Find the Inverse What is the inverse of f(x)=x^2-6x+9

What is the inverse of $f (x) = x^{2} - 6 x + 9$

Step 1

Interchange the variables.

$x = y^{2} - 6 y + 9$

Step 2

Solve for $y$ .

Rewrite the equation as $y^{2} - 6 y + 9 = x$ .

$y^{2} - 6 y + 9 = x$

Subtract $x$ from both sides of the equation.

$y^{2} - 6 y + 9 - x = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 6$ , and $c = 9 - x$ into the quadratic formula and solve for $y$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (1 \cdot (9 - x))}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$y = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot (9 - x)}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$y = \frac{6 \pm \sqrt{36 - 4 \cdot (9 - x)}}{2 \cdot 1}$

Apply the distributive property.

$y = \frac{6 \pm \sqrt{36 - 4 \cdot 9 - 4 (- x)}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$y = \frac{6 \pm \sqrt{36 - 36 - 4 (- x)}}{2 \cdot 1}$

Multiply $- 1$ by $- 4$ .

$y = \frac{6 \pm \sqrt{36 - 36 + 4 x}}{2 \cdot 1}$

Subtract $36$ from $36$ .

$y = \frac{6 \pm \sqrt{0 + 4 x}}{2 \cdot 1}$

Add $0$ and $4 x$ .

$y = \frac{6 \pm \sqrt{4 x}}{2 \cdot 1}$

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

$y = \frac{6 \pm \sqrt{2^{2} x}}{2 \cdot 1}$

Pull terms out from under the radical.

$y = \frac{6 \pm 2 \sqrt{x}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{6 \pm 2 \sqrt{x}}{2}$

Simplify $\frac{6 \pm 2 \sqrt{x}}{2}$ .

$y = 3 \pm \sqrt{x}$

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

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$y = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot (9 - x)}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$y = \frac{6 \pm \sqrt{36 - 4 \cdot (9 - x)}}{2 \cdot 1}$

Apply the distributive property.

$y = \frac{6 \pm \sqrt{36 - 4 \cdot 9 - 4 (- x)}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$y = \frac{6 \pm \sqrt{36 - 36 - 4 (- x)}}{2 \cdot 1}$

Multiply $- 1$ by $- 4$ .

$y = \frac{6 \pm \sqrt{36 - 36 + 4 x}}{2 \cdot 1}$

Subtract $36$ from $36$ .

$y = \frac{6 \pm \sqrt{0 + 4 x}}{2 \cdot 1}$

Add $0$ and $4 x$ .

$y = \frac{6 \pm \sqrt{4 x}}{2 \cdot 1}$

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

$y = \frac{6 \pm \sqrt{2^{2} x}}{2 \cdot 1}$

Pull terms out from under the radical.

$y = \frac{6 \pm 2 \sqrt{x}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{6 \pm 2 \sqrt{x}}{2}$

Simplify $\frac{6 \pm 2 \sqrt{x}}{2}$ .

$y = 3 \pm \sqrt{x}$

Change the $\pm$ to $+$ .

$y = 3 + \sqrt{x}$

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

Simplify the numerator.

Raise $- 6$ to the power of $2$ .

$y = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot (9 - x)}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$y = \frac{6 \pm \sqrt{36 - 4 \cdot (9 - x)}}{2 \cdot 1}$

Apply the distributive property.

$y = \frac{6 \pm \sqrt{36 - 4 \cdot 9 - 4 (- x)}}{2 \cdot 1}$

Multiply $- 4$ by $9$ .

$y = \frac{6 \pm \sqrt{36 - 36 - 4 (- x)}}{2 \cdot 1}$

Multiply $- 1$ by $- 4$ .

$y = \frac{6 \pm \sqrt{36 - 36 + 4 x}}{2 \cdot 1}$

Subtract $36$ from $36$ .

$y = \frac{6 \pm \sqrt{0 + 4 x}}{2 \cdot 1}$

Add $0$ and $4 x$ .

$y = \frac{6 \pm \sqrt{4 x}}{2 \cdot 1}$

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

$y = \frac{6 \pm \sqrt{2^{2} x}}{2 \cdot 1}$

Pull terms out from under the radical.

$y = \frac{6 \pm 2 \sqrt{x}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$y = \frac{6 \pm 2 \sqrt{x}}{2}$

Simplify $\frac{6 \pm 2 \sqrt{x}}{2}$ .

$y = 3 \pm \sqrt{x}$

Change the $\pm$ to $-$ .

$y = 3 - \sqrt{x}$

The final answer is the combination of both solutions.

$y = 3 + \sqrt{x}$

$y = 3 - \sqrt{x}$

$y = 3 + \sqrt{x}$

$y = 3 - \sqrt{x}$

Step 3

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

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

Step 4

Verify if $f^{- 1} (x) = 3 + \sqrt{x}, 3 - \sqrt{x}$ is the inverse of $f (x) = x^{2} - 6 x + 9$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = x^{2} - 6 x + 9$ and $f^{- 1} (x) = 3 + \sqrt{x}, 3 - \sqrt{x}$ and compare them.

Find the range of $f (x) = x^{2} - 6 x + 9$ .

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

Interval Notation:

$[0, \infty)$

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

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

$x \geq 0$

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

$[0, \infty)$

Find the domain of $f (x) = x^{2} - 6 x + 9$ .

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 $f^{- 1} (x) = 3 + \sqrt{x}, 3 - \sqrt{x}$ is the range of $f (x) = x^{2} - 6 x + 9$ and the range of $f^{- 1} (x) = 3 + \sqrt{x}, 3 - \sqrt{x}$ is the domain of $f (x) = x^{2} - 6 x + 9$ , then $f^{- 1} (x) = 3 + \sqrt{x}, 3 - \sqrt{x}$ is the inverse of $f (x) = x^{2} - 6 x + 9$ .

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

Step 5

Find the Inverse What is the inverse of f(x)=x^2-6x+9

Related Questions

Support LCMGCF.com by disabling your adblocker.