$\frac{4 y}{4} = \frac{7 x}{4} + \frac{3}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y}{4} = \frac{7 x}{4} + \frac{3}{4}$

Divide $y$ by $1$ .

$y = \frac{7 x}{4} + \frac{3}{4}$

Step 4

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

$f^{- 1} (x) = \frac{7 x}{4} + \frac{3}{4}$

Step 5

Verify if $f^{- 1} (x) = \frac{7 x}{4} + \frac{3}{4}$ is the inverse of $f (x) = \frac{4 x - 3}{7}$ .

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 $f^{- 1} (\frac{4 x - 3}{7})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\frac{4 x - 3}{7}) = \frac{7 (\frac{4 x - 3}{7})}{4} + \frac{3}{4}$

Combine the numerators over the common denominator.

$f^{- 1} (\frac{4 x - 3}{7}) = \frac{7 (\frac{4 x - 3}{7}) + 3}{4}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f^{- 1} (\frac{4 x - 3}{7}) = \frac{7 (\frac{4 x - 3}{7}) + 3}{4}$

Rewrite the expression.

$f^{- 1} (\frac{4 x - 3}{7}) = \frac{4 x - 3 + 3}{4}$

Combine the opposite terms in $4 x - 3 + 3$ .

Add $- 3$ and $3$ .

$f^{- 1} (\frac{4 x - 3}{7}) = \frac{4 x + 0}{4}$

Add $4 x$ and $0$ .

$f^{- 1} (\frac{4 x - 3}{7}) = \frac{4 x}{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

$f^{- 1} (\frac{4 x - 3}{7}) = \frac{4 x}{4}$

Divide $x$ by $1$ .

$f^{- 1} (\frac{4 x - 3}{7}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{7 x}{4} + \frac{3}{4})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{7 x}{4} + \frac{3}{4}) = \frac{4 (\frac{7 x}{4} + \frac{3}{4}) - 3}{7}$

Simplify the numerator.

Apply the distributive property.

$f (\frac{7 x}{4} + \frac{3}{4}) = \frac{4 (\frac{7 x}{4}) + 4 (\frac{3}{4}) - 3}{7}$

Cancel the common factor of $4$ .

Cancel the common factor.

$f (\frac{7 x}{4} + \frac{3}{4}) = \frac{4 (\frac{7 x}{4}) + 4 (\frac{3}{4}) - 3}{7}$

Rewrite the expression.

$f (\frac{7 x}{4} + \frac{3}{4}) = \frac{7 x + 4 (\frac{3}{4}) - 3}{7}$

Cancel the common factor of $4$ .

Cancel the common factor.

$f (\frac{7 x}{4} + \frac{3}{4}) = \frac{7 x + 4 (\frac{3}{4}) - 3}{7}$

Rewrite the expression.

$f (\frac{7 x}{4} + \frac{3}{4}) = \frac{7 x + 3 - 3}{7}$

Subtract $3$ from $3$ .

$f (\frac{7 x}{4} + \frac{3}{4}) = \frac{7 x + 0}{7}$

Add $7 x$ and $0$ .

$f (\frac{7 x}{4} + \frac{3}{4}) = \frac{7 x}{7}$

Cancel the common factor of $7$ .

Cancel the common factor.

$f (\frac{7 x}{4} + \frac{3}{4}) = \frac{7 x}{7}$

Divide $x$ by $1$ .

$f (\frac{7 x}{4} + \frac{3}{4}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{7 x}{4} + \frac{3}{4}$ is the inverse of $f (x) = \frac{4 x - 3}{7}$ .

$f^{- 1} (x) = \frac{7 x}{4} + \frac{3}{4}$

Find the Inverse f(x)=(4x-3)/7

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