Problem 3 - The Calculus with Analytic Geometry

Question

Do each of the following: (a) Draw a sketch of the graph of the function; (b) determine if \(f\) is continuous at \(x_{1} ;\) (c) find \(f^{\prime}-\left(x_{1}\right)\) and \(f_{+}^{\prime}\left(x_{1}\right)\) if they exist; (d) determine if \(f\) is differentiable at \(x_{1}\). $$ f(x)=|x-3| ; x_{1}=3 $$

Step-by-Step Solution

Answer

The function \( f(x) = |x - 3| \) is continuous at \( x_1 = 3 \), but it is not differentiable there.

Step 1 - Sketch the Graph

To sketch the graph of the function, note that the function given is a piecewise function: \( f(x) = |x - 3| \). The absolute value function is V-shaped, with a vertex at the point \( x = 3 \). The slope of the function is -1 to the left of \( x = 3 \) and 1 to the right of \( x = 3 \). Hence, the graph is symmetric around the vertex. Draw the V-shaped graph accordingly.

Step 2 - Determine Continuity at \( x_1 = 3 \)

To check if \( f \) is continuous at \( x_1 = 3 \), we need to verify if \( f(3) \) exists, if \( \lim_{{x \to 3}} f(x) \) exists, and if \( \lim_{{x \to 3}} f(x) = f(3) \). Since \( f(3) = |3 - 3| = 0 \) and \( \lim_{{x \to 3}} f(x) = 0 \) for both left-hand and right-hand limits, the function is continuous at \( x_1 = 3 \).

Step 3 - Find Left and Right Derivatives

To find the left-hand derivative \( f'_-(3) \) and the right-hand derivative \( f'_+(3) \), compute the limits of the difference quotient as follows: \[ f'_-(3) = \lim_{{h \to 0^-}} \frac{f(3 + h) - f(3)}{h} = \lim_{{h \to 0^-}} \frac{|3 + h - 3| - 0}{h} = \lim_{{h \to 0^-}} \frac{|h|}{h} = -1 \]. Similarly for the right derivative: \[ f'_+(3) = \lim_{{h \to 0^+}} \frac{f(3 + h) - f(3)}{h} = \lim_{{h \to 0^+}} \frac{|3 + h - 3| - 0}{h} = \lim_{{h \to 0^+}} \frac{|h|}{h} = 1 \].

Step 4 - Determine Differentiability at \( x_1 = 3 \)

For the function to be differentiable at \( x_1 = 3 \), the left-hand and right-hand derivatives must be equal. Since \( f'_-(3) = -1 \) and \( f'_+(3) = 1 \), they are not equal. Hence, the function is not differentiable at \( x_1 = 3 \).

Key Concepts

Continuity of Functions

To address the continuity of a function, one must ensure three key conditions are met. First, the function value at the point must exist. Second, the limit of the function as it approaches the point from both sides must exist. Lastly, the function value at the point must equal the limit value. In our case, for the given function \(f(x) = |x-3|\), we need to check continuity at \(x_1 = 3\). We find that \(f(3) = |3 - 3| = 0\). Next, the limit as \(x\) approaches 3 is \( \lim_{{x \to 3}} f(x) = 0 \). Both the left-hand limit and the right-hand limit are the same and match the function value at 3. Hence, the function is continuous at \(x_1 = 3\).

Differentiability

Differentiability of a function at a point requires that the function's derivative exists and is continuous at that point. The derivative, which represents the function's slope, must be the same when approaching the point from either side. For the absolute value function \(f(x) = |x - 3|\) at \(x_1 = 3\), we calculate the left-hand and right-hand derivatives. They are found to be \( f'_-(3) = -1 \) and \( f'_+(3) = 1 \) respectively. Since these derivatives are not equal, the function is not differentiable at \(x_1 = 3\). Therefore, the function has a 'corner' at \(x = 3\), which prevents it from being differentiable there.

Left and Right Derivatives

In general, to determine the differentiability at a point, it's essential to compute both the left and right derivatives. These are calculated using limits. For \( f(x) = |x-3| \), the left-hand derivative at \(x_1 = 3\) is calculated as: \[ f'_-(3) = \lim_{{h \to 0^-}} \frac{|3 + h - 3| - f(3)}{h} = \lim_{{h \to 0^-}} \frac{|h| - 0}{h} = \lim_{{h \to 0^-}} \frac{|h|}{h} = -1 \]. For the right-hand derivative: \[ f'_+(3) = \lim_{{h \to 0^+}} \frac{|3 + h - 3| - f(3)}{h} = \lim_{{h \to 0^+}} \frac{|h| - 0}{h} = \lim_{{h \to 0^+}} \frac{|h|}{h} = 1 \]. These derivatives are significant as they help determine the function's behavior just before and just after a point.

Piecewise Functions

Piecewise functions are defined by different expressions depending on the interval of the input. The absolute value function \( f(x) = |x - 3| \) is a classic example and can be written as a piecewise function: \[ f(x) = \begin{cases} -(x - 3), & \text{if } x < 3 \ (x - 3), & \text{if } x \geq 3 \end{cases} \]. This means for values of \(x\) less than 3, the function behaves like \( -(x - 3) \), and for values of \(x\) greater than or equal to 3, it behaves like \(x - 3\). Piecewise functions can often present challenges in analysis, particularly at the points where the definition switches, like \(x = 3\) in our example. These switch points may lead to potential discontinuities or non-differentiable corners.