Find the Inverse f(x)=4x^2 , x<=0

$f (x) = 4 x^{2}$ , $x \leq 0$

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.

$x = 4 y^{2}$

Solve for $y$ .

Rewrite the equation as $4 y^{2} = x$ .

$4 y^{2} = x$

Divide each term in $4 y^{2} = x$ by $4$ and simplify.

Divide each term in $4 y^{2} = x$ by $4$ .

$\frac{4 y^{2}}{4} = \frac{x}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 y^{2}}{4} = \frac{x}{4}$

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

$y^{2} = \frac{x}{4}$

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

$y = \pm \sqrt{\frac{x}{4}}$

Simplify $\pm \sqrt{\frac{x}{4}}$ .

Rewrite $\frac{x}{4}$ as ${(\frac{1}{2})}^{2} x$ .

Factor the perfect power $1^{2}$ out of $x$ .

$y = \pm \sqrt{\frac{1^{2} x}{4}}$

Factor the perfect power $2^{2}$ out of $4$ .

$y = \pm \sqrt{\frac{1^{2} x}{2^{2} \cdot 1}}$

Rearrange the fraction $\frac{1^{2} x}{2^{2} \cdot 1}$ .

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

Pull terms out from under the radical.

$y = \pm \frac{1}{2} \sqrt{x}$

Combine $\frac{1}{2}$ and $\sqrt{x}$ .

$y = \pm \frac{\sqrt{x}}{2}$

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.

$y = \frac{\sqrt{x}}{2}$

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

$y = - \frac{\sqrt{x}}{2}$

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

$y = \frac{\sqrt{x}}{2}$

$y = - \frac{\sqrt{x}}{2}$

$y = \frac{\sqrt{x}}{2}$

$y = - \frac{\sqrt{x}}{2}$

$y = \frac{\sqrt{x}}{2}$

$y = - \frac{\sqrt{x}}{2}$

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

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

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

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

Find the Inverse f(x)=4x^2 , x<=0

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