Find the Inverse y = square root of x+8 , y>=8

$y = \sqrt{x} + 8$ , $y \geq 8$

Write $y = \sqrt{x} + 8$ as a function.

$f (x) = \sqrt{x} + 8$

Find the range of the given function.

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

$[2 \sqrt{2} + 8, \infty)$

Convert $[2 \sqrt{2} + 8, \infty)$ to an inequality.

$y \geq 2 \sqrt{2} + 8$

Find the inverse.

Interchange the variables.

$x = \sqrt{y} + 8$

Solve for $y$ .

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

$\sqrt{y} + 8 = x$

Subtract $8$ from both sides of the equation.

$\sqrt{y} = x - 8$

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

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

Simplify each side of the equation.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

$y = {(x - 8)}^{2}$

Simplify the right side.

Simplify ${(x - 8)}^{2}$ .

Rewrite ${(x - 8)}^{2}$ as $(x - 8) (x - 8)$ .

$y = (x - 8) (x - 8)$

Expand $(x - 8) (x - 8)$ using the FOIL Method.

Apply the distributive property.

$y = x (x - 8) - 8 (x - 8)$

Apply the distributive property.

$y = x \cdot x + x \cdot - 8 - 8 (x - 8)$

Apply the distributive property.

$y = x \cdot x + x \cdot - 8 - 8 x - 8 \cdot - 8$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$y = x^{2} + x \cdot - 8 - 8 x - 8 \cdot - 8$

Move $- 8$ to the left of $x$ .

$y = x^{2} - 8 \cdot x - 8 x - 8 \cdot - 8$

Multiply $- 8$ by $- 8$ .

$y = x^{2} - 8 x - 8 x + 64$

Subtract $8 x$ from $- 8 x$ .

$y = x^{2} - 16 x + 64$

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

$f^{- 1} (x) = x^{2} - 16 x + 64$

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

$f^{- 1} (x) = x^{2} - 16 x + 64, x \geq 2 \sqrt{2} + 8$

Find the Inverse y = square root of x+8 , y>=8

Related Questions

Support LCMGCF.com by disabling your adblocker.