Let `f(x) = x^(2) and g (x) = 2 x + 1 ` be two real functions. Find `((f)/(g)) (x)` . Also, state its domain

Answer

To find \(\frac{f}{g}(x)\) where \(f(x) = x^2\) and \(g(x) = 2x + 1\), we will follow these steps:

### Step 1: Write the expression for \(\frac{f}{g}(x)\)
We start by substituting the functions into the expression:
\[ \frac{f}{g}(x) = \frac{f(x)}{g(x)} = \frac{x^2}{2x + 1} \]

### Step 2: Identify the domain of \(\frac{f}{g}(x)\)
The domain of a function is the set of all possible input values (x-values) for which the function is defined. For the function \(\frac{f}{g}(x)\), we need to ensure that the denominator \(g(x) = 2x + 1\) is not equal to zero.

Set the denominator equal to zero and solve for \(x\):
\[ 2x + 1 = 0 \]
\[ 2x = -1 \]
\[ x = -\frac{1}{2} \]

Thus, \(g(x)\) is zero when \(x = -\frac{1}{2}\). Therefore, the function \(\frac{f}{g}(x)\) is undefined at this point.

### Step 3: State the domain
The domain of \(\frac{f}{g}(x)\) is all real numbers except \(x = -\frac{1}{2}\). In interval notation, this can be expressed as:
\[ \text{Domain} = (-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, \infty) \]

### Final Answer
Thus, we have:
\[ \frac{f}{g}(x) = \frac{x^2}{2x + 1}, \quad \text{Domain} = (-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, \infty) \]

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