Find the Inverse square root of 48/x

$\sqrt{\frac{48}{x}}$

Step 1

Interchange the variables.

$x = \sqrt{\frac{48}{y}}$

Step 2

Solve for $y$ .

Rewrite the equation as $\sqrt{\frac{48}{y}} = x$ .

$\sqrt{\frac{48}{y}} = x$

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

${\sqrt{\frac{48}{y}}}^{2} = x^{2}$

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{\frac{48}{y}}$ as ${(\frac{48}{y})}^{\frac{1}{2}}$ .

${({(\frac{48}{y})}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(\frac{48}{y})}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(\frac{48}{y})}^{\frac{1}{2}})}^{2}$ .

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

${(\frac{48}{y})}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(\frac{48}{y})}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(\frac{48}{y})}^{1} = x^{2}$

Simplify.

$\frac{48}{y} = x^{2}$

Solve for $y$ .

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y, 1$

The LCM of one and any expression is the expression.

$y$

Multiply each term in $\frac{48}{y} = x^{2}$ by $y$ to eliminate the fractions.

Multiply each term in $\frac{48}{y} = x^{2}$ by $y$ .

$\frac{48}{y} y = x^{2} y$

Simplify the left side.

Cancel the common factor of $y$ .

Cancel the common factor.

$\frac{48}{y} y = x^{2} y$

Rewrite the expression.

$48 = x^{2} y$

Solve the equation.

Rewrite the equation as $x^{2} y = 48$ .

$x^{2} y = 48$

Divide each term in $x^{2} y = 48$ by $x^{2}$ and simplify.

Divide each term in $x^{2} y = 48$ by $x^{2}$ .

$\frac{x^{2} y}{x^{2}} = \frac{48}{x^{2}}$

Simplify the left side.

Cancel the common factor of $x^{2}$ .

Cancel the common factor.

$\frac{x^{2} y}{x^{2}} = \frac{48}{x^{2}}$

Divide $y$ by $1$ .

$y = \frac{48}{x^{2}}$

Step 3

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

$f^{- 1} (x) = \frac{48}{x^{2}}$

Step 4

Verify if $f^{- 1} (x) = \frac{48}{x^{2}}$ is the inverse of $f (x) = \sqrt{\frac{48}{x}}$ .

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 $f^{- 1} (\sqrt{\frac{48}{x}})$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\sqrt{\frac{48}{x}})}^{2}}$

Simplify the denominator.

Rewrite $\sqrt{\frac{48}{x}}$ as $\frac{\sqrt{48}}{\sqrt{x}}$ .

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{\sqrt{48}}{\sqrt{x}})}^{2}}$

Simplify the numerator.

Rewrite $48$ as $4^{2} \cdot 3$ .

Factor $16$ out of $48$ .

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{\sqrt{16 (3)}}{\sqrt{x}})}^{2}}$

Rewrite $16$ as $4^{2}$ .

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{\sqrt{4^{2} \cdot 3}}{\sqrt{x}})}^{2}}$

Pull terms out from under the radical.

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{3}}{\sqrt{x}})}^{2}}$

Multiply $\frac{4 \sqrt{3}}{\sqrt{x}}$ by $\frac{\sqrt{x}}{\sqrt{x}}$ .

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{3}}{\sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{x}})}^{2}}$

Combine and simplify the denominator.

Multiply $\frac{4 \sqrt{3}}{\sqrt{x}}$ by $\frac{\sqrt{x}}{\sqrt{x}}$ .

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{3} \sqrt{x}}{\sqrt{x} \sqrt{x}})}^{2}}$

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

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{3} \sqrt{x}}{\sqrt{x} \sqrt{x}})}^{2}}$

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

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{3} \sqrt{x}}{\sqrt{x} \sqrt{x}})}^{2}}$

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

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{3} \sqrt{x}}{{\sqrt{x}}^{1 + 1}})}^{2}}$

Add $1$ and $1$ .

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{3} \sqrt{x}}{{\sqrt{x}}^{2}})}^{2}}$

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

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

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{3} \sqrt{x}}{{(x^{\frac{1}{2}})}^{2}})}^{2}}$

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

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{3} \sqrt{x}}{x^{\frac{1}{2} \cdot 2}})}^{2}}$

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

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{3} \sqrt{x}}{x^{\frac{2}{2}}})}^{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{3} \sqrt{x}}{x^{\frac{2}{2}}})}^{2}}$

Rewrite the expression.

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{3} \sqrt{x}}{x})}^{2}}$

Simplify.

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{3} \sqrt{x}}{x})}^{2}}$

Combine using the product rule for radicals.

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{x \cdot 3}}{x})}^{2}}$

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

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{{(\frac{4 \sqrt{3 \cdot x}}{x})}^{2}}$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $\frac{4 \sqrt{3 x}}{x}$ .

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{{(4 \sqrt{3 x})}^{2}}{x^{2}}}$

Apply the product rule to $4 \sqrt{3 x}$ .

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{4^{2} {\sqrt{3 x}}^{2}}{x^{2}}}$

Simplify the numerator.

Raise $4$ to the power of $2$ .

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{16 {\sqrt{3 x}}^{2}}{x^{2}}}$

Rewrite ${\sqrt{3 x}}^{2}$ as $3 x$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 x}$ as ${(3 x)}^{\frac{1}{2}}$ .

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{16 {({(3 x)}^{\frac{1}{2}})}^{2}}{x^{2}}}$

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

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{16 {(3 x)}^{\frac{1}{2} \cdot 2}}{x^{2}}}$

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

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{16 {(3 x)}^{\frac{2}{2}}}{x^{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{16 {(3 x)}^{\frac{2}{2}}}{x^{2}}}$

Rewrite the expression.

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{16 (3 x)}{x^{2}}}$

Simplify.

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{16 (3 x)}{x^{2}}}$

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{16 \cdot (3 x)}{x^{2}}}$

Multiply $16$ by $3$ .

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{48 x}{x^{2}}}$

Cancel the common factor of $x$ and $x^{2}$ .

Factor $x$ out of $48 x$ .

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{x \cdot 48}{x^{2}}}$

Cancel the common factors.

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

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{x \cdot 48}{x \cdot x}}$

Cancel the common factor.

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{x \cdot 48}{x \cdot x}}$

Rewrite the expression.

$f^{- 1} (\sqrt{\frac{48}{x}}) = \frac{48}{\frac{48}{x}}$

Multiply the numerator by the reciprocal of the denominator.

$f^{- 1} (\sqrt{\frac{48}{x}}) = 48 (\frac{x}{48})$

Cancel the common factor of $48$ .

Cancel the common factor.

$f^{- 1} (\sqrt{\frac{48}{x}}) = 48 (\frac{x}{48})$

Rewrite the expression.

$f^{- 1} (\sqrt{\frac{48}{x}}) = x$

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

Set up the composite result function.

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

Evaluate $f (\frac{48}{x^{2}})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{48}{x^{2}}) = \sqrt{\frac{48}{\frac{48}{x^{2}}}}$

Multiply the numerator by the reciprocal of the denominator.

$f (\frac{48}{x^{2}}) = \sqrt{48 (\frac{x^{2}}{48})}$

Cancel the common factor of $48$ .

Cancel the common factor.

$f (\frac{48}{x^{2}}) = \sqrt{48 (\frac{x^{2}}{48})}$

Rewrite the expression.

$f (\frac{48}{x^{2}}) = \sqrt{x^{2}}$

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

$f (\frac{48}{x^{2}}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{48}{x^{2}}$ is the inverse of $f (x) = \sqrt{\frac{48}{x}}$ .

$f^{- 1} (x) = \frac{48}{x^{2}}$

Find the Inverse square root of 48/x

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