Find the Inverse f(x)=(x^4)/7+27

$f (x) = \frac{x^{4}}{7} + 27$

Step 1

Write $f (x) = \frac{x^{4}}{7} + 27$ as an equation.

$y = \frac{x^{4}}{7} + 27$

Step 2

Interchange the variables.

$x = \frac{y^{4}}{7} + 27$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{y^{4}}{7} + 27 = x$ .

$\frac{y^{4}}{7} + 27 = x$

Subtract $27$ from both sides of the equation.

$\frac{y^{4}}{7} = x - 27$

Multiply both sides of the equation by $7$ .

$7 \frac{y^{4}}{7} = 7 (x - 27)$

Simplify both sides of the equation.

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$7 \frac{y^{4}}{7} = 7 (x - 27)$

Rewrite the expression.

$y^{4} = 7 (x - 27)$

Simplify the right side.

Simplify $7 (x - 27)$ .

Apply the distributive property.

$y^{4} = 7 x + 7 \cdot - 27$

Multiply $7$ by $- 27$ .

$y^{4} = 7 x - 189$

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

$y = \pm \sqrt[4]{7 x - 189}$

Factor $7$ out of $7 x - 189$ .

Factor $7$ out of $7 x$ .

$y = \pm \sqrt[4]{7 (x) - 189}$

Factor $7$ out of $- 189$ .

$y = \pm \sqrt[4]{7 x + 7 \cdot - 27}$

Factor $7$ out of $7 x + 7 \cdot - 27$ .

$y = \pm \sqrt[4]{7 (x - 27)}$

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.

$y = \sqrt[4]{7 (x - 27)}$

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

$y = - \sqrt[4]{7 (x - 27)}$

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

$y = \sqrt[4]{7 (x - 27)}$

$y = - \sqrt[4]{7 (x - 27)}$

$y = \sqrt[4]{7 (x - 27)}$

$y = - \sqrt[4]{7 (x - 27)}$

$y = \sqrt[4]{7 (x - 27)}$

$y = - \sqrt[4]{7 (x - 27)}$

Step 4

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

$f^{- 1} (x) = \sqrt[4]{7 (x - 27)}, - \sqrt[4]{7 (x - 27)}$

Step 5

Verify if $f^{- 1} (x) = \sqrt[4]{7 (x - 27)}, - \sqrt[4]{7 (x - 27)}$ is the inverse of $f (x) = \frac{x^{4}}{7} + 27$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = \frac{x^{4}}{7} + 27$ and $f^{- 1} (x) = \sqrt[4]{7 (x - 27)}, - \sqrt[4]{7 (x - 27)}$ and compare them.

Find the range of $f (x) = \frac{x^{4}}{7} + 27$ .

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

Interval Notation:

$[27, \infty)$

Find the domain of $\sqrt[4]{7 (x - 27)}$ .

Set the radicand in $\sqrt[4]{7 (x - 27)}$ greater than or equal to $0$ to find where the expression is defined.

$7 (x - 27) \geq 0$

Solve for $x$ .

Divide each term in $7 (x - 27) \geq 0$ by $7$ and simplify.

Divide each term in $7 (x - 27) \geq 0$ by $7$ .

$\frac{7 (x - 27)}{7} \geq \frac{0}{7}$

Simplify the left side.

Cancel the common factor of $7$ .

Cancel the common factor.

$\frac{7 (x - 27)}{7} \geq \frac{0}{7}$

Divide $x - 27$ by $1$ .

$x - 27 \geq \frac{0}{7}$

Simplify the right side.

Divide $0$ by $7$ .

$x - 27 \geq 0$

Add $27$ to both sides of the inequality.

$x \geq 27$

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

$[27, \infty)$

Find the domain of $f (x) = \frac{x^{4}}{7} + 27$ .

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 $f^{- 1} (x) = \sqrt[4]{7 (x - 27)}, - \sqrt[4]{7 (x - 27)}$ is the range of $f (x) = \frac{x^{4}}{7} + 27$ and the range of $f^{- 1} (x) = \sqrt[4]{7 (x - 27)}, - \sqrt[4]{7 (x - 27)}$ is the domain of $f (x) = \frac{x^{4}}{7} + 27$ , then $f^{- 1} (x) = \sqrt[4]{7 (x - 27)}, - \sqrt[4]{7 (x - 27)}$ is the inverse of $f (x) = \frac{x^{4}}{7} + 27$ .

$f^{- 1} (x) = \sqrt[4]{7 (x - 27)}, - \sqrt[4]{7 (x - 27)}$

Step 6

Find the Inverse f(x)=(x^4)/7+27

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