Find the Inverse -3x^2-12

$- 3 x^{2} - 12$

Step 1

Interchange the variables.

$x = - 3 y^{2} - 12$

Step 2

Solve for $y$ .

Rewrite the equation as $- 3 y^{2} - 12 = x$ .

$- 3 y^{2} - 12 = x$

Add $12$ to both sides of the equation.

$- 3 y^{2} = x + 12$

Divide each term in $- 3 y^{2} = x + 12$ by $- 3$ and simplify.

Divide each term in $- 3 y^{2} = x + 12$ by $- 3$ .

$\frac{- 3 y^{2}}{- 3} = \frac{x}{- 3} + \frac{12}{- 3}$

Simplify the left side.

Cancel the common factor of $- 3$ .

Cancel the common factor.

$\frac{- 3 y^{2}}{- 3} = \frac{x}{- 3} + \frac{12}{- 3}$

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

$y^{2} = \frac{x}{- 3} + \frac{12}{- 3}$

Simplify the right side.

Simplify each term.

Move the negative in front of the fraction.

$y^{2} = - \frac{x}{3} + \frac{12}{- 3}$

Divide $12$ by $- 3$ .

$y^{2} = - \frac{x}{3} - 4$

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

$y = \pm \sqrt{- \frac{x}{3} - 4}$

Simplify $\pm \sqrt{- \frac{x}{3} - 4}$ .

Factor $- 1$ out of $- \frac{x}{3} - 4$ .

Factor $- 1$ out of $- \frac{x}{3}$ .

$y = \pm \sqrt{- (\frac{x}{3}) - 4}$

Rewrite $- 4$ as $- 1 (4)$ .

$y = \pm \sqrt{- (\frac{x}{3}) - 1 (4)}$

Factor $- 1$ out of $- (\frac{x}{3}) - 1 (4)$ .

$y = \pm \sqrt{- (\frac{x}{3} + 4)}$

To write $4$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$y = \pm \sqrt{- (\frac{x}{3} + 4 \cdot \frac{3}{3})}$

Combine $4$ and $\frac{3}{3}$ .

$y = \pm \sqrt{- (\frac{x}{3} + \frac{4 \cdot 3}{3})}$

Combine the numerators over the common denominator.

$y = \pm \sqrt{- \frac{x + 4 \cdot 3}{3}}$

Multiply $4$ by $3$ .

$y = \pm \sqrt{- \frac{x + 12}{3}}$

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.

$y = \sqrt{- \frac{x + 12}{3}}$

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

$y = - \sqrt{- \frac{x + 12}{3}}$

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

$y = \sqrt{- \frac{x + 12}{3}}$

$y = - \sqrt{- \frac{x + 12}{3}}$

$y = \sqrt{- \frac{x + 12}{3}}$

$y = - \sqrt{- \frac{x + 12}{3}}$

$y = \sqrt{- \frac{x + 12}{3}}$

$y = - \sqrt{- \frac{x + 12}{3}}$

Step 3

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

$f^{- 1} (x) = \sqrt{- \frac{x + 12}{3}}, - \sqrt{- \frac{x + 12}{3}}$

Step 4

Verify if $f^{- 1} (x) = \sqrt{- \frac{x + 12}{3}}, - \sqrt{- \frac{x + 12}{3}}$ is the inverse of $f (x) = - 3 x^{2} - 12$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = - 3 x^{2} - 12$ and $f^{- 1} (x) = \sqrt{- \frac{x + 12}{3}}, - \sqrt{- \frac{x + 12}{3}}$ and compare them.

Find the range of $f (x) = - 3 x^{2} - 12$ .

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

Interval Notation:

$(- \infty, - 12]$

Find the domain of $\sqrt{- \frac{x + 12}{3}}$ .

Set the radicand in $\sqrt{- \frac{x + 12}{3}}$ greater than or equal to $0$ to find where the expression is defined.

$- \frac{x + 12}{3} \geq 0$

Solve for $x$ .

Divide each term in $- \frac{x + 12}{3} \geq 0$ by $- 1$ and simplify.

Divide each term in $- \frac{x + 12}{3} \geq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- \frac{x + 12}{3}}{- 1} \leq \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\frac{x + 12}{3}}{1} \leq \frac{0}{- 1}$

Divide $\frac{x + 12}{3}$ by $1$ .

$\frac{x + 12}{3} \leq \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

$\frac{x + 12}{3} \leq 0$

Multiply both sides by $3$ .

$\frac{x + 12}{3} \cdot 3 \leq 0 \cdot 3$

Simplify.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{x + 12}{3} \cdot 3 \leq 0 \cdot 3$

Rewrite the expression.

$x + 12 \leq 0 \cdot 3$

Simplify the right side.

Multiply $0$ by $3$ .

$x + 12 \leq 0$

Subtract $12$ from both sides of the inequality.

$x \leq - 12$

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

$(- \infty, - 12]$

Find the domain of $f (x) = - 3 x^{2} - 12$ .

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 $f^{- 1} (x) = \sqrt{- \frac{x + 12}{3}}, - \sqrt{- \frac{x + 12}{3}}$ is the range of $f (x) = - 3 x^{2} - 12$ and the range of $f^{- 1} (x) = \sqrt{- \frac{x + 12}{3}}, - \sqrt{- \frac{x + 12}{3}}$ is the domain of $f (x) = - 3 x^{2} - 12$ , then $f^{- 1} (x) = \sqrt{- \frac{x + 12}{3}}, - \sqrt{- \frac{x + 12}{3}}$ is the inverse of $f (x) = - 3 x^{2} - 12$ .

$f^{- 1} (x) = \sqrt{- \frac{x + 12}{3}}, - \sqrt{- \frac{x + 12}{3}}$

Step 5

Find the Inverse -3x^2-12

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