If f(x) is equal to x+1 and g(x) is equal to 5x + 4 over x - 3, then figure out the domain and piecewise function of f over g

To find the domain and the piecewise function of \(\frac{f}{g}(x)\), we start by defining the functions \(f(x)\) and \(g(x)\):

\[ f(x) = x + 1 \]
\[ g(x) = \frac{5x + 4}{x - 3} \]

The function \(\frac{f}{g}(x)\) is defined as:
\[ \frac{f}{g}(x) = \frac{x + 1}{\frac{5x + 4}{x - 3}} \]

To simplify this, we multiply by the reciprocal of \(g(x)\):
\[ \frac{f}{g}(x) = \frac{x + 1}{\frac{5x + 4}{x - 3}} = (x + 1) \cdot \frac{x - 3}{5x + 4} \]

So, the simplified form is:
\[ \frac{f}{g}(x) = \frac{(x + 1)(x - 3)}{5x + 4} \]

### Domain of \(\frac{f}{g}(x)\)

To determine the domain, we need to identify the values of \(x\) for which \(\frac{f}{g}(x)\) is defined. The key points to consider are:

1. The domain of \(f(x)\): \(f(x) = x + 1\) is defined for all real numbers \(x\).
2. The domain of \(g(x)\): \(g(x) = \frac{5x + 4}{x - 3}\) is defined for all \(x \neq 3\), because \(x = 3\) would make the denominator zero.

Additionally, we need to ensure that the denominator of \(\frac{f}{g}(x)\), which is \(5x + 4\), is not zero:
\[ 5x + 4 \neq 0 \]
\[ x \neq -\

Explore the domain and piecewise function of f(x) over g(x) where f(x) = x + 1 and g(x) = (5x + 4) / (x - 3). Understand how to determine the domain restrictions and compute the piecewise function. Suitable for advanced high school students.

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