Find the Inverse f(x)=1/5(x-6)^2

$f (x) = \frac{1}{5} {(x - 6)}^{2}$

Step 1

Write $f (x) = \frac{1}{5} {(x - 6)}^{2}$ as an equation.

$y = \frac{1}{5} {(x - 6)}^{2}$

Step 2

Interchange the variables.

$x = \frac{1}{5} \cdot {(y - 6)}^{2}$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{1}{5} \cdot {(y - 6)}^{2} = x$ .

$\frac{1}{5} \cdot {(y - 6)}^{2} = x$

Multiply both sides of the equation by $5$ .

$5 (\frac{1}{5} \cdot {(y - 6)}^{2}) = 5 x$

Simplify the left side.

Simplify $5 (\frac{1}{5} \cdot {(y - 6)}^{2})$ .

Combine $\frac{1}{5}$ and ${(y - 6)}^{2}$ .

$5 \frac{{(y - 6)}^{2}}{5} = 5 x$

Cancel the common factor of $5$ .

Cancel the common factor.

$5 \frac{{(y - 6)}^{2}}{5} = 5 x$

Rewrite the expression.

${(y - 6)}^{2} = 5 x$

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

$y - 6 = \pm \sqrt{5 x}$

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.

$y - 6 = \sqrt{5 x}$

Add $6$ to both sides of the equation.

$y = \sqrt{5 x} + 6$

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

$y - 6 = - \sqrt{5 x}$

Add $6$ to both sides of the equation.

$y = - \sqrt{5 x} + 6$

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

$y = \sqrt{5 x} + 6$

$y = - \sqrt{5 x} + 6$

$y = \sqrt{5 x} + 6$

$y = - \sqrt{5 x} + 6$

$y = \sqrt{5 x} + 6$

$y = - \sqrt{5 x} + 6$

Step 4

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

$f^{- 1} (x) = \sqrt{5 x} + 6, - \sqrt{5 x} + 6$

Step 5

Verify if $f^{- 1} (x) = \sqrt{5 x} + 6, - \sqrt{5 x} + 6$ is the inverse of $f (x) = \frac{1}{5} \cdot {(x - 6)}^{2}$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = \frac{1}{5} \cdot {(x - 6)}^{2}$ and $f^{- 1} (x) = \sqrt{5 x} + 6, - \sqrt{5 x} + 6$ and compare them.

Find the range of $f (x) = \frac{1}{5} \cdot {(x - 6)}^{2}$ .

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 $\sqrt{5 x} + 6$ .

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

$5 x \geq 0$

Divide each term in $5 x \geq 0$ by $5$ and simplify.

Divide each term in $5 x \geq 0$ by $5$ .

$\frac{5 x}{5} \geq \frac{0}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} \geq \frac{0}{5}$

Divide $x$ by $1$ .

$x \geq \frac{0}{5}$

Simplify the right side.

Divide $0$ by $5$ .

$x \geq 0$

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

$[0, \infty)$

Find the domain of $f (x) = \frac{1}{5} \cdot {(x - 6)}^{2}$ .

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 $f^{- 1} (x) = \sqrt{5 x} + 6, - \sqrt{5 x} + 6$ is the range of $f (x) = \frac{1}{5} \cdot {(x - 6)}^{2}$ and the range of $f^{- 1} (x) = \sqrt{5 x} + 6, - \sqrt{5 x} + 6$ is the domain of $f (x) = \frac{1}{5} \cdot {(x - 6)}^{2}$ , then $f^{- 1} (x) = \sqrt{5 x} + 6, - \sqrt{5 x} + 6$ is the inverse of $f (x) = \frac{1}{5} \cdot {(x - 6)}^{2}$ .

$f^{- 1} (x) = \sqrt{5 x} + 6, - \sqrt{5 x} + 6$

Step 6

Find the Inverse f(x)=1/5(x-6)^2

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