Chapter 171: Problem 4
Die Funktion \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}^{2}\) und die
..Streifen" \(G_{1}, G_{2} \subset \mathbf{R}^{2}\) seien gegeben durch
$$
f(x, y):=\left(\begin{array}{l}
\sin x \cosh y \\
\cos x \sinh y
\end{array}\right) . \quad G_{1}:=\left\\{\left(\begin{array}{l}
x \\
y
\end{array}\right): 0
Short Answer
Step by step solution
Understand the function and sets
Compute \( f(x, y) \) for \( G_1 \)
Compute \( f(x, y) \) for \( G_2 \)
Analyze injectivity on \( G_1 \) and \( G_2 \)
Analyze non-injectivity on \( G_1 \cup G_2 \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Image of a Function
- For \( G_1 \), the image is in the first quadrant, since both function components are positive for the values between zero and \( \frac{\pi}{2} \).
- For \( G_2 \), the scenario shifts to the fourth quadrant due to the sign change in \( \cos x \) when \( x \) is between \( \frac{\pi}{2} \) and \( \pi \).
Trigonometric Functions
- The sine function, \( \sin x \), reaches its peak value at \( \frac{\pi}{2} \) and is positive as \( x \) increases from 0 to \( \pi \).
- The cosine function, \( \cos x \), decreases from 1 to 0 in the interval \( 0 \) to \( \frac{\pi}{2} \) and becomes negative beyond \( \frac{\pi}{2} \).
Hyperbolic Functions
- \( \sinh y = \frac{e^{y} - e^{-y}}{2} \): similar to the sine wave but not periodic.
- \( \cosh y = \frac{e^{y} + e^{-y}}{2} \): always positive, contributing to the function's range.
Plane Geometry
- \( G_1 \): a horizontal band with values \( 0 < x < \frac{\pi}{2} \).
- \( G_2 \): another band but shifted to \( \frac{\pi}{2} < x < \pi \).