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

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

Step 1

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

$y = \frac{1}{x^{2}}$

Step 2

Interchange the variables.

$x = \frac{1}{y^{2}}$

Step 3

Solve for $y$ .

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

$\frac{1}{y^{2}} = x$

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$y^{2}, 1$

The LCM of one and any expression is the expression.

$y^{2}$

Multiply each term in $\frac{1}{y^{2}} = x$ by $y^{2}$ to eliminate the fractions.

Multiply each term in $\frac{1}{y^{2}} = x$ by $y^{2}$ .

$\frac{1}{y^{2}} y^{2} = x y^{2}$

Simplify the left side.

Cancel the common factor of $y^{2}$ .

Cancel the common factor.

$\frac{1}{y^{2}} y^{2} = x y^{2}$

Rewrite the expression.

$1 = x y^{2}$

Solve the equation.

Rewrite the equation as $x y^{2} = 1$ .

$x y^{2} = 1$

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

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

$\frac{x y^{2}}{x} = \frac{1}{x}$

Simplify the left side.

Cancel the common factor of $x$ .

Cancel the common factor.

$\frac{x y^{2}}{x} = \frac{1}{x}$

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

$y^{2} = \frac{1}{x}$

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

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

Simplify $\pm \sqrt{\frac{1}{x}}$ .

Rewrite $\sqrt{\frac{1}{x}}$ as $\frac{\sqrt{1}}{\sqrt{x}}$ .

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

Any root of $1$ is $1$ .

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

Multiply $\frac{1}{\sqrt{x}}$ by $\frac{\sqrt{x}}{\sqrt{x}}$ .

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

Combine and simplify the denominator.

Multiply $\frac{1}{\sqrt{x}}$ by $\frac{\sqrt{x}}{\sqrt{x}}$ .

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

Raise $\sqrt{x}$ to the power of $1$ .

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

Raise $\sqrt{x}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt{x}}^{2}$ as $x$ .

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

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

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

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

Step 4

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

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

Step 5

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

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

Find the range of $f (x) = \frac{1}{x^{2}}$ .

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 $\frac{\sqrt{x}}{x}$ .

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

$x \geq 0$

Set the denominator in $\frac{\sqrt{x}}{x}$ equal to $0$ to find where the expression is undefined.

$x = 0$

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

$(0, \infty)$

Find the domain of $f (x) = \frac{1}{x^{2}}$ .

Set the denominator in $\frac{1}{x^{2}}$ equal to $0$ to find where the expression is undefined.

$x^{2} = 0$

Solve for $x$ .

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

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

$(- \infty, 0) \cup (0, \infty)$

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

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

Step 6

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

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