Find the Inverse F(x)=2x^2+2

$F (x) = 2 x^{2} + 2$

Write $F (x) = 2 x^{2} + 2$ as an equation.

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

Interchange the variables.

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

Solve for $y$ .

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

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

Subtract $2$ from both sides of the equation.

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

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

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

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

Simplify the right side.

Divide $- 2$ by $2$ .

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

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

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

Simplify $\pm \sqrt{\frac{x}{2} - 1}$ .

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Combine $- 1$ and $\frac{2}{2}$ .

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

Combine the numerators over the common denominator.

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

Multiply $- 1$ by $2$ .

$y = \pm \sqrt{\frac{x - 2}{2}}$

Rewrite $\sqrt{\frac{x - 2}{2}}$ as $\frac{\sqrt{x - 2}}{\sqrt{2}}$ .

$y = \pm \frac{\sqrt{x - 2}}{\sqrt{2}}$

Multiply $\frac{\sqrt{x - 2}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$y = \pm \frac{\sqrt{x - 2}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{x - 2}}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

$y = \pm \frac{\sqrt{x - 2} \sqrt{2}}{\sqrt{2} \sqrt{2}}$

Raise $\sqrt{2}$ to the power of $1$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

$y = \pm \frac{\sqrt{x - 2} \sqrt{2}}{{\sqrt{2}}^{2}}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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

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

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

$y = \pm \frac{\sqrt{x - 2} \sqrt{2}}{2^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$y = \pm \frac{\sqrt{x - 2} \sqrt{2}}{2^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$y = \pm \frac{\sqrt{x - 2} \sqrt{2}}{2^{\frac{2}{2}}}$

Rewrite the expression.

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

Evaluate the exponent.

$y = \pm \frac{\sqrt{x - 2} \sqrt{2}}{2}$

Combine using the product rule for radicals.

$y = \pm \frac{\sqrt{(x - 2) \cdot 2}}{2}$

Reorder factors in $\pm \frac{\sqrt{(x - 2) \cdot 2}}{2}$ .

$y = \pm \frac{\sqrt{2 (x - 2)}}{2}$

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.

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

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

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

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

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

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

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

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

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

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

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

$F^{- 1} (x) = \frac{\sqrt{2 (x - 2)}}{2}, - \frac{\sqrt{2 (x - 2)}}{2}$

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

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

Find the range of $F (x) = 2 x^{2} + 2$ .

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

Interval Notation:

$[2, \infty)$

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

Set the radicand in $\sqrt{2 (x - 2)}$ greater than or equal to $0$ to find where the expression is defined.

$2 (x - 2) \geq 0$

Solve for $x$ .

Divide each term in $2 (x - 2) \geq 0$ by $2$ and simplify.

Divide each term in $2 (x - 2) \geq 0$ by $2$ .

$\frac{2 (x - 2)}{2} \geq \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (x - 2)}{2} \geq \frac{0}{2}$

Divide $x - 2$ by $1$ .

$x - 2 \geq \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$x - 2 \geq 0$

Add $2$ to both sides of the inequality.

$x \geq 2$

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

$[2, \infty)$

Find the domain of $F (x) = 2 x^{2} + 2$ .

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

$F^{- 1} (x) = \frac{\sqrt{2 (x - 2)}}{2}, - \frac{\sqrt{2 (x - 2)}}{2}$

Find the Inverse F(x)=2x^2+2

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