Find the Inverse f(x)=25/(x^2)

$f (x) = \frac{25}{x^{2}}$

Replace $f (x)$ with $y$ .

$y = \frac{25}{x^{2}}$

Interchange the variables.

$x = \frac{25}{y^{2}}$

Solve for $y$ .

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Rewrite the equation as $\frac{25}{y^{2}} = x$ .

$\frac{25}{y^{2}} = x$

Solve for $y$ .

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Multiply each term by $y^{2}$ and simplify.

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Multiply each term in $\frac{25}{y^{2}} = x$ by $y^{2}$ .

$\frac{25}{y^{2}} \cdot y^{2} = x \cdot y^{2}$

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

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Cancel the common factor.

$\frac{25}{y^{2}} \cdot y^{2} = x \cdot y^{2}$

Rewrite the expression.

$25 = x \cdot y^{2}$

$25 = x y^{2}$

Rewrite the equation as $x y^{2} = 25$ .

$x y^{2} = 25$

Divide each term by $x$ and simplify.

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Divide each term in $x y^{2} = 25$ by $x$ .

$\frac{x y^{2}}{x} = \frac{25}{x}$

Cancel the common factor of $x$ .

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Cancel the common factor.

$\frac{x y^{2}}{x} = \frac{25}{x}$

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

$y^{2} = \frac{25}{x}$

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

$y = \pm \sqrt{\frac{25}{x}}$

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

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Simplify the right side of the equation.

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Rewrite $\sqrt{\frac{25}{x}}$ as $\frac{\sqrt{25}}{\sqrt{x}}$ .

$y = \pm \frac{\sqrt{25}}{\sqrt{x}}$

Simplify the numerator.

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Rewrite $25$ as $5^{2}$ .

$y = \pm \frac{\sqrt{5^{2}}}{\sqrt{x}}$

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

$y = \pm \frac{5}{\sqrt{x}}$

Multiply $\frac{5}{\sqrt{x}}$ by $\frac{\sqrt{x}}{\sqrt{x}}$ .

$y = \pm \frac{5}{\sqrt{x}} \cdot \frac{\sqrt{x}}{\sqrt{x}}$

Combine and simplify the denominator.

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Multiply $\frac{5}{\sqrt{x}}$ and $\frac{\sqrt{x}}{\sqrt{x}}$ .

$y = \pm \frac{5 \sqrt{x}}{\sqrt{x} \sqrt{x}}$

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

$y = \pm \frac{5 \sqrt{x}}{\sqrt{x} \sqrt{x}}$

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

$y = \pm \frac{5 \sqrt{x}}{\sqrt{x} \sqrt{x}}$

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

$y = \pm \frac{5 \sqrt{x}}{{\sqrt{x}}^{1 + 1}}$

Add $1$ and $1$ .

$y = \pm \frac{5 \sqrt{x}}{{\sqrt{x}}^{2}}$

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

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

$y = \pm \frac{5 \sqrt{x}}{{(x^{\frac{1}{2}})}^{2}}$

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

$y = \pm \frac{5 \sqrt{x}}{x^{\frac{1}{2} \cdot 2}}$

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

$y = \pm \frac{5 \sqrt{x}}{x^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$y = \pm \frac{5 \sqrt{x}}{x^{\frac{2}{2}}}$

Divide $1$ by $1$ .

$y = \pm \frac{5 \sqrt{x}}{x}$

Simplify.

$y = \pm \frac{5 \sqrt{x}}{x}$

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

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

$y = \frac{5 \sqrt{x}}{x}$

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

$y = - \frac{5 \sqrt{x}}{x}$

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

$y = \frac{5 \sqrt{x}}{x}$

$y = - \frac{5 \sqrt{x}}{x}$

$y = \frac{5 \sqrt{x}}{x}$

$y = - \frac{5 \sqrt{x}}{x}$

$y = \frac{5 \sqrt{x}}{x}$

$y = - \frac{5 \sqrt{x}}{x}$

$y = \frac{5 \sqrt{x}}{x}$

$y = - \frac{5 \sqrt{x}}{x}$

$y = \frac{5 \sqrt{x}}{x}$

$y = - \frac{5 \sqrt{x}}{x}$

Solve for $y$ and replace with $f^{- 1} (x)$ .

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

$f^{- 1} (x) = \frac{5 \sqrt{x}}{x}, - \frac{5 \sqrt{x}}{x}$

Set up the composite result function.

$g (f (x))$

Evaluate $g (f (x))$ by substituting in the value of $f$ into $g$ .

$g (\frac{25}{x^{2}}) = \frac{5 \sqrt{\frac{25}{x^{2}}}}{\frac{25}{x^{2}}}$

Multiply the numerator by the reciprocal of the denominator.

$g (\frac{25}{x^{2}}) = 5 \sqrt{\frac{25}{x^{2}}} (\frac{x^{2}}{25})$

Simplify terms.

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Cancel the common factor of $5$ .

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Factor $5$ out of $5 \sqrt{\frac{25}{x^{2}}}$ .

$g (\frac{25}{x^{2}}) = 5 (\sqrt{\frac{25}{x^{2}}}) (\frac{x^{2}}{25})$

Factor $5$ out of $25$ .

$g (\frac{25}{x^{2}}) = 5 (\sqrt{\frac{25}{x^{2}}}) (\frac{x^{2}}{5 (5)})$

Cancel the common factor.

$g (\frac{25}{x^{2}}) = 5 \sqrt{\frac{25}{x^{2}}} (\frac{x^{2}}{5 \cdot 5})$

Rewrite the expression.

$g (\frac{25}{x^{2}}) = \sqrt{\frac{25}{x^{2}}} (\frac{x^{2}}{5})$

Combine $\sqrt{\frac{25}{x^{2}}}$ and $\frac{x^{2}}{5}$ .

$g (\frac{25}{x^{2}}) = \frac{\sqrt{\frac{25}{x^{2}}} x^{2}}{5}$

Simplify the numerator.

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Rewrite $25$ as $5^{2}$ .

$g (\frac{25}{x^{2}}) = \frac{\sqrt{\frac{5^{2}}{x^{2}}} x^{2}}{5}$

Rewrite $\frac{5^{2}}{x^{2}}$ as ${(\frac{5}{x})}^{2}$ .

$g (\frac{25}{x^{2}}) = \frac{\sqrt{{(\frac{5}{x})}^{2}} x^{2}}{5}$

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

$g (\frac{25}{x^{2}}) = \frac{\frac{5}{x} \cdot x^{2}}{5}$

Combine $\frac{5}{x}$ and $x^{2}$ .

$g (\frac{25}{x^{2}}) = \frac{\frac{5 x^{2}}{x}}{5}$

Reduce the expression by cancelling the common factors.

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Reduce the expression $\frac{5 x^{2}}{x}$ by cancelling the common factors.

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Factor $x$ out of $5 x^{2}$ .

$g (\frac{25}{x^{2}}) = \frac{\frac{x (5 x)}{x}}{5}$

Raise $x$ to the power of $1$ .

$g (\frac{25}{x^{2}}) = \frac{\frac{x (5 x)}{x}}{5}$

Factor $x$ out of $x^{1}$ .

$g (\frac{25}{x^{2}}) = \frac{\frac{x (5 x)}{x \cdot 1}}{5}$

Cancel the common factor.

$g (\frac{25}{x^{2}}) = \frac{\frac{x (5 x)}{x \cdot 1}}{5}$

Rewrite the expression.

$g (\frac{25}{x^{2}}) = \frac{\frac{5 x}{1}}{5}$

Divide $5 x$ by $1$ .

$g (\frac{25}{x^{2}}) = \frac{5 x}{5}$

Cancel the common factor of $5$ .

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Cancel the common factor.

$g (\frac{25}{x^{2}}) = \frac{5 x}{5}$

Divide $x$ by $1$ .

$g (\frac{25}{x^{2}}) = x$

Since $g (f (x)) = x$ , $f^{- 1} (x) = \frac{5 \sqrt{x}}{x}, - \frac{5 \sqrt{x}}{x}$ is the inverse of $f (x) = \frac{25}{x^{2}}$ .

$f^{- 1} (x) = \frac{5 \sqrt{x}}{x}, - \frac{5 \sqrt{x}}{x}$

Find the Inverse f(x)=25/(x^2)

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