Find the Inverse y = square root of x^2-4

$y = \sqrt{x^{2} - 4}$

Interchange the variables.

$x = \sqrt{y^{2} - 4}$

Solve for $y$ .

Rewrite the equation as $\sqrt{y^{2} - 4} = x$ .

$\sqrt{y^{2} - 4} = x$

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

${\sqrt{y^{2} - 4}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y^{2} - 4}$ as ${(y^{2} - 4)}^{\frac{1}{2}}$ .

${({(y^{2} - 4)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(y^{2} - 4)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(y^{2} - 4)}^{\frac{1}{2}})}^{2}$ .

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

${(y^{2} - 4)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(y^{2} - 4)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(y^{2} - 4)}^{1} = x^{2}$

Simplify.

$y^{2} - 4 = x^{2}$

Solve for $y$ .

Add $4$ to both sides of the equation.

$y^{2} = x^{2} + 4$

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

$y = \pm \sqrt{x^{2} + 4}$

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.

$y = \sqrt{x^{2} + 4}$

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

$y = - \sqrt{x^{2} + 4}$

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

$y = \sqrt{x^{2} + 4}$

$y = - \sqrt{x^{2} + 4}$

$y = \sqrt{x^{2} + 4}$

$y = - \sqrt{x^{2} + 4}$

$y = \sqrt{x^{2} + 4}$

$y = - \sqrt{x^{2} + 4}$

$y = \sqrt{x^{2} + 4}$

$y = - \sqrt{x^{2} + 4}$

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

$f^{- 1} (x) = \sqrt{x^{2} + 4}, - \sqrt{x^{2} + 4}$

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

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

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

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 the inverse.

Find the domain of $\sqrt{x^{2} + 4}$ .

Set the radicand in $\sqrt{x^{2} + 4}$ greater than or equal to $0$ to find where the expression is defined.

$x^{2} + 4 \geq 0$

Solve for $x$ .

Subtract $4$ from both sides of the inequality.

$x^{2} \geq - 4$

Since the left side has an even power, it is always positive for all real numbers.

All real numbers

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.

$(- \infty, \infty)$

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

$(- \infty, \infty)$

Since the domain of $f^{- 1} (x) = \sqrt{x^{2} + 4}, - \sqrt{x^{2} + 4}$ is not equal to the range of $f (x) = \sqrt{x^{2} - 4}$ , then $f^{- 1} (x) = \sqrt{x^{2} + 4}, - \sqrt{x^{2} + 4}$ is not an inverse of $f (x) = \sqrt{x^{2} - 4}$ .

There is no inverse

Find the Inverse y = square root of x^2-4

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