Rowe Gough
10/01/2024 · Middle School
Ch. 8: Limits and Continuity EXAMPLE 17 . Compute the vertical and horizontal asymptote of \( f(x)=\frac{x}{\sqrt{x^{2}-4}} \). Hence, sketch the graph
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of the function.
要解决这个问题,我们需要找到函数 \( f(x) = \frac{x}{\sqrt{x^2 - 4}} \) 的垂直和水平渐近线,并绘制出该函数的图像。以下是具体的解题步骤:
### 步骤 1:确定垂直渐近线
垂直渐近线出现在函数分母为零而分子不为零的地方。我们需要找到使分母为零的 \( x \) 值。
1. 分母为 \( \sqrt{x^2 - 4} \),令其等于零:
\[
\sqrt{x^2 - 4} = 0
\]
2. 解方程:
\[
x^2 - 4 = 0
\]
\[
x^2 = 4
\]
\[
x = \pm 2
\]
因此,当 \( x = 2 \) 或 \( x = -2 \) 时,分母为零,所以 \( x = 2 \) 和 \( x = -2 \) 是垂直渐近线。
### 步骤 2:确定水平渐近线
水平渐近线出现在 \( x \) 趋向于无穷大或无穷小时,函数值趋向于某个常数的情况。我们需要计算 \( \lim_{x \to \infty} f(x) \) 和 \( \lim_{x \to -\infty} f(x) \)。
1. 计算 \( \lim_{x \to \infty} f(x) \):
\[
\lim_{x \to \infty} \frac{x}{\sqrt{x^2 - 4}} = \lim_{x \to \infty} \frac{x}{x\sqrt{1 - \frac{4}{x^2}}} = \lim_{x \to \infty} \frac{1}{\sqrt{1 - \frac{4}{x^2}}} = \frac{1}{\sqrt{1 - 0}} = 1
\]
2. 计算 \( \lim_{x \to -\infty} f(x) \):
\[
\lim_{x \to -\infty} \frac{x}{\sqrt{x^2 - 4}} = \lim_{x \to -\infty} \frac{x}{x\sqrt{1 - \frac{4}{x^2}}} = \lim_{x \to -\infty} \frac{1}{\sqrt{1 - \frac{4}{x^2}}} = \frac{1}{\sqrt{1 - 0}} = 1
\]
因此,水平渐近线为 \( y = 1 \)。
### 步骤 3:绘制函数图像
根据以上分析,我们可以绘制出函数 \( f(x) = \frac{x}{\sqrt{x^2 - 4}} \) 的图像:
1. 垂直渐近线:\( x = 2 \) 和 \( x = -2 \)
2. 水平渐近线:\( y = 1 \)
3. 当 \( x \) 接近 \( 2 \) 或 \( -2 \) 时,函数值趋向于无穷大或无穷小。
绘制图像时,注意这些关键点,并确保图像符合这些渐近线。
### 总结
- 垂直渐近线:\( x = 2 \) 和 \( x = -2 \)
- 水平渐近线:\( y = 1 \)
- 函数图像在 \( x = 2 \) 和 \( x = -2 \) 处有垂直渐近线,在 \( y = 1 \) 处有水平渐近线。
Quick Answer
- 垂直渐近线:\( x = 2 \) 和 \( x = -2 \)
- 水平渐近线:\( y = 1 \)
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