To find the inverse of the function [tex]\( f(x) = x^2 + 1 \)[/tex], where the domain of [tex]\( f(x) \)[/tex] is [tex]\([0, \infty)\)[/tex], we can follow these steps:
### Step-by-Step Solution:
1. Set up the equation to solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
We start by setting [tex]\( f(x) \)[/tex] equal to [tex]\( y \)[/tex]:
[tex]\[
y = x^2 + 1
\][/tex]
2. Isolate [tex]\( x^2 \)[/tex]:
To solve for [tex]\( x \)[/tex], we first isolate [tex]\( x^2 \)[/tex] by subtracting 1 from both sides of the equation:
[tex]\[
y - 1 = x^2
\][/tex]
3. Solve for [tex]\( x \)[/tex]:
Next, take the square root of both sides to solve for [tex]\( x \)[/tex]. Since the domain of the original function [tex]\( f(x) \)[/tex] is [tex]\([0, \infty)\)[/tex], [tex]\( x \)[/tex] must be non-negative. Hence, we consider only the positive square root:
[tex]\[
x = \sqrt{y - 1}
\][/tex]
4. Express the inverse function [tex]\( f^{-1}(x) \)[/tex]:
The inverse function [tex]\( f^{-1}(x) \)[/tex] can be written as:
[tex]\[
f^{-1}(x) = \sqrt{x - 1}
\][/tex]
5. Determine the domain of [tex]\( f^{-1}(x) \)[/tex]:
The domain of the inverse function [tex]\( f^{-1}(x) \)[/tex] is determined by the range of the original function [tex]\( f(x) \)[/tex]. Given [tex]\( f(x) = x^2 + 1 \)[/tex] and the domain [tex]\( [0, \infty) \)[/tex]:
- The minimum value of [tex]\( f(x) \)[/tex] is [tex]\( f(0) = 1 \)[/tex].
- As [tex]\( x \)[/tex] increases without bound, [tex]\( f(x) \)[/tex] also increases without bound.
Thus, the range of [tex]\( f(x) \)[/tex] is [tex]\([1, \infty)\)[/tex], which becomes the domain of [tex]\( f^{-1}(x) \)[/tex].
### Conclusion:
- The inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[
f^{-1}(x) = \sqrt{x - 1}
\][/tex]
- The domain of [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[
[1, \infty)
\][/tex]
