Find the Inverse f(x)=4x^(2/5)

$f (x) = 4 x^{\frac{2}{5}}$

Write $f (x) = 4 x^{\frac{2}{5}}$ as an equation.

$y = 4 x^{\frac{2}{5}}$

Interchange the variables.

$x = 4 y^{\frac{2}{5}}$

Solve for $y$ .

Rewrite the equation as $4 y^{\frac{2}{5}} = x$ .

$4 y^{\frac{2}{5}} = x$

Divide each term in $4 y^{\frac{2}{5}} = x$ by $4$ and simplify.

Divide each term in $4 y^{\frac{2}{5}} = x$ by $4$ .

$\frac{4 y^{\frac{2}{5}}}{4} = \frac{x}{4}$

Simplify the left side.

Cancel the common factor.

$\frac{4 y^{\frac{2}{5}}}{4} = \frac{x}{4}$

Divide $y^{\frac{2}{5}}$ by $1$ .

$y^{\frac{2}{5}} = \frac{x}{4}$

Raise each side of the equation to the power of $\frac{5}{2}$ to eliminate the fractional exponent on the left side.

${(y^{\frac{2}{5}})}^{\frac{5}{2}} = \pm {(\frac{x}{4})}^{\frac{5}{2}}$

Simplify the exponent.

Simplify the left side.

Simplify ${(y^{\frac{2}{5}})}^{\frac{5}{2}}$ .

Multiply the exponents in ${(y^{\frac{2}{5}})}^{\frac{5}{2}}$ .

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

$y^{\frac{2}{5} \cdot \frac{5}{2}} = \pm {(\frac{x}{4})}^{\frac{5}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$y^{\frac{2}{5} \cdot \frac{5}{2}} = \pm {(\frac{x}{4})}^{\frac{5}{2}}$

Rewrite the expression.

$y^{\frac{1}{5} \cdot 5} = \pm {(\frac{x}{4})}^{\frac{5}{2}}$

Cancel the common factor of $5$ .

Cancel the common factor.

$y^{\frac{1}{5} \cdot 5} = \pm {(\frac{x}{4})}^{\frac{5}{2}}$

Rewrite the expression.

$y^{1} = \pm {(\frac{x}{4})}^{\frac{5}{2}}$

Simplify.

$y = \pm {(\frac{x}{4})}^{\frac{5}{2}}$

Simplify the right side.

Simplify $\pm {(\frac{x}{4})}^{\frac{5}{2}}$ .

Apply the product rule to $\frac{x}{4}$ .

$y = \pm \frac{x^{\frac{5}{2}}}{4^{\frac{5}{2}}}$

Simplify the denominator.

Rewrite $4$ as $2^{2}$ .

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

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

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Raise $2$ to the power of $5$ .

$y = \pm \frac{x^{\frac{5}{2}}}{32}$

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.

$y = \frac{x^{\frac{5}{2}}}{32}$

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

$y = - \frac{x^{\frac{5}{2}}}{32}$

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

$y = \frac{x^{\frac{5}{2}}}{32}$

$y = - \frac{x^{\frac{5}{2}}}{32}$

$y = \frac{x^{\frac{5}{2}}}{32}$

$y = - \frac{x^{\frac{5}{2}}}{32}$

$y = \frac{x^{\frac{5}{2}}}{32}$

$y = - \frac{x^{\frac{5}{2}}}{32}$

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

$f^{- 1} (x) = \frac{x^{\frac{5}{2}}}{32}, - \frac{x^{\frac{5}{2}}}{32}$

Verify if $f^{- 1} (x) = \frac{x^{\frac{5}{2}}}{32}, - \frac{x^{\frac{5}{2}}}{32}$ is the inverse of $f (x) = 4 x^{\frac{2}{5}}$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = 4 x^{\frac{2}{5}}$ and $f^{- 1} (x) = \frac{x^{\frac{5}{2}}}{32}, - \frac{x^{\frac{5}{2}}}{32}$ and compare them.

Find the range of $f (x) = 4 x^{\frac{2}{5}}$ .

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 $\frac{x^{\frac{5}{2}}}{32}$ .

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$\frac{\sqrt{x^{5}}}{32}$

Set the radicand in $\sqrt{x^{5}}$ greater than or equal to $0$ to find where the expression is defined.

$x^{5} \geq 0$

Solve for $x$ .

Take the 5th root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt[5]{x^{5}} \geq \sqrt[5]{0}$

Simplify the equation.

Simplify the left side.

Pull terms out from under the radical.

$x \geq \sqrt[5]{0}$

Simplify the right side.

Simplify $\sqrt[5]{0}$ .

Rewrite $0$ as $0^{5}$ .

$x \geq \sqrt[5]{0^{5}}$

Pull terms out from under the radical.

$x \geq 0$

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

$[0, \infty)$

Find the domain of $f (x) = 4 x^{\frac{2}{5}}$ .

Apply the rule $x^{\frac{m}{n}} = \sqrt[n]{x^{m}}$ to rewrite the exponentiation as a radical.

$4 \sqrt[5]{x^{2}}$

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.

$(- \infty, \infty)$

Find the range of $\frac{x^{\frac{5}{2}}}{32}$ .

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

Interval Notation:

$(- \infty, \infty)$

Since the range of $f^{- 1} (x) = \frac{x^{\frac{5}{2}}}{32}, - \frac{x^{\frac{5}{2}}}{32}$ is not equal to the domain of $f (x) = 4 x^{\frac{2}{5}}$ , then $f^{- 1} (x) = \frac{x^{\frac{5}{2}}}{32}, - \frac{x^{\frac{5}{2}}}{32}$ is not an inverse of $f (x) = 4 x^{\frac{2}{5}}$ .

There is no inverse

Find the Inverse f(x)=4x^(2/5)

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