91Ó°ÊÓ

In calculus, one-sided derivatives are related to the concept of differentiability, focusing on the behavior of a function as it approaches a certain point from one side. When a function is not differentiable at a point, it may still have well-defined derivatives when approaching that point from the left or the right. Consider the exercise provided, which involves the function \( f(x) = \sqrt{1 - \cos x} \). The key point of interest is \( x = 2k\pi \). Approaching from the left, the left-hand derivative is calculated as the limit \( \lim_{h \to 0^-} \frac{f(2k\pi + h) - f(2k\pi)}{h} \), resulting in a value of \(-\frac{1}{\sqrt{2}}\). Approaching from the right, the right-hand derivative is given by \( \lim_{h \to 0^+} \frac{|h|}{h\sqrt{2}} = \frac{1}{\sqrt{2}}\).

• These one-sided derivatives provide insight into how the function behaves differently from each side at critical points.
• They are useful to understand why the function is not differentiable at \( x = 2k\pi \).

Continuity is a foundational concept in calculus concerning the smoothness and predictability of functions. For a function to be continuous at a point, three conditions must hold:

• The function value exists at that point.
• The limit of the function exists as the input approaches that point.
• The limit equals the function value at that point.

In the given exercise, the function \( f(x) = \sqrt{1 - \cos x} \) is examined for continuity at the points critical to its differentiability, namely \( x = 2k\pi \). At these points, \( \cos x = 1 \) makes \( 1 - \cos x = 0 \), and consequently, \( f(x) = \sqrt{0} = 0 \). Since this value exists, the function passes the first condition at these points.However, the limit of \( f(x) \) as \( x \to 2k\pi \) must also have a smooth transition for \( f(x) \) to be continuous. Upon checking, despite the limit existing, the abrupt transition from the left to the right hand leads to a breakdown in smoothness, thus impacting the differentiability at \( x = 2k\pi \).

Critical points in calculus are crucial for understanding where a function changes behavior. They occur where the derivative of the function is zero or undefined, and they help identify peaks, troughs, or points of inflection on a graph.In the case of \( f(x) = \sqrt{1 - \cos x} \), critical points are identified at \( x = 2k\pi \) due to the function's differentiability being compromised. The derivatives at these points either do not exist or point out significant changes as one-sided derivatives are unequal.

• Each critical point might suggest a location for potential maxima, minima, or sharp turns, which can significantly affect the graph's shape.
• By analyzing \( f(x) \) at these critical points one can determine where the function shifts from differentiable to non-differentiable.

Hence, these insights into critical points, alongside the calculations of one-sided derivatives, allow for a comprehensive understanding of the function's qualitative behavior at \( x = 2k\pi \).