Find the Inverse square root of 8x+9

$\sqrt{8 x + 9}$

Step 1

Interchange the variables.

$x = \sqrt{8 y + 9}$

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt{8 y + 9} = x$ .

$\sqrt{8 y + 9} = x$

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

${\sqrt{8 y + 9}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{8 y + 9}$ as ${(8 y + 9)}^{\frac{1}{2}}$ .

${({(8 y + 9)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(8 y + 9)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(8 y + 9)}^{\frac{1}{2}})}^{2}$ .

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

${(8 y + 9)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(8 y + 9)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(8 y + 9)}^{1} = x^{2}$

Simplify.

$8 y + 9 = x^{2}$

Solve for $y$ .

Subtract $9$ from both sides of the equation.

$8 y = x^{2} - 9$

Divide each term in $8 y = x^{2} - 9$ by $8$ and simplify.

Divide each term in $8 y = x^{2} - 9$ by $8$ .

$\frac{8 y}{8} = \frac{x^{2}}{8} + \frac{- 9}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 y}{8} = \frac{x^{2}}{8} + \frac{- 9}{8}$

Divide $y$ by $1$ .

$y = \frac{x^{2}}{8} + \frac{- 9}{8}$

Simplify the right side.

Move the negative in front of the fraction.

$y = \frac{x^{2}}{8} - \frac{9}{8}$

Step 3

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

$f^{- 1} (x) = \frac{x^{2}}{8} - \frac{9}{8}$

Step 4

Verify if $f^{- 1} (x) = \frac{x^{2}}{8} - \frac{9}{8}$ is the inverse of $f (x) = \sqrt{8 x + 9}$ .

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 $f^{- 1} (\sqrt{8 x + 9})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt{8 x + 9}) = \frac{{(\sqrt{8 x + 9})}^{2}}{8} - \frac{9}{8}$

Combine the numerators over the common denominator.

$f^{- 1} (\sqrt{8 x + 9}) = \frac{{(\sqrt{8 x + 9})}^{2} - 9}{8}$

Rewrite ${\sqrt{8 x + 9}}^{2}$ as $8 x + 9$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{8 x + 9}$ as ${(8 x + 9)}^{\frac{1}{2}}$ .

$f^{- 1} (\sqrt{8 x + 9}) = \frac{{({(8 x + 9)}^{\frac{1}{2}})}^{2} - 9}{8}$

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

$f^{- 1} (\sqrt{8 x + 9}) = \frac{{(8 x + 9)}^{\frac{1}{2} \cdot 2} - 9}{8}$

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

$f^{- 1} (\sqrt{8 x + 9}) = \frac{{(8 x + 9)}^{\frac{2}{2}} - 9}{8}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\sqrt{8 x + 9}) = \frac{{(8 x + 9)}^{\frac{2}{2}} - 9}{8}$

Rewrite the expression.

$f^{- 1} (\sqrt{8 x + 9}) = \frac{(8 x + 9) - 9}{8}$

Simplify.

$f^{- 1} (\sqrt{8 x + 9}) = \frac{8 x + 9 - 9}{8}$

Simplify terms.

Combine the opposite terms in $8 x + 9 - 9$ .

Subtract $9$ from $9$ .

$f^{- 1} (\sqrt{8 x + 9}) = \frac{8 x + 0}{8}$

Add $8 x$ and $0$ .

$f^{- 1} (\sqrt{8 x + 9}) = \frac{8 x}{8}$

Cancel the common factor of $8$ .

Cancel the common factor.

$f^{- 1} (\sqrt{8 x + 9}) = \frac{8 x}{8}$

Divide $x$ by $1$ .

$f^{- 1} (\sqrt{8 x + 9}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{x^{2}}{8} - \frac{9}{8})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{2}}{8} - \frac{9}{8}) = \sqrt{8 (\frac{x^{2}}{8} - \frac{9}{8}) + 9}$

Apply the distributive property.

$f (\frac{x^{2}}{8} - \frac{9}{8}) = \sqrt{8 (\frac{x^{2}}{8}) + 8 (- \frac{9}{8}) + 9}$

Cancel the common factor of $8$ .

Cancel the common factor.

$f (\frac{x^{2}}{8} - \frac{9}{8}) = \sqrt{8 (\frac{x^{2}}{8}) + 8 (- \frac{9}{8}) + 9}$

Rewrite the expression.

$f (\frac{x^{2}}{8} - \frac{9}{8}) = \sqrt{x^{2} + 8 (- \frac{9}{8}) + 9}$

Cancel the common factor of $8$ .

Move the leading negative in $- \frac{9}{8}$ into the numerator.

$f (\frac{x^{2}}{8} - \frac{9}{8}) = \sqrt{x^{2} + 8 (\frac{- 9}{8}) + 9}$

Cancel the common factor.

$f (\frac{x^{2}}{8} - \frac{9}{8}) = \sqrt{x^{2} + 8 (\frac{- 9}{8}) + 9}$

Rewrite the expression.

$f (\frac{x^{2}}{8} - \frac{9}{8}) = \sqrt{x^{2} - 9 + 9}$

Simplify by adding numbers.

Add $- 9$ and $9$ .

$f (\frac{x^{2}}{8} - \frac{9}{8}) = \sqrt{x^{2} + 0}$

Add $x^{2}$ and $0$ .

$f (\frac{x^{2}}{8} - \frac{9}{8}) = \sqrt{x^{2}}$

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

$f (\frac{x^{2}}{8} - \frac{9}{8}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{2}}{8} - \frac{9}{8}$ is the inverse of $f (x) = \sqrt{8 x + 9}$ .

$f^{- 1} (x) = \frac{x^{2}}{8} - \frac{9}{8}$

Find the Inverse square root of 8x+9

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