Find the Inverse f(x)=(x^2-1)/10

$f (x) = \frac{x^{2} - 1}{10}$

Step 1

Write $f (x) = \frac{x^{2} - 1}{10}$ as an equation.

$y = \frac{x^{2} - 1}{10}$

Step 2

Interchange the variables.

$x = \frac{y^{2} - 1}{10}$

Step 3

Solve for $y$ .

Rewrite the equation as $\frac{y^{2} - 1}{10} = x$ .

$\frac{y^{2} - 1}{10} = x$

Multiply both sides of the equation by $10$ .

$10 \frac{y^{2} - 1}{10} = 10 x$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$10 \frac{y^{2} - 1}{10} = 10 x$

Rewrite the expression.

$y^{2} - 1 = 10 x$

Add $1$ to both sides of the equation.

$y^{2} = 10 x + 1$

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

$y = \pm \sqrt{10 x + 1}$

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.

$y = \sqrt{10 x + 1}$

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

$y = - \sqrt{10 x + 1}$

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

$y = \sqrt{10 x + 1}$

$y = - \sqrt{10 x + 1}$

$y = \sqrt{10 x + 1}$

$y = - \sqrt{10 x + 1}$

$y = \sqrt{10 x + 1}$

$y = - \sqrt{10 x + 1}$

Step 4

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

$f^{- 1} (x) = \sqrt{10 x + 1}, - \sqrt{10 x + 1}$

Step 5

Verify if $f^{- 1} (x) = \sqrt{10 x + 1}, - \sqrt{10 x + 1}$ is the inverse of $f (x) = \frac{x^{2} - 1}{10}$ .

The domain of the inverse is the range of the original function and vice versa. Find the domain and the range of $f (x) = \frac{x^{2} - 1}{10}$ and $f^{- 1} (x) = \sqrt{10 x + 1}, - \sqrt{10 x + 1}$ and compare them.

Find the range of $f (x) = \frac{x^{2} - 1}{10}$ .

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 $\sqrt{10 x + 1}$ .

Set the radicand in $\sqrt{10 x + 1}$ greater than or equal to $0$ to find where the expression is defined.

$10 x + 1 \geq 0$

Solve for $x$ .

Subtract $1$ from both sides of the inequality.

$10 x \geq - 1$

Divide each term in $10 x \geq - 1$ by $10$ and simplify.

Divide each term in $10 x \geq - 1$ by $10$ .

$\frac{10 x}{10} \geq \frac{- 1}{10}$

Simplify the left side.

Cancel the common factor of $10$ .

Cancel the common factor.

$\frac{10 x}{10} \geq \frac{- 1}{10}$

Divide $x$ by $1$ .

$x \geq \frac{- 1}{10}$

Simplify the right side.

Move the negative in front of the fraction.

$x \geq - \frac{1}{10}$

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

$[- \frac{1}{10}, \infty)$

Since the domain of $f^{- 1} (x) = \sqrt{10 x + 1}, - \sqrt{10 x + 1}$ is not equal to the range of $f (x) = \frac{x^{2} - 1}{10}$ , then $f^{- 1} (x) = \sqrt{10 x + 1}, - \sqrt{10 x + 1}$ is not an inverse of $f (x) = \frac{x^{2} - 1}{10}$ .

There is no inverse

Step 6

Find the Inverse f(x)=(x^2-1)/10

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