Problem 8 - Advanced Calculus

Question

Investigate the existence of local and global inverses for the function \(f(x)=A x-\sin x\), for various values of \(A\).

Step-by-Step Solution

Answer

The function \(f(x)=A x-\sin x\) is globally invertible for \(A>=2\) or \(A<=0\). For \(0

Step 1: Find the derivative of function \(f(x)\)

The derivative of the function \(f(x)=A x-\sin x\) is \(f'(x)=A-\cos x\)

Step 2: Find the value of \(A\) for which \(|f'(x)|>=1\)

The function \(f(x)\) will be invertible if \(|f'(x)|>=1\), which means \(|A-\cos x|>=1\). Since \(|\cos x|\leq 1\), the function will be globally invertible when \(A>=2\) or \(A<=0\). When \(0

Step 3: Conclusion

In conclusion, the function \(f(x)=A x-\sin x\) is globally invertible for \(A>=2\) or \(A<=0\)\. If \(0

Key Concepts

Global Inverse

Understanding the global inverse of a function is crucial for developing a comprehensive mathematical foundation. A global inverse exists when a function is one-to-one (bijective) across its entire domain, meaning every output is associated with exactly one input.

For the function in our exercise, \( f(x) = Ax - \sin x \), determining a global inverse hinges on the constant \( A \). When \( A \) is greater than or equal to 2, or less than or equal to 0, \( f(x) \) has a global inverse. This is because the function becomes one-to-one throughout the real numbers, ensuring that each input maps to a unique output without any overlaps.

Local Inverse

In contrast, a local inverse pertains to a specific region of the function's domain where it maintains a one-to-one relationship.

For the given function, if \( 0 < A < 2 \), there are certain intervals in which the function is one-to-one and hence has a local inverse. It implies that for any selected piece of the function's graph, one can find a unique inverse function for that portion. However, this does not extend globally across the entire domain, which is why the function lacks a global inverse in this case.

Derivative of Function

The derivative of a function provides information about the function's rate of change at any given point. To examine invertibility, the derivative plays a pivotal role as it can indicate whether a function is increasing or decreasing, and therefore whether it has a chance of being invertible at all.

The function \( f(x) = Ax - \sin x \) has a derivative of \( f'(x) = A - \cos x \). A constant and non-zero derivative suggests that the function is either strictly increasing or decreasing, which is a necessary condition for invertibility. By studying the function's derivative, we understand that if the absolute value of the derivative, \( |f'(x)| \), is greater than or equal to 1 for all \( x \), the function maintains its monotonicity and hence, is invertible.

Invertibility Conditions

Identifying when a function is invertible involves checking specific conditions for its derivative. For the function in our exercise, the condition \( |f'(x)| >= 1 \) must be fulfilled for the function to be invertible.

This condition means that for every point \( x \) in the domain, the slope of the function must not be zero and must have a magnitude of at least 1, preventing the function from being flat or having a turning point. These invertibility conditions allow us to conclude definitively that for \( A >= 2 \) or \( A <= 0 \), the function \( f(x) \) is globally invertible and for \( 0 < A < 2 \), it is only invertible locally, not globally.

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Problem 8

Other exercises in this chapter

Problem 8

Use the example \(f(x)=x^{2} /\left(1+x^{2}\right)\) to show that a continuous function does not always have to map a closed set onto a closed set.

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Problem 8

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The function $$ \exp \left(\frac{x^{2}+y^{2}-x y}{x^{2}+y^{2}}\right)=f(x, y) $$ is continuous on the open first quadrant. (a) Is it bounded there? (b) Can \(f\

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Problem 8

Show that any function that is locally constant on an open connected set \(D\) is in fact constant on \(D\).

View solution

Key Concepts

Global Inverse

Understanding the global inverse of a function is crucial for developing a comprehensive mathematical foundation. A global inverse exists when a function is one-to-one (bijective) across its entire domain, meaning every output is associated with exactly one input.

For the function in our exercise, \( f(x) = Ax - \sin x \), determining a global inverse hinges on the constant \( A \). When \( A \) is greater than or equal to 2, or less than or equal to 0, \( f(x) \) has a global inverse. This is because the function becomes one-to-one throughout the real numbers, ensuring that each input maps to a unique output without any overlaps.

Local Inverse

In contrast, a local inverse pertains to a specific region of the function's domain where it maintains a one-to-one relationship.

For the given function, if \( 0 < A < 2 \), there are certain intervals in which the function is one-to-one and hence has a local inverse. It implies that for any selected piece of the function's graph, one can find a unique inverse function for that portion. However, this does not extend globally across the entire domain, which is why the function lacks a global inverse in this case.

Derivative of Function

The derivative of a function provides information about the function's rate of change at any given point. To examine invertibility, the derivative plays a pivotal role as it can indicate whether a function is increasing or decreasing, and therefore whether it has a chance of being invertible at all.

The function \( f(x) = Ax - \sin x \) has a derivative of \( f'(x) = A - \cos x \). A constant and non-zero derivative suggests that the function is either strictly increasing or decreasing, which is a necessary condition for invertibility. By studying the function's derivative, we understand that if the absolute value of the derivative, \( |f'(x)| \), is greater than or equal to 1 for all \( x \), the function maintains its monotonicity and hence, is invertible.

Invertibility Conditions

Identifying when a function is invertible involves checking specific conditions for its derivative. For the function in our exercise, the condition \( |f'(x)| >= 1 \) must be fulfilled for the function to be invertible.

This condition means that for every point \( x \) in the domain, the slope of the function must not be zero and must have a magnitude of at least 1, preventing the function from being flat or having a turning point. These invertibility conditions allow us to conclude definitively that for \( A >= 2 \) or \( A <= 0 \), the function \( f(x) \) is globally invertible and for \( 0 < A < 2 \), it is only invertible locally, not globally.