Find the Inverse y = square root of 2+5x

$y = \sqrt{2 + 5 x}$

Step 1

Interchange the variables.

$x = \sqrt{2 + 5 y}$

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt{2 + 5 y} = x$ .

$\sqrt{2 + 5 y} = x$

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

${\sqrt{2 + 5 y}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 + 5 y}$ as ${(2 + 5 y)}^{\frac{1}{2}}$ .

${({(2 + 5 y)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(2 + 5 y)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(2 + 5 y)}^{\frac{1}{2}})}^{2}$ .

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

${(2 + 5 y)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(2 + 5 y)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(2 + 5 y)}^{1} = x^{2}$

Simplify.

$2 + 5 y = x^{2}$

Solve for $y$ .

Subtract $2$ from both sides of the equation.

$5 y = x^{2} - 2$

Divide each term in $5 y = x^{2} - 2$ by $5$ and simplify.

Divide each term in $5 y = x^{2} - 2$ by $5$ .

$\frac{5 y}{5} = \frac{x^{2}}{5} + \frac{- 2}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 y}{5} = \frac{x^{2}}{5} + \frac{- 2}{5}$

Divide $y$ by $1$ .

$y = \frac{x^{2}}{5} + \frac{- 2}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$y = \frac{x^{2}}{5} - \frac{2}{5}$

Step 3

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

$f^{- 1} (x) = \frac{x^{2}}{5} - \frac{2}{5}$

Step 4

Verify if $f^{- 1} (x) = \frac{x^{2}}{5} - \frac{2}{5}$ is the inverse of $f (x) = \sqrt{2 + 5 x}$ .

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 $f^{- 1} (\sqrt{2 + 5 x})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt{2 + 5 x}) = \frac{{(\sqrt{2 + 5 x})}^{2}}{5} - \frac{2}{5}$

Combine the numerators over the common denominator.

$f^{- 1} (\sqrt{2 + 5 x}) = \frac{{(\sqrt{2 + 5 x})}^{2} - 2}{5}$

Rewrite ${\sqrt{2 + 5 x}}^{2}$ as $2 + 5 x$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 + 5 x}$ as ${(2 + 5 x)}^{\frac{1}{2}}$ .

$f^{- 1} (\sqrt{2 + 5 x}) = \frac{{({(2 + 5 x)}^{\frac{1}{2}})}^{2} - 2}{5}$

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

$f^{- 1} (\sqrt{2 + 5 x}) = \frac{{(2 + 5 x)}^{\frac{1}{2} \cdot 2} - 2}{5}$

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

$f^{- 1} (\sqrt{2 + 5 x}) = \frac{{(2 + 5 x)}^{\frac{2}{2}} - 2}{5}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\sqrt{2 + 5 x}) = \frac{{(2 + 5 x)}^{\frac{2}{2}} - 2}{5}$

Rewrite the expression.

$f^{- 1} (\sqrt{2 + 5 x}) = \frac{(2 + 5 x) - 2}{5}$

Simplify.

$f^{- 1} (\sqrt{2 + 5 x}) = \frac{2 + 5 x - 2}{5}$

Simplify terms.

Combine the opposite terms in $2 + 5 x - 2$ .

Subtract $2$ from $2$ .

$f^{- 1} (\sqrt{2 + 5 x}) = \frac{5 x + 0}{5}$

Add $5 x$ and $0$ .

$f^{- 1} (\sqrt{2 + 5 x}) = \frac{5 x}{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f^{- 1} (\sqrt{2 + 5 x}) = \frac{5 x}{5}$

Divide $x$ by $1$ .

$f^{- 1} (\sqrt{2 + 5 x}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{x^{2}}{5} - \frac{2}{5})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{x^{2}}{5} - \frac{2}{5}) = \sqrt{2 + 5 (\frac{x^{2}}{5} - \frac{2}{5})}$

Apply the distributive property.

$f (\frac{x^{2}}{5} - \frac{2}{5}) = \sqrt{2 + 5 (\frac{x^{2}}{5}) + 5 (- \frac{2}{5})}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (\frac{x^{2}}{5} - \frac{2}{5}) = \sqrt{2 + 5 (\frac{x^{2}}{5}) + 5 (- \frac{2}{5})}$

Rewrite the expression.

$f (\frac{x^{2}}{5} - \frac{2}{5}) = \sqrt{2 + x^{2} + 5 (- \frac{2}{5})}$

Cancel the common factor of $5$ .

Move the leading negative in $- \frac{2}{5}$ into the numerator.

$f (\frac{x^{2}}{5} - \frac{2}{5}) = \sqrt{2 + x^{2} + 5 (\frac{- 2}{5})}$

Cancel the common factor.

$f (\frac{x^{2}}{5} - \frac{2}{5}) = \sqrt{2 + x^{2} + 5 (\frac{- 2}{5})}$

Rewrite the expression.

$f (\frac{x^{2}}{5} - \frac{2}{5}) = \sqrt{2 + x^{2} - 2}$

Simplify by subtracting numbers.

Subtract $2$ from $2$ .

$f (\frac{x^{2}}{5} - \frac{2}{5}) = \sqrt{x^{2} + 0}$

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

$f (\frac{x^{2}}{5} - \frac{2}{5}) = \sqrt{x^{2}}$

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

$f (\frac{x^{2}}{5} - \frac{2}{5}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{x^{2}}{5} - \frac{2}{5}$ is the inverse of $f (x) = \sqrt{2 + 5 x}$ .

$f^{- 1} (x) = \frac{x^{2}}{5} - \frac{2}{5}$

Find the Inverse y = square root of 2+5x

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