Find the Inverse f(b)=(b+6)(b-2)

$f (b) = (b + 6) (b - 2)$

Step 1

Write $f (b) = (b + 6) (b - 2)$ as an equation.

$y = (b + 6) (b - 2)$

Step 2

Interchange the variables.

$b = (y + 6) (y - 2)$

Step 3

Solve for $y$ .

Rewrite the equation as $(y + 6) (y - 2) = b$ .

$(y + 6) (y - 2) = b$

Simplify $(y + 6) (y - 2)$ .

Expand $(y + 6) (y - 2)$ using the FOIL Method.

Apply the distributive property.

$y (y - 2) + 6 (y - 2) = b$

Apply the distributive property.

$y \cdot y + y \cdot - 2 + 6 (y - 2) = b$

Apply the distributive property.

$y \cdot y + y \cdot - 2 + 6 y + 6 \cdot - 2 = b$

Simplify and combine like terms.

Simplify each term.

Multiply $y$ by $y$ .

$y^{2} + y \cdot - 2 + 6 y + 6 \cdot - 2 = b$

Move $- 2$ to the left of $y$ .

$y^{2} - 2 \cdot y + 6 y + 6 \cdot - 2 = b$

Multiply $6$ by $- 2$ .

$y^{2} - 2 y + 6 y - 12 = b$

Add $- 2 y$ and $6 y$ .

$y^{2} + 4 y - 12 = b$

Subtract $b$ from both sides of the equation.

$y^{2} + 4 y - 12 - b = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 4$ , and $c = - 12 - b$ into the quadratic formula and solve for $y$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot (- 12 - b))}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot (- 12 - b)}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot (- 12 - b)}}{2 \cdot 1}$

Apply the distributive property.

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 12 - 4 (- b)}}{2 \cdot 1}$

Multiply $- 4$ by $- 12$ .

$y = \frac{- 4 \pm \sqrt{16 + 48 - 4 (- b)}}{2 \cdot 1}$

Multiply $- 1$ by $- 4$ .

$y = \frac{- 4 \pm \sqrt{16 + 48 + 4 b}}{2 \cdot 1}$

Add $16$ and $48$ .

$y = \frac{- 4 \pm \sqrt{64 + 4 b}}{2 \cdot 1}$

Factor $4$ out of $64 + 4 b$ .

Factor $4$ out of $64$ .

$y = \frac{- 4 \pm \sqrt{4 \cdot 16 + 4 b}}{2 \cdot 1}$

Factor $4$ out of $4 \cdot 16 + 4 b$ .

$y = \frac{- 4 \pm \sqrt{4 (16 + b)}}{2 \cdot 1}$

Rewrite $4 (16 + b)$ as $2^{2} (4^{2} + b)$ .

Rewrite $4$ as $2^{2}$ .

$y = \frac{- 4 \pm \sqrt{2^{2} (16 + b)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

$y = \frac{- 4 \pm 2 \sqrt{4^{2} + b}}{2 \cdot 1}$

Raise $4$ to the power of $2$ .

$y = \frac{- 4 \pm 2 \sqrt{16 + b}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{- 4 \pm 2 \sqrt{16 + b}}{2}$

Simplify $\frac{- 4 \pm 2 \sqrt{16 + b}}{2}$ .

$y = - 2 \pm \sqrt{16 + b}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $4$ to the power of $2$ .

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot (- 12 - b)}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot (- 12 - b)}}{2 \cdot 1}$

Apply the distributive property.

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 12 - 4 (- b)}}{2 \cdot 1}$

Multiply $- 4$ by $- 12$ .

$y = \frac{- 4 \pm \sqrt{16 + 48 - 4 (- b)}}{2 \cdot 1}$

Multiply $- 1$ by $- 4$ .

$y = \frac{- 4 \pm \sqrt{16 + 48 + 4 b}}{2 \cdot 1}$

Add $16$ and $48$ .

$y = \frac{- 4 \pm \sqrt{64 + 4 b}}{2 \cdot 1}$

Factor $4$ out of $64 + 4 b$ .

Factor $4$ out of $64$ .

$y = \frac{- 4 \pm \sqrt{4 \cdot 16 + 4 b}}{2 \cdot 1}$

Factor $4$ out of $4 \cdot 16 + 4 b$ .

$y = \frac{- 4 \pm \sqrt{4 (16 + b)}}{2 \cdot 1}$

Rewrite $4 (16 + b)$ as $2^{2} (4^{2} + b)$ .

Rewrite $4$ as $2^{2}$ .

$y = \frac{- 4 \pm \sqrt{2^{2} (16 + b)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

$y = \frac{- 4 \pm 2 \sqrt{4^{2} + b}}{2 \cdot 1}$

Raise $4$ to the power of $2$ .

$y = \frac{- 4 \pm 2 \sqrt{16 + b}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{- 4 \pm 2 \sqrt{16 + b}}{2}$

Simplify $\frac{- 4 \pm 2 \sqrt{16 + b}}{2}$ .

$y = - 2 \pm \sqrt{16 + b}$

Change the $\pm$ to $+$ .

$y = - 2 + \sqrt{16 + b}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $4$ to the power of $2$ .

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot (- 12 - b)}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot (- 12 - b)}}{2 \cdot 1}$

Apply the distributive property.

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot - 12 - 4 (- b)}}{2 \cdot 1}$

Multiply $- 4$ by $- 12$ .

$y = \frac{- 4 \pm \sqrt{16 + 48 - 4 (- b)}}{2 \cdot 1}$

Multiply $- 1$ by $- 4$ .

$y = \frac{- 4 \pm \sqrt{16 + 48 + 4 b}}{2 \cdot 1}$

Add $16$ and $48$ .

$y = \frac{- 4 \pm \sqrt{64 + 4 b}}{2 \cdot 1}$

Factor $4$ out of $64 + 4 b$ .

Factor $4$ out of $64$ .

$y = \frac{- 4 \pm \sqrt{4 \cdot 16 + 4 b}}{2 \cdot 1}$

Factor $4$ out of $4 \cdot 16 + 4 b$ .

$y = \frac{- 4 \pm \sqrt{4 (16 + b)}}{2 \cdot 1}$

Rewrite $4 (16 + b)$ as $2^{2} (4^{2} + b)$ .

Rewrite $4$ as $2^{2}$ .

$y = \frac{- 4 \pm \sqrt{2^{2} (16 + b)}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

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

Pull terms out from under the radical.

$y = \frac{- 4 \pm 2 \sqrt{4^{2} + b}}{2 \cdot 1}$

Raise $4$ to the power of $2$ .

$y = \frac{- 4 \pm 2 \sqrt{16 + b}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{- 4 \pm 2 \sqrt{16 + b}}{2}$

Simplify $\frac{- 4 \pm 2 \sqrt{16 + b}}{2}$ .

$y = - 2 \pm \sqrt{16 + b}$

Change the $\pm$ to $-$ .

$y = - 2 - \sqrt{16 + b}$

The final answer is the combination of both solutions.

$y = - 2 + \sqrt{16 + b}$

$y = - 2 - \sqrt{16 + b}$

$y = - 2 + \sqrt{16 + b}$

$y = - 2 - \sqrt{16 + b}$

Step 4

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

$f^{- 1} (b) = - 2 + \sqrt{16 + b}, - 2 - \sqrt{16 + b}$

Step 5

Verify if $f^{- 1} (b) = - 2 + \sqrt{16 + b}, - 2 - \sqrt{16 + b}$ is the inverse of $f (b) = (b + 6) (b - 2)$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (b) = (b + 6) (b - 2)$ and $f^{- 1} (b) = - 2 + \sqrt{16 + b}, - 2 - \sqrt{16 + b}$ and compare them.

Find the range of $f (b) = (b + 6) (b - 2)$ .

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

Interval Notation:

$[- 16, \infty)$

Find the domain of $- 2 + \sqrt{16 + b}$ .

Set the radicand in $\sqrt{16 + b}$ greater than or equal to $0$ to find where the expression is defined.

$16 + b \geq 0$

Subtract $16$ from both sides of the inequality.

$b \geq - 16$

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

$[- 16, \infty)$

Find the domain of $f (b) = (b + 6) (b - 2)$ .

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 $f^{- 1} (b) = - 2 + \sqrt{16 + b}, - 2 - \sqrt{16 + b}$ is the range of $f (b) = (b + 6) (b - 2)$ and the range of $f^{- 1} (b) = - 2 + \sqrt{16 + b}, - 2 - \sqrt{16 + b}$ is the domain of $f (b) = (b + 6) (b - 2)$ , then $f^{- 1} (b) = - 2 + \sqrt{16 + b}, - 2 - \sqrt{16 + b}$ is the inverse of $f (b) = (b + 6) (b - 2)$ .

$f^{- 1} (b) = - 2 + \sqrt{16 + b}, - 2 - \sqrt{16 + b}$

Step 6

Find the Inverse f(b)=(b+6)(b-2)

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