Find the Inverse 1/2x^2-5

$\frac{1}{2} x^{2} - 5$

Step 1

Interchange the variables.

$x = \frac{1}{2} y^{2} - 5$

Step 2

Solve for $y$ .

Rewrite the equation as $\frac{1}{2} y^{2} - 5 = x$ .

$\frac{1}{2} y^{2} - 5 = x$

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

$\frac{y^{2}}{2} - 5 = x$

Add $5$ to both sides of the equation.

$\frac{y^{2}}{2} = x + 5$

Multiply both sides of the equation by $2$ .

$2 \frac{y^{2}}{2} = 2 (x + 5)$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$2 \frac{y^{2}}{2} = 2 (x + 5)$

Rewrite the expression.

$y^{2} = 2 (x + 5)$

Simplify the right side.

Simplify $2 (x + 5)$ .

Apply the distributive property.

$y^{2} = 2 x + 2 \cdot 5$

Multiply $2$ by $5$ .

$y^{2} = 2 x + 10$

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

$y = \pm \sqrt{2 x + 10}$

Factor $2$ out of $2 x + 10$ .

Factor $2$ out of $2 x$ .

$y = \pm \sqrt{2 (x) + 10}$

Factor $2$ out of $10$ .

$y = \pm \sqrt{2 x + 2 \cdot 5}$

Factor $2$ out of $2 x + 2 \cdot 5$ .

$y = \pm \sqrt{2 (x + 5)}$

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.

$y = \sqrt{2 (x + 5)}$

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

$y = - \sqrt{2 (x + 5)}$

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

$y = \sqrt{2 (x + 5)}$

$y = - \sqrt{2 (x + 5)}$

$y = \sqrt{2 (x + 5)}$

$y = - \sqrt{2 (x + 5)}$

$y = \sqrt{2 (x + 5)}$

$y = - \sqrt{2 (x + 5)}$

Step 3

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

$f^{- 1} (x) = \sqrt{2 (x + 5)}, - \sqrt{2 (x + 5)}$

Step 4

Verify if $f^{- 1} (x) = \sqrt{2 (x + 5)}, - \sqrt{2 (x + 5)}$ is the inverse of $f (x) = \frac{1}{2} x^{2} - 5$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = \frac{1}{2} x^{2} - 5$ and $f^{- 1} (x) = \sqrt{2 (x + 5)}, - \sqrt{2 (x + 5)}$ and compare them.

Find the range of $f (x) = \frac{1}{2} x^{2} - 5$ .

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

Interval Notation:

$[- 5, \infty)$

Find the domain of $\sqrt{2 (x + 5)}$ .

Set the radicand in $\sqrt{2 (x + 5)}$ greater than or equal to $0$ to find where the expression is defined.

$2 (x + 5) \geq 0$

Solve for $x$ .

Divide each term in $2 (x + 5) \geq 0$ by $2$ and simplify.

Divide each term in $2 (x + 5) \geq 0$ by $2$ .

$\frac{2 (x + 5)}{2} \geq \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (x + 5)}{2} \geq \frac{0}{2}$

Divide $x + 5$ by $1$ .

$x + 5 \geq \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$x + 5 \geq 0$

Subtract $5$ from both sides of the inequality.

$x \geq - 5$

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

$[- 5, \infty)$

Find the domain of $f (x) = \frac{1}{2} x^{2} - 5$ .

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 $f^{- 1} (x) = \sqrt{2 (x + 5)}, - \sqrt{2 (x + 5)}$ is the range of $f (x) = \frac{1}{2} x^{2} - 5$ and the range of $f^{- 1} (x) = \sqrt{2 (x + 5)}, - \sqrt{2 (x + 5)}$ is the domain of $f (x) = \frac{1}{2} x^{2} - 5$ , then $f^{- 1} (x) = \sqrt{2 (x + 5)}, - \sqrt{2 (x + 5)}$ is the inverse of $f (x) = \frac{1}{2} x^{2} - 5$ .

$f^{- 1} (x) = \sqrt{2 (x + 5)}, - \sqrt{2 (x + 5)}$

Step 5

Find the Inverse 1/2x^2-5

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