Find the Inverse f(x)=1/2(x+3)^2-4

$f (x) = \frac{1}{2} {(x + 3)}^{2} - 4$

Step 1

Write $f (x) = \frac{1}{2} {(x + 3)}^{2} - 4$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{1}{2} \cdot {(y + 3)}^{2} - 4 = x$ .

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

Add $4$ to both sides of the equation.

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

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

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

Multiply both sides of the equation by $2$ .

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

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

${(y + 3)}^{2} = 2 (x + 4)$

Simplify the right side.

Simplify $2 (x + 4)$ .

Apply the distributive property.

${(y + 3)}^{2} = 2 x + 2 \cdot 4$

Multiply $2$ by $4$ .

${(y + 3)}^{2} = 2 x + 8$

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

$y + 3 = \pm \sqrt{2 x + 8}$

Factor $2$ out of $2 x + 8$ .

Factor $2$ out of $2 x$ .

$y + 3 = \pm \sqrt{2 (x) + 8}$

Factor $2$ out of $8$ .

$y + 3 = \pm \sqrt{2 x + 2 \cdot 4}$

Factor $2$ out of $2 x + 2 \cdot 4$ .

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

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.

$y + 3 = \sqrt{2 (x + 4)}$

Subtract $3$ from both sides of the equation.

$y = \sqrt{2 (x + 4)} - 3$

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

$y + 3 = - \sqrt{2 (x + 4)}$

Subtract $3$ from both sides of the equation.

$y = - \sqrt{2 (x + 4)} - 3$

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

$y = \sqrt{2 (x + 4)} - 3$

$y = - \sqrt{2 (x + 4)} - 3$

$y = \sqrt{2 (x + 4)} - 3$

$y = - \sqrt{2 (x + 4)} - 3$

$y = \sqrt{2 (x + 4)} - 3$

$y = - \sqrt{2 (x + 4)} - 3$

Step 4

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

$f^{- 1} (x) = \sqrt{2 (x + 4)} - 3, - \sqrt{2 (x + 4)} - 3$

Step 5

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

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

Find the range of $f (x) = \frac{1}{2} \cdot {(x + 3)}^{2} - 4$ .

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 $\sqrt{2 (x + 4)} - 3$ .

Set the radicand in $\sqrt{2 (x + 4)}$ greater than or equal to $0$ to find where the expression is defined.

$2 (x + 4) \geq 0$

Solve for $x$ .

Divide each term in $2 (x + 4) \geq 0$ by $2$ and simplify.

Divide each term in $2 (x + 4) \geq 0$ by $2$ .

$\frac{2 (x + 4)}{2} \geq \frac{0}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 (x + 4)}{2} \geq \frac{0}{2}$

Divide $x + 4$ by $1$ .

$x + 4 \geq \frac{0}{2}$

Simplify the right side.

Divide $0$ by $2$ .

$x + 4 \geq 0$

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 $f (x) = \frac{1}{2} \cdot {(x + 3)}^{2} - 4$ .

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

$f^{- 1} (x) = \sqrt{2 (x + 4)} - 3, - \sqrt{2 (x + 4)} - 3$

Step 6

Find the Inverse f(x)=1/2(x+3)^2-4

Related Questions

Support LCMGCF.com by disabling your adblocker.