Find the Inverse y = square root of x+2-3

$y = \sqrt{x + 2} - 3$

Step 1

Interchange the variables.

$x = \sqrt{y + 2} - 3$

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt{y + 2} - 3 = x$ .

$\sqrt{y + 2} - 3 = x$

Add $3$ to both sides of the equation.

$\sqrt{y + 2} = x + 3$

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

${\sqrt{y + 2}}^{2} = {(x + 3)}^{2}$

Simplify each side of the equation.

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

${({(y + 2)}^{\frac{1}{2}})}^{2} = {(x + 3)}^{2}$

Simplify the left side.

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

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

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

${(y + 2)}^{\frac{1}{2} \cdot 2} = {(x + 3)}^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(y + 2)}^{\frac{1}{2} \cdot 2} = {(x + 3)}^{2}$

Rewrite the expression.

${(y + 2)}^{1} = {(x + 3)}^{2}$

Simplify.

$y + 2 = {(x + 3)}^{2}$

Simplify the right side.

Simplify ${(x + 3)}^{2}$ .

Rewrite ${(x + 3)}^{2}$ as $(x + 3) (x + 3)$ .

$y + 2 = (x + 3) (x + 3)$

Expand $(x + 3) (x + 3)$ using the FOIL Method.

Apply the distributive property.

$y + 2 = x (x + 3) + 3 (x + 3)$

Apply the distributive property.

$y + 2 = x \cdot x + x \cdot 3 + 3 (x + 3)$

Apply the distributive property.

$y + 2 = x \cdot x + x \cdot 3 + 3 x + 3 \cdot 3$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$y + 2 = x^{2} + x \cdot 3 + 3 x + 3 \cdot 3$

Move $3$ to the left of $x$ .

$y + 2 = x^{2} + 3 \cdot x + 3 x + 3 \cdot 3$

Multiply $3$ by $3$ .

$y + 2 = x^{2} + 3 x + 3 x + 9$

Add $3 x$ and $3 x$ .

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

Move all terms not containing $y$ to the right side of the equation.

Subtract $2$ from both sides of the equation.

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

Subtract $2$ from $9$ .

$y = x^{2} + 6 x + 7$

Step 3

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

$f^{- 1} (x) = x^{2} + 6 x + 7$

Step 4

Verify if $f^{- 1} (x) = x^{2} + 6 x + 7$ is the inverse of $f (x) = \sqrt{x + 2} - 3$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (\sqrt{x + 2} - 3)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt{x + 2} - 3) = {(\sqrt{x + 2} - 3)}^{2} + 6 (\sqrt{x + 2} - 3) + 7$

Simplify each term.

Rewrite ${(\sqrt{x + 2} - 3)}^{2}$ as $(\sqrt{x + 2} - 3) (\sqrt{x + 2} - 3)$ .

$f^{- 1} (\sqrt{x + 2} - 3) = (\sqrt{x + 2} - 3) (\sqrt{x + 2} - 3) + 6 (\sqrt{x + 2} - 3) + 7$

Expand $(\sqrt{x + 2} - 3) (\sqrt{x + 2} - 3)$ using the FOIL Method.

Apply the distributive property.

$f^{- 1} (\sqrt{x + 2} - 3) = \sqrt{x + 2} (\sqrt{x + 2} - 3) - 3 (\sqrt{x + 2} - 3) + 6 (\sqrt{x + 2} - 3) + 7$

Apply the distributive property.

$f^{- 1} (\sqrt{x + 2} - 3) = \sqrt{x + 2} \sqrt{x + 2} + \sqrt{x + 2} \cdot - 3 - 3 (\sqrt{x + 2} - 3) + 6 (\sqrt{x + 2} - 3) + 7$

Apply the distributive property.

$f^{- 1} (\sqrt{x + 2} - 3) = \sqrt{x + 2} \sqrt{x + 2} + \sqrt{x + 2} \cdot - 3 - 3 \sqrt{x + 2} - 3 \cdot - 3 + 6 (\sqrt{x + 2} - 3) + 7$

Simplify and combine like terms.

Simplify each term.

Multiply $\sqrt{x + 2} \sqrt{x + 2}$ .

Raise $\sqrt{x + 2}$ to the power of $1$ .

$f^{- 1} (\sqrt{x + 2} - 3) = \sqrt{x + 2} \sqrt{x + 2} + \sqrt{x + 2} \cdot - 3 - 3 \sqrt{x + 2} - 3 \cdot - 3 + 6 (\sqrt{x + 2} - 3) + 7$

Raise $\sqrt{x + 2}$ to the power of $1$ .

$f^{- 1} (\sqrt{x + 2} - 3) = \sqrt{x + 2} \sqrt{x + 2} + \sqrt{x + 2} \cdot - 3 - 3 \sqrt{x + 2} - 3 \cdot - 3 + 6 (\sqrt{x + 2} - 3) + 7$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f^{- 1} (\sqrt{x + 2} - 3) = {\sqrt{x + 2}}^{1 + 1} + \sqrt{x + 2} \cdot - 3 - 3 \sqrt{x + 2} - 3 \cdot - 3 + 6 (\sqrt{x + 2} - 3) + 7$

Add $1$ and $1$ .

$f^{- 1} (\sqrt{x + 2} - 3) = {\sqrt{x + 2}}^{2} + \sqrt{x + 2} \cdot - 3 - 3 \sqrt{x + 2} - 3 \cdot - 3 + 6 (\sqrt{x + 2} - 3) + 7$

Rewrite ${\sqrt{x + 2}}^{2}$ as $x + 2$ .

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

$f^{- 1} (\sqrt{x + 2} - 3) = {({(x + 2)}^{\frac{1}{2}})}^{2} + \sqrt{x + 2} \cdot - 3 - 3 \sqrt{x + 2} - 3 \cdot - 3 + 6 (\sqrt{x + 2} - 3) + 7$

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

$f^{- 1} (\sqrt{x + 2} - 3) = {(x + 2)}^{\frac{1}{2} \cdot 2} + \sqrt{x + 2} \cdot - 3 - 3 \sqrt{x + 2} - 3 \cdot - 3 + 6 (\sqrt{x + 2} - 3) + 7$

Combine $\frac{1}{2}$ and $2$ .

$f^{- 1} (\sqrt{x + 2} - 3) = {(x + 2)}^{\frac{2}{2}} + \sqrt{x + 2} \cdot - 3 - 3 \sqrt{x + 2} - 3 \cdot - 3 + 6 (\sqrt{x + 2} - 3) + 7$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\sqrt{x + 2} - 3) = {(x + 2)}^{\frac{2}{2}} + \sqrt{x + 2} \cdot - 3 - 3 \sqrt{x + 2} - 3 \cdot - 3 + 6 (\sqrt{x + 2} - 3) + 7$

Rewrite the expression.

$f^{- 1} (\sqrt{x + 2} - 3) = (x + 2) + \sqrt{x + 2} \cdot - 3 - 3 \sqrt{x + 2} - 3 \cdot - 3 + 6 (\sqrt{x + 2} - 3) + 7$

Simplify.

$f^{- 1} (\sqrt{x + 2} - 3) = x + 2 + \sqrt{x + 2} \cdot - 3 - 3 \sqrt{x + 2} - 3 \cdot - 3 + 6 (\sqrt{x + 2} - 3) + 7$

Move $- 3$ to the left of $\sqrt{x + 2}$ .

$f^{- 1} (\sqrt{x + 2} - 3) = x + 2 - 3 \cdot \sqrt{x + 2} - 3 \sqrt{x + 2} - 3 \cdot - 3 + 6 (\sqrt{x + 2} - 3) + 7$

Multiply $- 3$ by $- 3$ .

$f^{- 1} (\sqrt{x + 2} - 3) = x + 2 - 3 \sqrt{x + 2} - 3 \sqrt{x + 2} + 9 + 6 (\sqrt{x + 2} - 3) + 7$

Add $2$ and $9$ .

$f^{- 1} (\sqrt{x + 2} - 3) = x + 11 - 3 \sqrt{x + 2} - 3 \sqrt{x + 2} + 6 (\sqrt{x + 2} - 3) + 7$

Subtract $3 \sqrt{x + 2}$ from $- 3 \sqrt{x + 2}$ .

$f^{- 1} (\sqrt{x + 2} - 3) = x + 11 - 6 \sqrt{x + 2} + 6 (\sqrt{x + 2} - 3) + 7$

Apply the distributive property.

$f^{- 1} (\sqrt{x + 2} - 3) = x + 11 - 6 \sqrt{x + 2} + 6 \sqrt{x + 2} + 6 \cdot - 3 + 7$

Multiply $6$ by $- 3$ .

$f^{- 1} (\sqrt{x + 2} - 3) = x + 11 - 6 \sqrt{x + 2} + 6 \sqrt{x + 2} - 18 + 7$

Simplify by adding terms.

Combine the opposite terms in $x + 11 - 6 \sqrt{x + 2} + 6 \sqrt{x + 2} - 18 + 7$ .

Add $- 6 \sqrt{x + 2}$ and $6 \sqrt{x + 2}$ .

$f^{- 1} (\sqrt{x + 2} - 3) = x + 11 + 0 - 18 + 7$

Add $x + 11$ and $0$ .

$f^{- 1} (\sqrt{x + 2} - 3) = x + 11 - 18 + 7$

Subtract $18$ from $11$ .

$f^{- 1} (\sqrt{x + 2} - 3) = x - 7 + 7$

Combine the opposite terms in $x - 7 + 7$ .

Add $- 7$ and $7$ .

$f^{- 1} (\sqrt{x + 2} - 3) = x + 0$

Add $x$ and $0$ .

$f^{- 1} (\sqrt{x + 2} - 3) = x$

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (x^{2} + 6 x + 7)$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (x^{2} + 6 x + 7) = \sqrt{(x^{2} + 6 x + 7) + 2} - 3$

Simplify each term.

Add $7$ and $2$ .

$f (x^{2} + 6 x + 7) = \sqrt{x^{2} + 6 x + 9} - 3$

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$f (x^{2} + 6 x + 7) = \sqrt{x^{2} + 6 x + 3^{2}} - 3$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot x \cdot 3$

Rewrite the polynomial.

$f (x^{2} + 6 x + 7) = \sqrt{x^{2} + 2 \cdot x \cdot 3 + 3^{2}} - 3$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 3$ .

$f (x^{2} + 6 x + 7) = \sqrt{{(x + 3)}^{2}} - 3$

Pull terms out from under the radical, assuming positive real numbers.

$f (x^{2} + 6 x + 7) = x + 3 - 3$

Combine the opposite terms in $x + 3 - 3$ .

Subtract $3$ from $3$ .

$f (x^{2} + 6 x + 7) = x + 0$

Add $x$ and $0$ .

$f (x^{2} + 6 x + 7) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = x^{2} + 6 x + 7$ is the inverse of $f (x) = \sqrt{x + 2} - 3$ .

$f^{- 1} (x) = x^{2} + 6 x + 7$

Find the Inverse y = square root of x+2-3

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