Simmons Rodriguez
05/14/2023 · Primary School
4.2 Determine whether Rolle's Theorem applies to the given function on the given interval. If so, find the point(s) that are guaranteed to exist by Rolle's Theorem. \( f(x)=-\sin 4 x\left[\left[\frac{\pi}{4}, \frac{\pi}{2}\right]\right. \) Select the correct choice and, if necessary, fill in the answer box to complete your choice. A. Rolle's. Thenrem annlies and the moint(c)
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要确定罗尔定理是否适用于给定的函数和区间,我们需要检查以下三个条件:
1. 函数在闭区间 \(\left[\frac{\pi}{4}, \frac{\pi}{2}\right]\) 上连续。
2. 函数在开区间 \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\) 上可导。
3. 函数在区间的两端点取相同的值。
### 步骤 1:检查连续性
函数 \( f(x) = -\sin(4x) \) 是一个三角函数,而三角函数在其定义域内是连续的。因此,\( f(x) \) 在闭区间 \(\left[\frac{\pi}{4}, \frac{\pi}{2}\right]\) 上是连续的。
### 步骤 2:检查可导性
同样,三角函数在其定义域内是可导的。因此,\( f(x) \) 在开区间 \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\) 上是可导的。
### 步骤 3:检查两端点值
我们需要计算 \( f(x) \) 在区间两端点 \(\frac{\pi}{4}\) 和 \(\frac{\pi}{2}\) 的值。
\[
f\left(\frac{\pi}{4}\right) = -\sin\left(4 \cdot \frac{\pi}{4}\right) = -\sin(\pi) = 0
\]
\[
f\left(\frac{\pi}{2}\right) = -\sin\left(4 \cdot \frac{\pi}{2}\right) = -\sin(2\pi) = 0
\]
由于 \( f\left(\frac{\pi}{4}\right) = f\left(\frac{\pi}{2}\right) = 0 \),函数在区间的两端点取相同的值。
### 结论
由于函数 \( f(x) = -\sin(4x) \) 在闭区间 \(\left[\frac{\pi}{4}, \frac{\pi}{2}\right]\) 上连续,在开区间 \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\) 上可导,并且在两端点取相同的值,因此罗尔定理适用于这个函数和区间。
根据罗尔定理,存在至少一个 \( c \) 在 \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\) 内,使得 \( f'(c) = 0 \)。
因此,正确答案是:
A. Rolle's theorem applies, and there exists at least one \( c \) in \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\) such that \( f'(c) = 0 \).
Quick Answer
Rolle's theorem applies, and there exists at least one \( c \) in \(\left(\frac{\pi}{4}, \frac{\pi}{2}\right)\) such that \( f'(c) = 0 \).
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