Find the Inverse x=y^2

$x = y^{2}$

Step 1

Rewrite the equation as $y^{2} = x$ .

$y^{2} = x$

Step 2

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

$y = \pm \sqrt{x}$

Step 3

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.

$y = \sqrt{x}$

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

$y = - \sqrt{x}$

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

$y = \sqrt{x}$

$y = - \sqrt{x}$

$y = \sqrt{x}$

$y = - \sqrt{x}$

Step 4

Interchange the variables. Create an equation for each expression.

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

Step 5

Solve for $y$ .

Rewrite the equation as $\sqrt{y} = x$ .

$\sqrt{y} = x$

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{y}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y}$ as $y^{\frac{1}{2}}$ .

${(y^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${(y^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${(y^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

$y^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

$y^{1} = x^{2}$

Simplify.

$y = x^{2}$

Step 6

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

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

Step 7

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

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

Find the range of $f (x) = \sqrt{x}, - \sqrt{x}$ .

Find the range of $\sqrt{x}$ .

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

Interval Notation:

$[0, \infty)$

Find the range of $- \sqrt{x}$ .

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

Interval Notation:

$(- \infty, 0]$

The union consists of all of the elements that are contained in each interval.

$(- \infty, \infty)$

Find the domain of $\sqrt{x}$ .

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.

$[0, \infty)$

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

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

Step 8

Find the Inverse x=y^2

Related Questions

Support LCMGCF.com by disabling your adblocker.