Question

Let \(\mathrm{f}(\mathrm{x})\) satisfy the requirement of lag ranger mean value theorm in \([0,2]\). If \(\mathrm{f}(0)=0\) and \(|\mathrm{f}(\mathrm{x})| \leq(1 / 2)\) for all \(\mathrm{x}\) in \(|0,2|\) then (a) \(\left|\mathrm{f}^{\prime}(\mathrm{x})\right| \leq 2\) (b) \(|\mathrm{f}(\mathrm{x})| \leq 1\) (c) \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}\) (d) \(\mathrm{f}(\mathrm{x})=3\) for at least one \(\mathrm{x}\) in \((0,2)\)

Short answer

The correct answer is that only statement (b) is true, as we have shown that \(|\mathrm{f}(\mathrm{x})| \leq 1\) for all \(x\) in \([0,2]\). Statements (a), (c), and (d) are false based on our analysis using Lagrange's Mean Value Theorem and the given conditions for the function \(\mathrm{f}(\mathrm{x})\).

Step-by-step solution

Determine the upper bound of the derivative

Using the given information, we have that \(|\mathrm{f}(\mathrm{x})| \leq \frac{1}{2}\) on the interval \([0, 2]\). The LMVT states that there exists a value \(c\) such that \(\frac{\mathrm{f}(2) - \mathrm{f}(0)}{2-0} = \mathrm{f}^{\prime}(c)\). Given that \(\mathrm{f}(0) = 0\), we have: \(\left|\frac{\mathrm{f}(2)}{2}\right| = \left|\mathrm{f}^{\prime}(c)\right|\) is the maximum absolute value of the derivative on the interval. We already know that \(|\mathrm{f}(x)|\leq \frac{1}{2}\), so it follows that: $$\left|f'(c)\right| = \left|\frac{f(2)}{2}\right| \leq \frac{1}{2}$$ Since \(|\mathrm{f}^{\prime}(c)| \leq \frac{1}{2}\), we can conclude that option (a) is false.

Prove or disprove remaining statements

(b) The statement claims that \(|\mathrm{f}(\mathrm{x})| \leq 1\) for all \(x\) in \([0,2]\), which is true as we already have \(|\mathrm{f}(\mathrm{x})| \leq \frac{1}{2}\), which is indeed less than or equal to \(1\). So, option (b) is true. (c) To prove or disprove this statement, we need to show whether \(\mathrm{f}(\mathrm{x})=2 \mathrm{x}\) on the interval \([0,2]\). However, we already know that the absolute value of the function is bounded by \(\frac{1}{2}\) on the interval \([0, 2]\). Thus, this function cannot satisfy the given condition, and option (c) is false. (d) To prove or disprove this statement, we must determine whether there exists at least one \(x\) in \((0,2)\) for which \(\mathrm{f}(\mathrm{x})=3\). Although we know that \(|\mathrm{f}(\mathrm{x})| \leq \frac{1}{2}\) for all \(x\) in this interval, we can directly show that \(|\mathrm{f}(x)|\leq 3\) cannot hold given the established bounds of the absolute value of the function. Therefore, option (d) is false. Various aspects have been briefly touched upon in the above steps, but based on our analysis we can preliminarily conclude as follows: (a) False (b) True (c) False (d) False

Key concepts

Derivative Bound

The Lagrange Mean Value Theorem (LMVT) provides insight into the behavior of a function's derivative across a given interval. The theorem establishes that there exists at least one point, labeled as \( c \), in an interval \([a, b]\) where the function \( f(x) \) behaves as if it were a linear function at that point. In essence, the derivative at \( c \), \( f'(c) \), equals the average rate of change of the function over that interval:\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]This average provides us an important clue about the maximum potential slope of the function over the entire interval. In the exercise, using LMVT, it's determined that:\[ \left|f'(c)\right| = \left|\frac{f(2)}{2}\right| \leq \frac{1}{2} \]Since \( |f'(c)| \) is capped at \( \frac{1}{2} \) in this interval, any notion that \( |f'(x)| \leq 2 \) must be false, directly disapproving this point.

Function Properties

The properties of a function outline how the function behaves, often confining it within certain bounds. In our case, stated properties specify that \( |f(x)| \leq \frac{1}{2} \) throughout the interval \([0, 2]\). Such a bound is crucial because it not only limits the highest and lowest possible values \( f(x) \) can reach but also constrains how much the function can oscillate or change in value over the interval.

• Given \( f(0) = 0 \), the function starts at zero and remains within \(-\frac{1}{2} \) to \( \frac{1}{2} \) by the interval's end.
• Henceforth, any statement suggesting otherwise, like \( f(x) = 3 \), is false, as \( f(x) \) cannot suddenly spike to such a high value.
• Furthermore, for \( f(x) = 2x \), matching the bounds of this new function to the original confines, proves this claim false. The function \( f(x) = 2x \) is unachievable since it is not restricted by the same properties.

Interval Analysis

Interval Analysis is crucial as it highlights the specific range \([0, 2]\) in which we apply our exploration of the function \( f(x) \). With LMVT in play, the behavior of \( f(x) \) is evaluated only in this interval.Understanding this interval is fundamental to interpreting any points \( c \) or function values \( f'(x) \):

• The analysis ensures that all x-values resulting in \( f(x) \) adhere to stated bounds. The LMVT operates precisely in this span, fortifying assumptions made solely with these-contained x-values.
• Through this lens, interpreting which intervals and segments the functions adhere to help evaluate the validity of each statement about \( f(x) \) in the problem.

In summary, the whole integrity of \( f(x) \) is confirmed solely in \([0, 2]\). Facts like \( |f(x)| \leq \frac{1}{2} \) make miraculous jumps to values like \( 3 \) impossible in this interval, grounding the theoretical exploration in concrete reality.