- The composite function \((g \circ h)(x)\) is found by substituting \(h(x)\) into \(g(x)\), resulting in \(\frac{2x-1}{2x+4}\).
- The domain is determined by identifying all \(x\) for which \(h(x)\) is defined and \(h(x)
eq -3\).
- Setting \(h(x) = -3\) yields \(x = -2\), so this value is excluded from the domain.
- The final composite function is \((g \circ h)(x) = \frac{2x-1}{2x+4}\) and the domain is \((-\infty, -2) \cup (-2, \infty)\).
\((g \circ h)(x) = \boxed{\frac{2x-1}{2x+4}}\)
Domain of \(g \circ h\): \(\boxed{(-\infty, -2) \cup (-2, \infty)}\)
Explanation
- Understanding the Problem and Definitions
We are given the functions \(g(x) = \frac{x-2}{x+3}\) and \(h(x) = 2x+1\). The goal is to find the composite function \((g \circ h)(x)\) and determine its domain. A composite function \((g \circ h)(x)\) is defined as \(g(h(x))\). The domain of this composite function consists of all \(x\) in the domain of \(h(x)\) such that \(h(x)\) is in the domain of \(g(x)\). Here, \(h(x) = 2x+1\) is a polynomial and is defined for all real numbers. For \(g(x) = \frac{x-2}{x+3}\), the function is defined for all \(x\) except where the denominator is zero, which is \(x+3=0 \Rightarrow x=-3\). Therefore, the domain of \(g\) is \(\mathbb{R} \setminus \{-3\}\).
- Finding the Composite Function (g ∘ h)(x)
To find the composite function, we substitute \(h(x)\) into \(g(x)\):
\[(g \circ h)(x) = g(h(x)) = g(2x+1)\]
Replacing \(x\) with \(2x+1\) in the expression for \(g(x)\):
\[g(2x+1) = \frac{(2x+1)-2}{(2x+1)+3}\]
Now, we simplify the numerator and the denominator:
- Numerator: \(2x + 1 - 2 = 2x - 1\)
- Denominator: \(2x + 1 + 3 = 2x + 4\)
Thus, the composite function is \((g \circ h)(x) = \frac{2x-1}{2x+4}\).
- Determining the Domain of (g ∘ h)(x)
The domain of the composite function \((g \circ h)(x)\) includes all \(x\) in the domain of \(h(x)\) such that \(h(x)\) is not equal to \(-3\).
We solve the equation \(h(x) = -3\):
\[2x + 1 = -3\]
\[2x = -4\]
\[x = -2\]
This means that \(x = -2\) must be excluded from the domain of the composite function. Alternatively, looking at our simplified expression \(\frac{2x-1}{2x+4}\), the denominator is zero when \(2x + 4 = 0\), which also gives \(x = -2\). Therefore, the domain of \((g \circ h)(x)\) is all real numbers except \(-2\). In interval notation, this is \((-\infty, -2) \cup (-2, \infty)\).
Examples
Composite functions are essential in modeling real-world processes where one operation depends on the outcome of another. For instance, consider a store that offers a standard discount on its items, and then an additional coupon discount at the register. If \(h(x)\) represents the price after the first store discount, and \(g(x)\) represents the final price after applying the coupon to that discounted value, the total discount process is represented by the composition \(g(h(x))\). This mathematical tool allows businesses and consumers to accurately calculate final costs across multiple sequential steps.