Find the Inverse x^2+x

$x^{2} + x$

Step 1

Interchange the variables.

$x = y^{2} + y$

Step 2

Solve for $y$ .

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

$y^{2} + y = x$

Subtract $x$ from both sides of the equation.

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

One to any power is one.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Multiply $2$ by $1$ .

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

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

Simplify the numerator.

One to any power is one.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Multiply $2$ by $1$ .

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

Change the $\pm$ to $+$ .

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

Rewrite $- 1$ as $- 1 (1)$ .

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

Factor $- 1$ out of $\sqrt{1 + 4 x}$ .

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

Factor $- 1$ out of $- 1 (1) - 1 (- \sqrt{1 + 4 x})$ .

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

Move the negative in front of the fraction.

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

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

Simplify the numerator.

One to any power is one.

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

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Multiply $2$ by $1$ .

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

Change the $\pm$ to $-$ .

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

Factor $- 1$ out of $- 1 - \sqrt{1 + 4 x}$ .

Reorder $1$ and $4 x$ .

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

Rewrite $- 1$ as $- 1 (1)$ .

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

Factor $- 1$ out of $- \sqrt{4 x + 1}$ .

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

Factor $- 1$ out of $- 1 (1) - (\sqrt{4 x + 1})$ .

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

Rewrite $- 1 (1 + \sqrt{4 x + 1})$ as $- (1 + \sqrt{4 x + 1})$ .

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

Move the negative in front of the fraction.

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

The final answer is the combination of both solutions.

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

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

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

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

Step 3

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

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

Step 4

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

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

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

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

Interval Notation:

$[- \frac{1}{4}, \infty)$

Find the domain of $- \frac{1 - \sqrt{1 + 4 x}}{2}$ .

Set the radicand in $\sqrt{1 + 4 x}$ greater than or equal to $0$ to find where the expression is defined.

$1 + 4 x \geq 0$

Solve for $x$ .

Subtract $1$ from both sides of the inequality.

$4 x \geq - 1$

Divide each term in $4 x \geq - 1$ by $4$ and simplify.

Divide each term in $4 x \geq - 1$ by $4$ .

$\frac{4 x}{4} \geq \frac{- 1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} \geq \frac{- 1}{4}$

Divide $x$ by $1$ .

$x \geq \frac{- 1}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{1}{4}$

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

$[- \frac{1}{4}, \infty)$

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

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

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

Step 5

Find the Inverse x^2+x

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