Find the Inverse f(x)=16x^2

$f (x) = 16 x^{2}$

Step 1

Write $f (x) = 16 x^{2}$ as an equation.

$y = 16 x^{2}$

Step 2

Interchange the variables.

$x = 16 y^{2}$

Step 3

Solve for $y$ .

Rewrite the equation as $16 y^{2} = x$ .

$16 y^{2} = x$

Divide each term in $16 y^{2} = x$ by $16$ and simplify.

Divide each term in $16 y^{2} = x$ by $16$ .

$\frac{16 y^{2}}{16} = \frac{x}{16}$

Simplify the left side.

Cancel the common factor of $16$ .

Cancel the common factor.

$\frac{16 y^{2}}{16} = \frac{x}{16}$

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

$y^{2} = \frac{x}{16}$

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

$y = \pm \sqrt{\frac{x}{16}}$

Simplify $\pm \sqrt{\frac{x}{16}}$ .

Rewrite $\frac{x}{16}$ as ${(\frac{1}{4})}^{2} x$ .

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

$y = \pm \sqrt{\frac{1^{2} x}{16}}$

Factor the perfect power $4^{2}$ out of $16$ .

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

Rearrange the fraction $\frac{1^{2} x}{4^{2} \cdot 1}$ .

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

Pull terms out from under the radical.

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

Combine $\frac{1}{4}$ and $\sqrt{x}$ .

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

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.

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

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

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

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

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

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

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

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

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

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

Step 4

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

$f^{- 1} (x) = \frac{\sqrt{x}}{4}, - \frac{\sqrt{x}}{4}$

Step 5

Verify if $f^{- 1} (x) = \frac{\sqrt{x}}{4}, - \frac{\sqrt{x}}{4}$ is the inverse of $f (x) = 16 x^{2}$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = 16 x^{2}$ and $f^{- 1} (x) = \frac{\sqrt{x}}{4}, - \frac{\sqrt{x}}{4}$ and compare them.

Find the range of $f (x) = 16 x^{2}$ .

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

Interval Notation:

$[0, \infty)$

Find the domain of $\frac{\sqrt{x}}{4}$ .

Set the radicand in $\sqrt{x}$ greater than or equal to $0$ to find where the expression is defined.

$x \geq 0$

The domain is all values of $x$ that make the expression defined.

$[0, \infty)$

Find the domain of $f (x) = 16 x^{2}$ .

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

$(- \infty, \infty)$

Since the domain of $f^{- 1} (x) = \frac{\sqrt{x}}{4}, - \frac{\sqrt{x}}{4}$ is the range of $f (x) = 16 x^{2}$ and the range of $f^{- 1} (x) = \frac{\sqrt{x}}{4}, - \frac{\sqrt{x}}{4}$ is the domain of $f (x) = 16 x^{2}$ , then $f^{- 1} (x) = \frac{\sqrt{x}}{4}, - \frac{\sqrt{x}}{4}$ is the inverse of $f (x) = 16 x^{2}$ .

$f^{- 1} (x) = \frac{\sqrt{x}}{4}, - \frac{\sqrt{x}}{4}$

Step 6

Find the Inverse f(x)=16x^2

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