Find the Inverse f(x)=3-x^2 , x>=0

$f (x) = 3 - x^{2}$ , $x \geq 0$

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.

$(- \infty, 3]$

Convert $(- \infty, 3]$ to an inequality.

$y \leq 3$

Step 2

Find the inverse.

Interchange the variables.

$x = 3 - y^{2}$

Solve for $y$ .

Rewrite the equation as $3 - y^{2} = x$ .

$3 - y^{2} = x$

Subtract $3$ from both sides of the equation.

$- y^{2} = x - 3$

Divide each term in $- y^{2} = x - 3$ by $- 1$ and simplify.

Divide each term in $- y^{2} = x - 3$ by $- 1$ .

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $y^{2}$ by $1$ .

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

Simplify the right side.

Simplify each term.

Move the negative one from the denominator of $\frac{x}{- 1}$ .

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

Rewrite $- 1 \cdot x$ as $- x$ .

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

Divide $- 3$ by $- 1$ .

$y^{2} = - x + 3$

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{- x + 3}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$y = \sqrt{- x + 3}$

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \sqrt{- x + 3}$

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt{- x + 3}$

$y = - \sqrt{- x + 3}$

$y = \sqrt{- x + 3}$

$y = - \sqrt{- x + 3}$

$y = \sqrt{- x + 3}$

$y = - \sqrt{- x + 3}$

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

$f^{- 1} (x) = \sqrt{- x + 3}, - \sqrt{- x + 3}$

Step 3

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

$f^{- 1} (x) = \sqrt{- x + 3}, x \leq 3$

Step 4

Find the Inverse f(x)=3-x^2 , x>=0

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