Find the Inverse y=1/3(x+2)^2-3

$y = \frac{1}{3} {(x + 2)}^{2} - 3$

Step 1

Interchange the variables.

$x = \frac{1}{3} \cdot {(y + 2)}^{2} - 3$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{1}{3} \cdot {(y + 2)}^{2} - 3 = x$ .

$\frac{1}{3} \cdot {(y + 2)}^{2} - 3 = x$

Add $3$ to both sides of the equation.

$\frac{1}{3} \cdot {(y + 2)}^{2} = x + 3$

Combine $\frac{1}{3}$ and ${(y + 2)}^{2}$ .

$\frac{{(y + 2)}^{2}}{3} = x + 3$

Multiply both sides of the equation by $3$ .

$3 \frac{{(y + 2)}^{2}}{3} = 3 (x + 3)$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$3 \frac{{(y + 2)}^{2}}{3} = 3 (x + 3)$

Rewrite the expression.

${(y + 2)}^{2} = 3 (x + 3)$

Simplify the right side.

Simplify $3 (x + 3)$ .

Apply the distributive property.

${(y + 2)}^{2} = 3 x + 3 \cdot 3$

Multiply $3$ by $3$ .

${(y + 2)}^{2} = 3 x + 9$

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

$y + 2 = \pm \sqrt{3 x + 9}$

Factor $3$ out of $3 x + 9$ .

Factor $3$ out of $3 x$ .

$y + 2 = \pm \sqrt{3 (x) + 9}$

Factor $3$ out of $9$ .

$y + 2 = \pm \sqrt{3 x + 3 \cdot 3}$

Factor $3$ out of $3 x + 3 \cdot 3$ .

$y + 2 = \pm \sqrt{3 (x + 3)}$

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.

$y + 2 = \sqrt{3 (x + 3)}$

Subtract $2$ from both sides of the equation.

$y = \sqrt{3 (x + 3)} - 2$

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

$y + 2 = - \sqrt{3 (x + 3)}$

Subtract $2$ from both sides of the equation.

$y = - \sqrt{3 (x + 3)} - 2$

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

$y = \sqrt{3 (x + 3)} - 2$

$y = - \sqrt{3 (x + 3)} - 2$

$y = \sqrt{3 (x + 3)} - 2$

$y = - \sqrt{3 (x + 3)} - 2$

$y = \sqrt{3 (x + 3)} - 2$

$y = - \sqrt{3 (x + 3)} - 2$

Step 3

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

$f^{- 1} (x) = \sqrt{3 (x + 3)} - 2, - \sqrt{3 (x + 3)} - 2$

Step 4

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

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

Find the range of $f (x) = \frac{1}{3} \cdot {(x + 2)}^{2} - 3$ .

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

Interval Notation:

$[- 3, \infty)$

Find the domain of $\sqrt{3 (x + 3)} - 2$ .

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

$3 (x + 3) \geq 0$

Solve for $x$ .

Divide each term in $3 (x + 3) \geq 0$ by $3$ and simplify.

Divide each term in $3 (x + 3) \geq 0$ by $3$ .

$\frac{3 (x + 3)}{3} \geq \frac{0}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 (x + 3)}{3} \geq \frac{0}{3}$

Divide $x + 3$ by $1$ .

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

Simplify the right side.

Divide $0$ by $3$ .

$x + 3 \geq 0$

Subtract $3$ from both sides of the inequality.

$x \geq - 3$

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

$[- 3, \infty)$

Find the domain of $f (x) = \frac{1}{3} \cdot {(x + 2)}^{2} - 3$ .

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

$f^{- 1} (x) = \sqrt{3 (x + 3)} - 2, - \sqrt{3 (x + 3)} - 2$

Step 5

Find the Inverse y=1/3(x+2)^2-3

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