Find the Inverse f(x)=x^2+1

$f (x) = x^{2} + 1$

Write $f (x) = x^{2} + 1$ as an equation.

$y = x^{2} + 1$

Interchange the variables.

$x = y^{2} + 1$

Solve for $y$ .

Rewrite the equation as $y^{2} + 1 = x$ .

$y^{2} + 1 = x$

Subtract $1$ from both sides of the equation.

$y^{2} = x - 1$

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

$y = \pm \sqrt{x - 1}$

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.

$y = \sqrt{x - 1}$

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

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

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

$y = \sqrt{x - 1}$

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

$y = \sqrt{x - 1}$

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

$y = \sqrt{x - 1}$

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

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

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

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

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

Find the range of $f (x) = x^{2} + 1$ .

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

Interval Notation:

$[1, \infty)$

Find the domain of $\sqrt{x - 1}$ .

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

$x - 1 \geq 0$

Add $1$ to both sides of the inequality.

$x \geq 1$

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

$[1, \infty)$

Find the domain of $f (x) = x^{2} + 1$ .

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

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

Find the Inverse f(x)=x^2+1

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