Find the Inverse ( square root of x)/5

$\frac{\sqrt{x}}{5}$

Interchange the variables.

$x = \frac{\sqrt{y}}{5}$

Solve for $y$ .

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

$\frac{\sqrt{y}}{5} = x$

Multiply both sides of the equation by $5$ .

$5 \frac{\sqrt{y}}{5} = 5 x$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$5 \frac{\sqrt{y}}{5} = 5 x$

Rewrite the expression.

$\sqrt{y} = 5 x$

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

${\sqrt{y}}^{2} = {(5 x)}^{2}$

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(5 x)}^{2}$ .

Apply the product rule to $5 x$ .

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

Raise $5$ to the power of $2$ .

$y = 25 x^{2}$

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

$f^{- 1} (x) = 25 x^{2}$

Verify if $f^{- 1} (x) = 25 x^{2}$ is the inverse of $f (x) = \frac{\sqrt{x}}{5}$ .

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

$f^{- 1} (\frac{\sqrt{x}}{5}) = 25 {(\frac{\sqrt{x}}{5})}^{2}$

Apply the product rule to $\frac{\sqrt{x}}{5}$ .

$f^{- 1} (\frac{\sqrt{x}}{5}) = 25 (\frac{{\sqrt{x}}^{2}}{5^{2}})$

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

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

$f^{- 1} (\frac{\sqrt{x}}{5}) = 25 (\frac{{(x^{\frac{1}{2}})}^{2}}{5^{2}})$

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

$f^{- 1} (\frac{\sqrt{x}}{5}) = 25 (\frac{x^{\frac{1}{2} \cdot 2}}{5^{2}})$

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

$f^{- 1} (\frac{\sqrt{x}}{5}) = 25 (\frac{x^{\frac{2}{2}}}{5^{2}})$

Cancel the common factor of $2$ .

Cancel the common factor.

$f^{- 1} (\frac{\sqrt{x}}{5}) = 25 (\frac{x^{\frac{2}{2}}}{5^{2}})$

Rewrite the expression.

$f^{- 1} (\frac{\sqrt{x}}{5}) = 25 (\frac{x}{5^{2}})$

Simplify.

$f^{- 1} (\frac{\sqrt{x}}{5}) = 25 (\frac{x}{5^{2}})$

Raise $5$ to the power of $2$ .

$f^{- 1} (\frac{\sqrt{x}}{5}) = 25 (\frac{x}{25})$

Cancel the common factor of $25$ .

Cancel the common factor.

$f^{- 1} (\frac{\sqrt{x}}{5}) = 25 (\frac{x}{25})$

Rewrite the expression.

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

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

Set up the composite result function.

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

Evaluate $f (25 x^{2})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (25 x^{2}) = \frac{\sqrt{25 x^{2}}}{5}$

Simplify the numerator.

Rewrite $25 x^{2}$ as ${(5 x)}^{2}$ .

$f (25 x^{2}) = \frac{\sqrt{{(5 x)}^{2}}}{5}$

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

$f (25 x^{2}) = \frac{5 x}{5}$

Cancel the common factor of $5$ .

Cancel the common factor.

$f (25 x^{2}) = \frac{5 x}{5}$

Divide $x$ by $1$ .

$f (25 x^{2}) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = 25 x^{2}$ is the inverse of $f (x) = \frac{\sqrt{x}}{5}$ .

$f^{- 1} (x) = 25 x^{2}$

Find the Inverse ( square root of x)/5

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