Combining functions: Advanced
Suppose that the functions $f$ and $g$ are defined as follows.
\[
\begin{array}{l}
f(x)=\sqrt{4 x-5} \\
g(x)=2 x-1
\end{array}
\]

Find $f+g$ and $f \cdot g$. Then, give their domains using interval notation.
\[
(f+g)(x)=\square
\]

Domain of $f+g: \square$
\[
(f \cdot g)(x)=\square
\]

Domain of $f \cdot g: \square$

Step 1 ：The sum of the functions \(f\) and \(g\) is given by \((f+g)(x) = f(x) + g(x)\). Substituting the given functions into the equation, we get \((f+g)(x) = \sqrt{4x - 5} + (2x - 1)\).

Step 2 ：The product of the functions \(f\) and \(g\) is given by \((f \cdot g)(x) = f(x) \cdot g(x)\). Substituting the given functions into the equation, we get \((f \cdot g)(x) = \sqrt{4x - 5} \cdot (2x - 1)\).

Step 3 ：For the function \(f+g\), the domain is the set of all x-values that make the expression under the square root in \(f(x)\) non-negative, and also make the expression in \(g(x)\) defined. Since \(g(x)\) is a linear function, it is defined for all real numbers. However, for \(f(x)\), we must have \(4x - 5 \geq 0\). Solving this inequality gives \(x \geq \frac{5}{4}\). So, the domain of \(f+g\) is \([\frac{5}{4}, \infty)\).

Step 4 ：For the function \(f \cdot g\), the domain is also the set of all x-values that make the expression under the square root in \(f(x)\) non-negative, and also make the expression in \(g(x)\) defined. Again, since \(g(x)\) is a linear function, it is defined for all real numbers. And for \(f(x)\), we must have \(4x - 5 \geq 0\). Solving this inequality gives \(x \geq \frac{5}{4}\). So, the domain of \(f \cdot g\) is also \([\frac{5}{4}, \infty)\).