Find the Inverse f(x)=x^2-2x , x<=1

$f (x) = x^{2} - 2 x$ , $x \leq 1$

Step 1

Find the range of the given function.

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

$[- 1, \infty)$

Convert $[- 1, \infty)$ to an inequality.

$y \geq - 1$

Step 2

Find the inverse.

Interchange the variables.

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

Solve for $y$ .

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

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

Subtract $x$ from both sides of the equation.

$y^{2} - 2 y - x = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 2$ , and $c = - x$ into the quadratic formula and solve for $y$ .

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

Simplify.

Simplify the numerator.

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

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

Multiply $- 4 \cdot 1 \cdot - 1$ .

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Factor $4$ out of $4 + 4 x$ .

Factor $4$ out of $4$ .

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

Factor $4$ out of $4 \cdot 1 + 4 x$ .

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

Rewrite $4 (1 + x)$ as $2^{2} (1^{2} + x)$ .

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

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

Rewrite $1$ as $1^{2}$ .

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

Pull terms out from under the radical.

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

One to any power is one.

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

Multiply $2$ by $1$ .

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

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

$y = 1 \pm \sqrt{1 + x}$

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

Simplify the numerator.

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

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

Multiply $- 4 \cdot 1 \cdot - 1$ .

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Factor $4$ out of $4 + 4 x$ .

Factor $4$ out of $4$ .

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

Factor $4$ out of $4 \cdot 1 + 4 x$ .

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

Rewrite $4 (1 + x)$ as $2^{2} (1^{2} + x)$ .

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

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

Rewrite $1$ as $1^{2}$ .

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

Pull terms out from under the radical.

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

One to any power is one.

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

Multiply $2$ by $1$ .

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

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

$y = 1 \pm \sqrt{1 + x}$

Change the $\pm$ to $+$ .

$y = 1 + \sqrt{1 + x}$

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

Simplify the numerator.

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

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

Multiply $- 4 \cdot 1 \cdot - 1$ .

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Factor $4$ out of $4 + 4 x$ .

Factor $4$ out of $4$ .

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

Factor $4$ out of $4 \cdot 1 + 4 x$ .

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

Rewrite $4 (1 + x)$ as $2^{2} (1^{2} + x)$ .

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

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

Rewrite $1$ as $1^{2}$ .

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

Pull terms out from under the radical.

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

One to any power is one.

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

Multiply $2$ by $1$ .

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

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

$y = 1 \pm \sqrt{1 + x}$

Change the $\pm$ to $-$ .

$y = 1 - \sqrt{1 + x}$

The final answer is the combination of both solutions.

$y = 1 + \sqrt{1 + x}$

$y = 1 - \sqrt{1 + x}$

$y = 1 + \sqrt{1 + x}$

$y = 1 - \sqrt{1 + x}$

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

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

Step 3

Find the inverse using the domain and the range of the original function.

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

Step 4

Find the Inverse f(x)=x^2-2x , x<=1

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