Find the Inverse y=(3x-4)^2

$y = {(3 x - 4)}^{2}$

Step 1

Interchange the variables.

$x = {(3 y - 4)}^{2}$

Step 2

Solve for $y$ .

Rewrite the equation as ${(3 y - 4)}^{2} = x$ .

${(3 y - 4)}^{2} = x$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$3 y - 4 = \pm \sqrt{x}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$3 y - 4 = \sqrt{x}$

Add $4$ to both sides of the equation.

$3 y = \sqrt{x} + 4$

Divide each term in $3 y = \sqrt{x} + 4$ by $3$ and simplify.

Divide each term in $3 y = \sqrt{x} + 4$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Next, use the negative value of the $\pm$ to find the second solution.

$3 y - 4 = - \sqrt{x}$

Add $4$ to both sides of the equation.

$3 y = - \sqrt{x} + 4$

Divide each term in $3 y = - \sqrt{x} + 4$ by $3$ and simplify.

Divide each term in $3 y = - \sqrt{x} + 4$ by $3$ .

$\frac{3 y}{3} = \frac{- \sqrt{x}}{3} + \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 y}{3} = \frac{- \sqrt{x}}{3} + \frac{4}{3}$

Divide $y$ by $1$ .

$y = \frac{- \sqrt{x}}{3} + \frac{4}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$y = - \frac{\sqrt{x}}{3} + \frac{4}{3}$

The complete solution is the result of both the positive and negative portions of the solution.

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

$y = - \frac{\sqrt{x}}{3} + \frac{4}{3}$

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

$y = - \frac{\sqrt{x}}{3} + \frac{4}{3}$

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

$y = - \frac{\sqrt{x}}{3} + \frac{4}{3}$

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt{x}}{3} + \frac{4}{3}, - \frac{\sqrt{x}}{3} + \frac{4}{3}$ is the inverse of $f (x) = {(3 x - 4)}^{2}$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = {(3 x - 4)}^{2}$ and $f^{- 1} (x) = \frac{\sqrt{x}}{3} + \frac{4}{3}, - \frac{\sqrt{x}}{3} + \frac{4}{3}$ and compare them.

Find the range of $f (x) = {(3 x - 4)}^{2}$ .

The range is the set of all valid $y$ values. Use the graph to find the range.

Interval Notation:

$[0, \infty)$

Find the domain of $\frac{\sqrt{x}}{3} + \frac{4}{3}$ .

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

The domain is all values of $x$ that make the expression defined.

$[0, \infty)$

Find the domain of $f (x) = {(3 x - 4)}^{2}$ .

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

$(- \infty, \infty)$

Since the domain of $f^{- 1} (x) = \frac{\sqrt{x}}{3} + \frac{4}{3}, - \frac{\sqrt{x}}{3} + \frac{4}{3}$ is the range of $f (x) = {(3 x - 4)}^{2}$ and the range of $f^{- 1} (x) = \frac{\sqrt{x}}{3} + \frac{4}{3}, - \frac{\sqrt{x}}{3} + \frac{4}{3}$ is the domain of $f (x) = {(3 x - 4)}^{2}$ , then $f^{- 1} (x) = \frac{\sqrt{x}}{3} + \frac{4}{3}, - \frac{\sqrt{x}}{3} + \frac{4}{3}$ is the inverse of $f (x) = {(3 x - 4)}^{2}$ .

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

Step 5

Find the Inverse y=(3x-4)^2

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