Find the Inverse f(x)=x+ square root of x

$f (x) = x + \sqrt{x}$

Step 1

Write $f (x) = x + \sqrt{x}$ as an equation.

$y = x + \sqrt{x}$

Step 2

Interchange the variables.

$x = y + \sqrt{y}$

Step 3

Solve for $y$ .

Rewrite the equation as $y + \sqrt{y} = x$ .

$y + \sqrt{y} = x$

Subtract $y$ from both sides of the equation.

$\sqrt{y} = x - y$

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{y}}^{2} = {(x - y)}^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{y}$ as $y^{\frac{1}{2}}$ .

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

Simplify the left side.

Simplify ${(y^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${(y^{\frac{1}{2}})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$y^{1} = {(x - y)}^{2}$

Simplify.

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

Simplify the right side.

Simplify ${(x - y)}^{2}$ .

Rewrite ${(x - y)}^{2}$ as $(x - y) (x - y)$ .

$y = (x - y) (x - y)$

Expand $(x - y) (x - y)$ using the FOIL Method.

Apply the distributive property.

$y = x (x - y) - y (x - y)$

Apply the distributive property.

$y = x \cdot x + x (- y) - y (x - y)$

Apply the distributive property.

$y = x \cdot x + x (- y) - y x - y (- y)$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$y = x^{2} + x (- y) - y x - y (- y)$

Rewrite using the commutative property of multiplication.

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

Rewrite using the commutative property of multiplication.

$y = x^{2} - x y - y x - 1 \cdot - 1 y \cdot y$

Multiply $y$ by $y$ by adding the exponents.

Move $y$ .

$y = x^{2} - x y - y x - 1 \cdot - 1 (y \cdot y)$

Multiply $y$ by $y$ .

$y = x^{2} - x y - y x - 1 \cdot - 1 y^{2}$

Multiply $- 1$ by $- 1$ .

$y = x^{2} - x y - y x + 1 y^{2}$

Multiply $y^{2}$ by $1$ .

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

Subtract $y x$ from $- x y$ .

Move $y$ .

$y = x^{2} - x y - 1 x y + y^{2}$

Subtract $x y$ from $- x y$ .

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

Solve for $y$ .

Since $y$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

Subtract $y$ from both sides of the equation.

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

Apply the distributive property.

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

Multiply $- 2$ by $- 1$ .

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

Multiply $- 1$ by $- 1$ .

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

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = - 2 x - 1$ and $b = 2 \cdot 1 x$ .

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

Simplify.

Multiply $2$ by $1$ .

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

Add $- 2 x$ and $2 x$ .

$y = \frac{2 x + 1 \pm \sqrt{(0 - 1) (- 2 x - 1 - (2 \cdot (1 x)))}}{2 \cdot 1}$

Subtract $1$ from $0$ .

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

Multiply $2$ by $1$ .

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

Multiply $2$ by $- 1$ .

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

Subtract $2 x$ from $- 2 x$ .

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

Multiply $2$ by $1$ .

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

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $- 2$ by $- 1$ .

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

Multiply $- 1$ by $- 1$ .

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

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = - 2 x - 1$ and $b = 2 \cdot 1 x$ .

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

Simplify.

Multiply $2$ by $1$ .

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

Add $- 2 x$ and $2 x$ .

$y = \frac{2 x + 1 \pm \sqrt{(0 - 1) (- 2 x - 1 - (2 \cdot (1 x)))}}{2 \cdot 1}$

Subtract $1$ from $0$ .

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

Multiply $2$ by $1$ .

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

Multiply $2$ by $- 1$ .

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

Subtract $2 x$ from $- 2 x$ .

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

Multiply $2$ by $1$ .

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

Change the $\pm$ to $+$ .

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

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

Simplify the numerator.

Apply the distributive property.

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

Multiply $- 2$ by $- 1$ .

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

Multiply $- 1$ by $- 1$ .

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

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = - 2 x - 1$ and $b = 2 \cdot 1 x$ .

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

Simplify.

Multiply $2$ by $1$ .

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

Add $- 2 x$ and $2 x$ .

$y = \frac{2 x + 1 \pm \sqrt{(0 - 1) (- 2 x - 1 - (2 \cdot (1 x)))}}{2 \cdot 1}$

Subtract $1$ from $0$ .

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

Multiply $2$ by $1$ .

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

Multiply $2$ by $- 1$ .

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

Subtract $2 x$ from $- 2 x$ .

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

Multiply $2$ by $1$ .

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

Change the $\pm$ to $-$ .

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

The final answer is the combination of both solutions.

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

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

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

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

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

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

Step 4

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

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

Step 5

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

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

Find the range of $f (x) = x + \sqrt{x}$ .

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

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

$- 1 (- 4 x - 1) \geq 0$

Solve for $x$ .

Divide each term in $- 1 (- 4 x - 1) \geq 0$ by $- 1$ and simplify.

Divide each term in $- 1 (- 4 x - 1) \geq 0$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 1 (- 4 x - 1)}{- 1} \leq \frac{0}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{- 4 x - 1}{1} \leq \frac{0}{- 1}$

Divide $- 4 x - 1$ by $1$ .

$- 4 x - 1 \leq \frac{0}{- 1}$

Simplify the right side.

Divide $0$ by $- 1$ .

$- 4 x - 1 \leq 0$

Add $1$ to both sides of the inequality.

$- 4 x \leq 1$

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

Divide each term in $- 4 x \leq 1$ by $- 4$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

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.

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

Since the domain of $f^{- 1} (x) = \frac{2 x + 1 + \sqrt{- 1 (- 4 x - 1)}}{2}, \frac{2 x + 1 - \sqrt{- 1 (- 4 x - 1)}}{2}$ is not equal to the range of $f (x) = x + \sqrt{x}$ , then $f^{- 1} (x) = \frac{2 x + 1 + \sqrt{- 1 (- 4 x - 1)}}{2}, \frac{2 x + 1 - \sqrt{- 1 (- 4 x - 1)}}{2}$ is not an inverse of $f (x) = x + \sqrt{x}$ .

There is no inverse

Step 6

Find the Inverse f(x)=x+ square root of x

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