Find the Inverse -x^2-7

$- x^{2} - 7$

Step 1

Interchange the variables.

$x = - y^{2} - 7$

Step 2

Solve for $y$ .

Rewrite the equation as $- y^{2} - 7 = x$ .

$- y^{2} - 7 = x$

Add $7$ to both sides of the equation.

$- y^{2} = x + 7$

Divide each term in $- y^{2} = x + 7$ by $- 1$ and simplify.

Divide each term in $- y^{2} = x + 7$ by $- 1$ .

$\frac{- y^{2}}{- 1} = \frac{x}{- 1} + \frac{7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{y^{2}}{1} = \frac{x}{- 1} + \frac{7}{- 1}$

Divide $y^{2}$ by $1$ .

$y^{2} = \frac{x}{- 1} + \frac{7}{- 1}$

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{x}{- 1}$ .

$y^{2} = - 1 \cdot x + \frac{7}{- 1}$

Rewrite $- 1 \cdot x$ as $- x$ .

$y^{2} = - x + \frac{7}{- 1}$

Divide $7$ by $- 1$ .

$y^{2} = - x - 7$

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

$y = \pm \sqrt{- x - 7}$

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.

$y = \sqrt{- x - 7}$

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

$y = - \sqrt{- x - 7}$

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

$y = \sqrt{- x - 7}$

$y = - \sqrt{- x - 7}$

$y = \sqrt{- x - 7}$

$y = - \sqrt{- x - 7}$

$y = \sqrt{- x - 7}$

$y = - \sqrt{- x - 7}$

Step 3

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

$f^{- 1} (x) = \sqrt{- x - 7}, - \sqrt{- x - 7}$

Step 4

Verify if $f^{- 1} (x) = \sqrt{- x - 7}, - \sqrt{- x - 7}$ is the inverse of $f (x) = - x^{2} - 7$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = - x^{2} - 7$ and $f^{- 1} (x) = \sqrt{- x - 7}, - \sqrt{- x - 7}$ and compare them.

Find the range of $f (x) = - x^{2} - 7$ .

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

Interval Notation:

$(- \infty, - 7]$

Find the domain of $\sqrt{- x - 7}$ .

Set the radicand in $\sqrt{- x - 7}$ greater than or equal to $0$ to find where the expression is defined.

$- x - 7 \geq 0$

Solve for $x$ .

Add $7$ to both sides of the inequality.

$- x \geq 7$

Divide each term in $- x \geq 7$ by $- 1$ and simplify.

Divide each term in $- x \geq 7$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- x}{- 1} \leq \frac{7}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} \leq \frac{7}{- 1}$

Divide $x$ by $1$ .

$x \leq \frac{7}{- 1}$

Simplify the right side.

Divide $7$ by $- 1$ .

$x \leq - 7$

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

$(- \infty, - 7]$

Find the domain of $f (x) = - x^{2} - 7$ .

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 $f^{- 1} (x) = \sqrt{- x - 7}, - \sqrt{- x - 7}$ is the range of $f (x) = - x^{2} - 7$ and the range of $f^{- 1} (x) = \sqrt{- x - 7}, - \sqrt{- x - 7}$ is the domain of $f (x) = - x^{2} - 7$ , then $f^{- 1} (x) = \sqrt{- x - 7}, - \sqrt{- x - 7}$ is the inverse of $f (x) = - x^{2} - 7$ .

$f^{- 1} (x) = \sqrt{- x - 7}, - \sqrt{- x - 7}$

Step 5

Find the Inverse -x^2-7

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