Answer:
Third choice
![(√(2x))/(2) + 5, x\ge0](https://img.qammunity.org/2023/formulas/mathematics/college/ejzbn23xmxw28lk7uruabi63nssch2d0k5.png)
Explanation:
Definition
An inverse function is defined as a function, which can reverse into another function.
Thus, if we have a function f(x) which evaluates to k at x = a or in other words f(a) = k, then the inverse of that function indicated by f⁻¹(x) will be such that f⁻¹(k) = a. This is the same as stating that f⁻¹(f(a)) = a
In order the find the in verse of a function f(x) the following are the steps
- Set y = f(x)
- Switch x and y
- Solve for y
Given
, set it equal to y
Swap the variables x and y
==>
![(x - 5)^2 = (x)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/6d1mbeed2gx6z2fg3sdkhwvw60kh0wgeel.png)
![x - 5 = \pm \sqrt(x)/(2)}\\\\x = \pm \sqrt(x)/(2)} + 5\\\\](https://img.qammunity.org/2023/formulas/mathematics/college/bid72wzxs55oa26sbk9h87788cvvqqndts.png)
Solve for y. There are two possible solutions
![y= (√(2) √(x))/(2) + 5 = (√(2x))/(2) + 5](https://img.qammunity.org/2023/formulas/mathematics/college/ubkngce5q9vf6g8r6i595grw5y3e5nrhig.png)
=
![- (√(2x))/(2) + 5](https://img.qammunity.org/2023/formulas/mathematics/college/x2jay9t33j9ewmq0jyszf6h0mfsl4rwenh.png)
The only matching choice are choices 1 and 3. In order to determine if it is the first or the third, plug in 0 and see if the inverse value results in a real number
Substituting 0 for x in
gives us 0 + 5 = 5 which is a valid real
So the domain of the function are all values ≥ 0
Hence choice 3, not choice 1