Find the Inverse y=9x^2-4

$y = 9 x^{2} - 4$

Step 1

Interchange the variables.

$x = 9 y^{2} - 4$

Step 2

Solve for $y$ .

Rewrite the equation as $9 y^{2} - 4 = x$ .

$9 y^{2} - 4 = x$

Add $4$ to both sides of the equation.

$9 y^{2} = x + 4$

Divide each term in $9 y^{2} = x + 4$ by $9$ and simplify.

Divide each term in $9 y^{2} = x + 4$ by $9$ .

$\frac{9 y^{2}}{9} = \frac{x}{9} + \frac{4}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 y^{2}}{9} = \frac{x}{9} + \frac{4}{9}$

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

$y^{2} = \frac{x}{9} + \frac{4}{9}$

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

$y = \pm \sqrt{\frac{x}{9} + \frac{4}{9}}$

Simplify $\pm \sqrt{\frac{x}{9} + \frac{4}{9}}$ .

Combine the numerators over the common denominator.

$y = \pm \sqrt{\frac{x + 4}{9}}$

Rewrite $\frac{x + 4}{9}$ as ${(\frac{1}{3})}^{2} (x + 4)$ .

Factor the perfect power $1^{2}$ out of $x + 4$ .

$y = \pm \sqrt{\frac{1^{2} (x + 4)}{9}}$

Factor the perfect power $3^{2}$ out of $9$ .

$y = \pm \sqrt{\frac{1^{2} (x + 4)}{3^{2} \cdot 1}}$

Rearrange the fraction $\frac{1^{2} (x + 4)}{3^{2} \cdot 1}$ .

$y = \pm \sqrt{{(\frac{1}{3})}^{2} (x + 4)}$

Pull terms out from under the radical.

$y = \pm \frac{1}{3} \sqrt{x + 4}$

Combine $\frac{1}{3}$ and $\sqrt{x + 4}$ .

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

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.

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

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

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

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

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

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

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

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

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

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

Step 3

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

$f^{- 1} (x) = \frac{\sqrt{x + 4}}{3}, - \frac{\sqrt{x + 4}}{3}$

Step 4

Verify if $f^{- 1} (x) = \frac{\sqrt{x + 4}}{3}, - \frac{\sqrt{x + 4}}{3}$ is the inverse of $f (x) = 9 x^{2} - 4$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = 9 x^{2} - 4$ and $f^{- 1} (x) = \frac{\sqrt{x + 4}}{3}, - \frac{\sqrt{x + 4}}{3}$ and compare them.

Find the range of $f (x) = 9 x^{2} - 4$ .

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

Interval Notation:

$[- 4, \infty)$

Find the domain of $\frac{\sqrt{x + 4}}{3}$ .

Set the radicand in $\sqrt{x + 4}$ greater than or equal to $0$ to find where the expression is defined.

$x + 4 \geq 0$

Subtract $4$ from both sides of the inequality.

$x \geq - 4$

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

$[- 4, \infty)$

Find the domain of $f (x) = 9 x^{2} - 4$ .

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 $f^{- 1} (x) = \frac{\sqrt{x + 4}}{3}, - \frac{\sqrt{x + 4}}{3}$ is the range of $f (x) = 9 x^{2} - 4$ and the range of $f^{- 1} (x) = \frac{\sqrt{x + 4}}{3}, - \frac{\sqrt{x + 4}}{3}$ is the domain of $f (x) = 9 x^{2} - 4$ , then $f^{- 1} (x) = \frac{\sqrt{x + 4}}{3}, - \frac{\sqrt{x + 4}}{3}$ is the inverse of $f (x) = 9 x^{2} - 4$ .

$f^{- 1} (x) = \frac{\sqrt{x + 4}}{3}, - \frac{\sqrt{x + 4}}{3}$

Step 5

Find the Inverse y=9x^2-4

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