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Chapter 3: Problem 4

If \(f(x)\) is defined and \(f^{\prime}(x)\) exists for \(x, a0\) for \(a

Short Answer

Expert verified
The exercise is solved by applying the Mean Value Theorem to prove that if the derivative of a function is greater than or equal to zero in an interval, the function is non-decreasing in the interval. If the derivative is strictly positive in the interval, the function is strictly increasing.

Step by step solution

01

Prove the first part

Assume that \(f^{\prime}(x) \geq 0\) for \(a<x<b\), and consider any two points \(x_1\) and \(x_2\) in this interval where \(x_1 < x_2\). Then applying the Mean Value Theorem, we can say that there exists a point \(c\) in the interval between \(x_1\) and \(x_2\) such that \(f^{\prime}(c) = (f(x_2) - f(x_1)) / (x_2 - x_1)\). Since \(f^{\prime}(c) \geq 0\) by our assumption, it implies that \(f(x_2) - f(x_1) \geq 0\), which means \(f(x_2) \geq f(x_1)\). This is valid for any \(x_1, x_2\) in the interval \(a<x<b\), which means that the function \(f(x)\) is non-decreasing in this interval.
02

Prove the second part

In a similar way, if we assume that \(f^{\prime}(x) > 0\) for \(a 0\) by our assumption, it implies that \(f(x_2) - f(x_1) > 0\), which means \(f(x_2) > f(x_1)\). This is valid for any \(x_1, x_2\) in the interval \(a<x<b\), which means that the function \(f(x)\) is strictly increasing in this interval.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Monotonic Functions
Monotonic functions are mathematical functions that either never increase or never decrease. This quality is dependent on the behavior of the function's derivative:
  • If the derivative \(f'(x)\) is greater than or equal to zero, the function is said to be non-decreasing, or monotonic.
  • If \(f'(x)\) is strictly greater than zero, the function is strictly increasing.
This means that the function does not decrease as \(x\) moves from left to right. It's important to understand the implications of this when looking at graphs or analyzing function behavior.
To visualize, imagine climbing a hill without ever descending鈥攊f you can accomplish that, then the function is monotonically increasing. However, stepping up onto level ground still qualifies as non-decreasing.
Mean Value Theorem
The Mean Value Theorem (MVT) provides a formal way to connect the derivative of a function to its overall increase or decrease on an interval.To apply MVT, there must be specific conditions:
  • The function \(f(x)\) must be continuous over a closed interval \([a, b]\).
  • The function must be differentiable over the open interval \((a, b)\).
If these conditions are met, then there exists at least one point \(c\) in the interval where the instantaneous rate of change (the derivative) is equal to the average rate of change over the whole interval:\[ f'(c) = \frac{f(b) - f(a)}{b - a}\]This theorem helps in proving monotonicity by showing that if \(f'(x)\) is always positive or non-negative, then the function has a consistent increasing or non-decreasing trend.
Derivatives
Derivatives are a fundamental concept in calculus, representing the rate at which a function is changing at any given point. When evaluating the monotonicity of a function, derivatives are crucial:
  • A positive derivative \(f'(x) > 0\) indicates the function is increasing at that point.
  • A zero derivative \(f'(x) = 0\) indicates a possible point of inflection or constant behavior.
  • A negative derivative \(f'(x) < 0\) suggests the function is decreasing at that point.
Understanding derivatives allows us to draw real-world connections, like acceleration in physics, or profit changes in economics. By analyzing these changes, we can understand whether a function remains steady, increases, or decreases.
In the context of the exercise, using the derivative helps determine if the function remains non-decreasing or strictly increasing over the given interval \((a,b)\), and this observation aligns with our understanding of monotonic behavior.

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Most popular questions from this chapter

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