Find the Inverse f(x)=(x-1)^2

$f (x) = {(x - 1)}^{2}$

Step 1

Write $f (x) = {(x - 1)}^{2}$ as an equation.

$y = {(x - 1)}^{2}$

Step 2

Interchange the variables.

$x = {(y - 1)}^{2}$

Step 3

Solve for $y$ .

Rewrite the equation as ${(y - 1)}^{2} = x$ .

${(y - 1)}^{2} = x$

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

$y - 1 = \pm \sqrt{x}$

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.

$y - 1 = \sqrt{x}$

Add $1$ to both sides of the equation.

$y = \sqrt{x} + 1$

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

$y - 1 = - \sqrt{x}$

Add $1$ to both sides of the equation.

$y = - \sqrt{x} + 1$

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

$y = \sqrt{x} + 1$

$y = - \sqrt{x} + 1$

$y = \sqrt{x} + 1$

$y = - \sqrt{x} + 1$

$y = \sqrt{x} + 1$

$y = - \sqrt{x} + 1$

Step 4

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

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

Step 5

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

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

Find the range of $f (x) = {(x - 1)}^{2}$ .

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 $\sqrt{x} + 1$ .

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 $f (x) = {(x - 1)}^{2}$ .

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

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

Step 6

Find the Inverse f(x)=(x-1)^2

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