Davey Harmon
06/16/2024 · Junior High School
\( p ( u ) = e ^ { u } \cos u + e ^ { u } \sin u - \frac { 1 } { 2 } , a = - \pi \)
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Step-by-step Solution
To analyze the function \( p(u) = e^u \cos u + e^u \sin u - \frac{1}{2} \) at the point \( a = -\pi \), we can evaluate \( p(-\pi) \).
First, we substitute \( u = -\pi \) into the function:
\[
p(-\pi) = e^{-\pi} \cos(-\pi) + e^{-\pi} \sin(-\pi) - \frac{1}{2}
\]
Next, we calculate the values of \( \cos(-\pi) \) and \( \sin(-\pi) \):
\[
\cos(-\pi) = -1 \quad \text{and} \quad \sin(-\pi) = 0
\]
Now substituting these values back into the function:
\[
p(-\pi) = e^{-\pi} (-1) + e^{-\pi} (0) - \frac{1}{2}
\]
This simplifies to:
\[
p(-\pi) = -e^{-\pi} - \frac{1}{2}
\]
Thus, the value of \( p(-\pi) \) is:
\[
p(-\pi) = -e^{-\pi} - \frac{1}{2}
\]
If you need further analysis, such as finding the derivative or evaluating limits, please let me know!
Quick Answer
\( p(-\pi) = -e^{-\pi} - \frac{1}{2} \)
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