Thornton Kirk
09/04/2023 · High School
Composition of two functions: Advanced the real-valued functions \( g(x)=2 x-3 \) and \( h(x)=\sqrt{x-3} \), find the composition \( g \circ h \) and specify its domain usir rval notation.
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为了求解复合函数 \( g \circ h \) 并确定其定义域，我们可以按照以下步骤进行：
### 步骤 1：确定复合函数
复合函数 \( g \circ h \) 的定义是 \( g(h(x)) \)。首先，我们需要将 \( h(x) \) 代入 \( g(x) \) 中。
已知：
\[ g(x) = 2x - 3 \]
\[ h(x) = \sqrt{x - 3} \]
所以：
\[ g(h(x)) = g(\sqrt{x - 3}) \]
### 步骤 2：代入并简化
将 \( h(x) \) 代入 \( g(x) \)：
\[ g(\sqrt{x - 3}) = 2\sqrt{x - 3} - 3 \]
因此，复合函数 \( g \circ h \) 为：
\[ g \circ h = 2\sqrt{x - 3} - 3 \]
### 步骤 3：确定定义域
复合函数的定义域是使得 \( h(x) \) 有意义的 \( x \) 值的集合，即 \( h(x) \) 的定义域。
由于 \( h(x) = \sqrt{x - 3} \)，根号下的表达式必须非负：
\[ x - 3 \geq 0 \]
解这个不等式：
\[ x \geq 3 \]
因此，复合函数 \( g \circ h \) 的定义域是：
\[ [3, \infty) \]
### 最终答案
复合函数 \( g \circ h \) 为：
\[ g \circ h = 2\sqrt{x - 3} - 3 \]
其定义域为：
\[ [3, \infty) \]
Quick Answer
复合函数 \( g \circ h = 2\sqrt{x - 3} - 3 \)，定义域为 \( [3, \infty) \)。
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