Find the Inverse f(x)=8 square root of x ; for x>=0

$f (x) = 8 \sqrt{x}$ ; for $x \geq 0$

Find the range of the given function.

The range is the set of all valid $y$ values. Use the graph to find the range.

$[0, \infty)$

Convert $[0, \infty)$ to an inequality.

$y \geq 0$

Find the inverse.

Interchange the variables.

$x = 8 \sqrt{y}$

Solve for $y$ .

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

$8 \sqrt{y} = x$

Divide each term in $8 \sqrt{y} = x$ by $8$ and simplify.

Divide each term in $8 \sqrt{y} = x$ by $8$ .

$\frac{8 \sqrt{y}}{8} = \frac{x}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 \sqrt{y}}{8} = \frac{x}{8}$

Divide $\sqrt{y}$ by $1$ .

$\sqrt{y} = \frac{x}{8}$

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

${\sqrt{y}}^{2} = {(\frac{x}{8})}^{2}$

Simplify each side of the equation.

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

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

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(\frac{x}{8})}^{2}$ .

Apply the product rule to $\frac{x}{8}$ .

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

Raise $8$ to the power of $2$ .

$y = \frac{x^{2}}{64}$

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

$f^{- 1} (x) = \frac{x^{2}}{64}$

Find the inverse using the domain and the range of the original function.

$f^{- 1} (x) = \frac{x^{2}}{64}, x \geq 0$

Find the Inverse f(x)=8 square root of x ; for x>=0

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