Question

find the inverse of
f
(
x
)
=
√
x
2
+
13

Answer 1

To find the inverse of the function \( f(x) = \sqrt{x^2 + 13} \), we can follow these steps:

1. **Replace \( f(x) \) with \( y \)**:
\[
y = \sqrt{x^2 + 13}
\]

2. **Solve for \( x \) in terms of \( y \)**:
First, we square both sides to eliminate the square root:
\[
y^2 = x^2 + 13
\]

Next, isolate \( x^2 \):
\[
x^2 = y^2 - 13
\]

Now, take the square root of both sides. We get:
\[
x = \pm \sqrt{y^2 - 13}
\]

Since the original function \( f(x) \) is defined for all real \( x \) and produces non-negative outputs (due to the square root), we will only consider the positive branch of the square root for the inverse:
\[
x = \sqrt{y^2 - 13}
\]

3. **Express the inverse function**:
We now replace \( y \) with \( x \) to express the inverse function:
\[
f^{-1}(x) = \sqrt{x^2 - 13}
\]

4. **Specify the domain and range**:
The original function \( f(x) = \sqrt{x^2 + 13} \) has a range of \( [\sqrt{13}, \infty) \) since \( \sqrt{x^2 + 13} \) is always greater than or equal to \( \sqrt{13} \). Therefore, the domain of the inverse function will be \( [\sqrt{13}, \infty) \).

Thus, the inverse of the function \( f(x) = \sqrt{x^2 + 13} \) is:

\[
f^{-1}(x) = \sqrt{x^2 - 13} \quad \text{for } x \geq \sqrt{13}
\]