Find the Inverse Find the inverse of f(x)=x^2-5

Find the inverse of $f (x) = x^{2} - 5$

Interchange the variables.

$x = y^{2} - 5$

Solve for $y$ .

Rewrite the equation as $y^{2} - 5 = x$ .

$y^{2} - 5 = x$

Add $5$ to both sides of the equation.

$y^{2} = x + 5$

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

$y = \pm \sqrt{x + 5}$

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.

$y = \sqrt{x + 5}$

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

$y = - \sqrt{x + 5}$

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

$y = \sqrt{x + 5}$

$y = - \sqrt{x + 5}$

$y = \sqrt{x + 5}$

$y = - \sqrt{x + 5}$

$y = \sqrt{x + 5}$

$y = - \sqrt{x + 5}$

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

$f^{- 1} (x) = \sqrt{x + 5}, - \sqrt{x + 5}$

Verify if $f^{- 1} (x) = \sqrt{x + 5}, - \sqrt{x + 5}$ is the inverse of $f (x) = x^{2} - 5$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = x^{2} - 5$ and $f^{- 1} (x) = \sqrt{x + 5}, - \sqrt{x + 5}$ and compare them.

Find the range of $f (x) = x^{2} - 5$ .

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

Interval Notation:

$[- 5, \infty)$

Find the domain of $\sqrt{x + 5}$ .

Set the radicand in $\sqrt{x + 5}$ greater than or equal to $0$ to find where the expression is defined.

$x + 5 \geq 0$

Subtract $5$ from both sides of the inequality.

$x \geq - 5$

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

$[- 5, \infty)$

Find the domain of $f (x) = x^{2} - 5$ .

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 $f^{- 1} (x) = \sqrt{x + 5}, - \sqrt{x + 5}$ is the range of $f (x) = x^{2} - 5$ and the range of $f^{- 1} (x) = \sqrt{x + 5}, - \sqrt{x + 5}$ is the domain of $f (x) = x^{2} - 5$ , then $f^{- 1} (x) = \sqrt{x + 5}, - \sqrt{x + 5}$ is the inverse of $f (x) = x^{2} - 5$ .

$f^{- 1} (x) = \sqrt{x + 5}, - \sqrt{x + 5}$

Find the Inverse Find the inverse of f(x)=x^2-5

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