Find the Inverse f(x)=(x-3)^(1/2)

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

Write $f (x) = {(x - 3)}^{\frac{1}{2}}$ as an equation.

$y = {(x - 3)}^{\frac{1}{2}}$

Interchange the variables.

$x = {(y - 3)}^{\frac{1}{2}}$

Solve for $y$ .

Rewrite the equation as ${(y - 3)}^{\frac{1}{2}} = x$ .

${(y - 3)}^{\frac{1}{2}} = x$

Raise each side of the equation to the power of $2$ to eliminate the fractional exponent on the left side.

${({(y - 3)}^{\frac{1}{2}})}^{2} = x^{2}$

Simplify the left side.

Simplify ${({(y - 3)}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(y - 3)}^{\frac{1}{2}})}^{2}$ .

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

${(y - 3)}^{\frac{1}{2} \cdot 2} = x^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(y - 3)}^{\frac{1}{2} \cdot 2} = x^{2}$

Rewrite the expression.

${(y - 3)}^{1} = x^{2}$

Simplify.

$y - 3 = x^{2}$

Add $3$ to both sides of the equation.

$y = x^{2} + 3$

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

$f^{- 1} (x) = x^{2} + 3$

Verify if $f^{- 1} (x) = x^{2} + 3$ is the inverse of $f (x) = {(x - 3)}^{\frac{1}{2}}$ .

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 $f^{- 1} ({(x - 3)}^{\frac{1}{2}})$ by substituting in the value of $f$ into $f^{- 1}$ .

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

Simplify each term.

Multiply the exponents in ${({(x - 3)}^{\frac{1}{2}})}^{2}$ .

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Combine the opposite terms in $x - 3 + 3$ .

Add $- 3$ and $3$ .

$f^{- 1} ({(x - 3)}^{\frac{1}{2}}) = x + 0$

Add $x$ and $0$ .

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

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

Set up the composite result function.

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

Evaluate $f (x^{2} + 3)$ by substituting in the value of $f^{- 1}$ into $f$ .

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

Combine the opposite terms in $(x^{2} + 3) - 3$ .

Subtract $3$ from $3$ .

$f (x^{2} + 3) = {(x^{2} + 0)}^{\frac{1}{2}}$

Add $x^{2}$ and $0$ .

$f (x^{2} + 3) = {(x^{2})}^{\frac{1}{2}}$

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

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

$f (x^{2} + 3) = x^{2 (\frac{1}{2})}$

Cancel the common factor of $2$ .

Cancel the common factor.

$f (x^{2} + 3) = x^{2 (\frac{1}{2})}$

Rewrite the expression.

$f (x^{2} + 3) = x$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = x^{2} + 3$ is the inverse of $f (x) = {(x - 3)}^{\frac{1}{2}}$ .

$f^{- 1} (x) = x^{2} + 3$

Find the Inverse f(x)=(x-3)^(1/2)

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