Find the Inverse f(x) = square root of -1+x , x>=1

$f (x) = \sqrt{- 1 + x}$ , $x \geq 1$

Step 1

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.

$y \geq 0$

Step 2

Find the inverse.

Interchange the variables.

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

Solve for $y$ .

Rewrite the equation as $\sqrt{- 1 + y} = x$ .

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

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

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

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Add $1$ to both sides of the equation.

$y = x^{2} + 1$

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

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

Step 3

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

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

Step 4