Find the Inverse f(x)=- square root of x+2 ; for x>=-2

$f (x) = - \sqrt{x + 2}$ ; for $x \geq - 2$

Find the range of the given function.

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

$(- \infty, 0]$

Convert $(- \infty, 0]$ to an inequality.

$y \leq 0$

Find the inverse.

Interchange the variables.

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

Solve for $y$ .

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

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

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $\sqrt{y + 2}$ by $1$ .

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

Simplify the right side.

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

$\sqrt{y + 2} = - 1 \cdot x$

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

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(- x)}^{2}$ .

Apply the product rule to $- x$ .

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

Raise $- 1$ to the power of $2$ .

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

Multiply $x^{2}$ by $1$ .

$y + 2 = x^{2}$

Subtract $2$ from both sides of the equation.

$y = x^{2} - 2$

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

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

Find the inverse using the domain and the range of the original function.

$f^{- 1} (x) = x^{2} - 2, x \leq 0$

Find the Inverse f(x)=- square root of x+2 ; for x>=-2

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