Find the Inverse f(x)=1/2 square root of 2x+1

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

Step 1

Write $f (x) = \frac{1}{2} \sqrt{2 x + 1}$ as an equation.

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

Step 2

Interchange the variables.

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

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{1}{2} \cdot \sqrt{2 y + 1} = x$ .

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

Multiply both sides of the equation by $2$ .

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

Simplify the left side.

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

Combine $\frac{1}{2}$ and $\sqrt{2 y + 1}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

$\sqrt{2 y + 1} = 2 x$

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

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

Apply the product rule to $2 x$ .

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

Raise $2$ to the power of $2$ .

$2 y + 1 = 4 x^{2}$

Solve for $y$ .

Subtract $1$ from both sides of the equation.

$2 y = 4 x^{2} - 1$

Divide each term in $2 y = 4 x^{2} - 1$ by $2$ and simplify.

Divide each term in $2 y = 4 x^{2} - 1$ by $2$ .

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $y$ by $1$ .

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

Simplify the right side.

Simplify each term.

Cancel the common factor of $4$ and $2$ .

Factor $2$ out of $4 x^{2}$ .

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

Cancel the common factors.

Factor $2$ out of $2$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Divide $2 x^{2}$ by $1$ .

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

Move the negative in front of the fraction.

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

Step 4

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

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

Step 5

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

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

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

Simplify each term.

Combine $\frac{1}{2}$ and $\sqrt{2 x + 1}$ .

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

Apply the product rule to $\frac{\sqrt{2 x + 1}}{2}$ .

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

Rewrite ${\sqrt{2 x + 1}}^{2}$ as $2 x + 1$ .

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Raise $2$ to the power of $2$ .

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

Cancel the common factor of $2$ .

Factor $2$ out of $4$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify terms.

Combine the numerators over the common denominator.

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

Combine the opposite terms in $2 x + 1 - 1$ .

Subtract $1$ from $1$ .

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

Add $2 x$ and $0$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

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

Set up the composite result function.

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

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

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

Apply the distributive property.

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

Multiply $2$ by $2$ .

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

Cancel the common factor of $2$ .

Move the leading negative in $- \frac{1}{2}$ into the numerator.

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify by adding numbers.

Add $- 1$ and $1$ .

$f (2 x^{2} - \frac{1}{2}) = \frac{1}{2} \cdot \sqrt{4 x^{2} + 0}$

Add $4 x^{2}$ and $0$ .

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

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

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

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

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

Cancel the common factor of $2$ .

Factor $2$ out of $2 x$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

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

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

Find the Inverse f(x)=1/2 square root of 2x+1

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