Certainly! Let's break down the problem step by step.
### Step 1: Define the functions
We are given two functions:
[tex]\[ f(x) = 3x \][/tex]
[tex]\[ g(x) = \sqrt{x - 1} \][/tex]
### Step 2: Combine the functions
We need to find the formula for [tex]\((f + g)(x)\)[/tex]. This is done by adding the two functions together:
[tex]\[ (f + g)(x) = f(x) + g(x) \][/tex]
[tex]\[ (f + g)(x) = 3x + \sqrt{x - 1} \][/tex]
So, the combined function is:
[tex]\[ (f + g)(x) = 3x + \sqrt{x - 1} \][/tex]
This is the simplified formula for the combined function.
### Step 3: Determine the domain
To find the domain of [tex]\((f + g)(x)\)[/tex], we need to consider the domains of both [tex]\(f(x)\)[/tex] and [tex]\(g(x)\)[/tex].
1. Domain of [tex]\(f(x) = 3x\)[/tex]:
- The function [tex]\(f(x)\)[/tex] is a linear function and is defined for all real numbers. Therefore, the domain of [tex]\(f(x)\)[/tex] is all real numbers, [tex]\( \mathbb{R} \)[/tex].
2. Domain of [tex]\(g(x) = \sqrt{x - 1}\)[/tex]:
- The square root function [tex]\(\sqrt{x - 1}\)[/tex] is defined only when the expression inside the square root is non-negative. Hence, [tex]\( x - 1 \geq 0 \)[/tex].
- Solving [tex]\( x - 1 \geq 0 \)[/tex], we get [tex]\( x \geq 1 \)[/tex].
Since [tex]\(f(x)\)[/tex] is defined for all [tex]\(x\)[/tex] and [tex]\(g(x)\)[/tex] is defined for [tex]\(x \geq 1\)[/tex], the domain of the combined function [tex]\((f + g)(x)\)[/tex] is where both functions are defined.
Therefore, the domain of [tex]\((f + g)(x)\)[/tex] is:
[tex]\[ x \geq 1 \][/tex]
### Final Answer
1. The formula for [tex]\((f + g)(x)\)[/tex] is:
[tex]\[ (f + g)(x) = 3x + \sqrt{x - 1} \][/tex]
2. The domain for [tex]\((f + g)(x)\)[/tex] is:
[tex]\[ x \geq 1 \][/tex]
So the complete solution to the problem is:
[tex]\[ (f + g)(x) = 3x + \sqrt{x - 1} \text{, Domain: } x \geq 1 \][/tex]
If you have any more questions, feel free to ask!
