Problem 77 - Calculus Volume 1

Question

For the following functions, a. sketch the graph and b. use the definition of a derivative to show that the function is not differentiable at \(x=1.\) $$f(x)=\left\\{\begin{array}{l}{2 x, x \leq 1} \\ {\frac{2}{x}, x>1}\end{array}\right.$$

Step-by-Step Solution

Answer

The function is not differentiable at \( x = 1 \) since left and right derivatives differ.

Step 1: Understand the Piecewise Function

The function \( f(x) \) is defined differently based on the value of \( x \). For \( x \leq 1 \), \( f(x) = 2x \). For \( x > 1 \), \( f(x) = \frac{2}{x} \). We need to analyze these two parts separately around \( x = 1 \).

Step 2: Sketch the Graph

Plot the linear part \( f(x) = 2x \) for \( x \leq 1 \), which is a line through the origin with a slope of 2, stopping at \( x = 1 \). For \( x > 1 \), plot \( f(x) = \frac{2}{x} \), which is a hyperbola starting from the right side of \( x = 1 \). Notice they don't connect smoothly at \( x = 1 \).

Step 3: Check Continuity at \( x = 1 \)

To check if \( f(x) \) is continuous at \( x = 1 \), compute \( f(1) \) from both the left and the right. For \( x \leq 1 \), \( f(1) = 2 \times 1 = 2 \). For \( x > 1 \), as \( x \to 1^+ \), \( f(x) = \frac{2}{1} = 2 \). Since both limits equal 2, the function is continuous at \( x = 1 \).

Step 4: Apply the Definition of Derivative

A function is differentiable at \( x = c \) if the derivative limit \( \lim_{{h \to 0}} \frac{f(c+h) - f(c)}{h} \) exists. Evaluate the derivative from the left: \( \lim_{{h \to 0^-}} \frac{2(1+h) - 2}{h} = \lim_{{h \to 0^-}} \frac{2h}{h} = 2 \).

Step 5: Evaluate the Right-side Derivative

For the right-side derivative, use \( f(x) = \frac{2}{x} \): \( \lim_{{h \to 0^+}} \frac{\frac{2}{1+h} - 2}{h} = \lim_{{h \to 0^+}} \frac{\frac{2 - 2(1+h)}{1+h}}{h} = \lim_{{h \to 0^+}} \frac{-2h}{h(1+h)} = \lim_{{h \to 0^+}} \frac{-2}{1+h} = -2 \).

Step 6: Conclude Differentiability

Since the left-hand and right-hand derivatives at \( x = 1 \) are different (2 and -2 respectively), the derivative does not exist at \( x = 1 \). Therefore, \( f(x) \) is not differentiable at \( x = 1 \).

Key Concepts

Piecewise Functions

In mathematics, a piecewise function is a function that applies different expressions to different parts of its domain. This means the function's definition changes based on the input value. Our example function, \( f(x) \), displays this property by having two expressions depending on the value of \( x \):

• For \( x \leq 1 \), \( f(x) = 2x \) which represents a linear component.
• For \( x > 1 \), \( f(x) = \frac{2}{x} \) which represents a hyperbolic component.

These components are stitched together at \( x = 1 \), but they have to be evaluated separately when determining other characteristics, such as continuity and differentiability. Understanding how these individual pieces work alone and at their joining point is crucial when analyzing piecewise-defined functions mathematically.

Limits

The concept of limits allows us to understand the behavior of a function as it approaches a specific point. Calculating limits is necessary for evaluating continuity and differentiability, especially at points where the function is defined piecewise. In the case of \( x=1 \) for the function \( f(x) \), we calculate

• The left-hand limit (as \( x \to 1^- \)) for \( 2x \) which is \( 2 \times 1 = 2 \).
• The right-hand limit (as \( x \to 1^+ \)) for \( \frac{2}{x} \), which also approaches \( 2 \).

Since both one-sided limits equal the value of the function at \( x=1 \), the overall limit exists and matches the function's value, thus suggesting continuity. Limits help prove whether parts of the function join seamlessly despite being defined by different expressions.

Continuity

A function is continuous at a point if the limit of the function as it approaches the point from either side is equal to the function’s value at that point. For the function \( f(x) \), continuity at \( x=1 \) is assessed by ensuring the function value \( f(1) \) matches the calculated limits from both sides. In this case: - \( f(1) = 2 \) for both parts.Since the left-hand limit, right-hand limit, and \( f(1)\) are all equal to 2, \( f(x) \) is continuous at \( x=1 \). Although the pieces join continuously, continuity alone does not guarantee differentiability, as different slopes might exist on each side of the point.

Derivatives

The derivative of a function at a point provides the slope of the tangent line to the function's graph at that point. If the function is differentiable, changes occur smoothly. However, for piecewise functions, differentiability can fail, particularly at the junction where the pieces meet, as happened here with \( f(x) \) at \( x=1 \). We calculate:

• The left-hand derivative is \( \lim_{{h \to 0^-}} \frac{2(1+h) - 2}{h} = 2 \).
• The right-hand derivative is \( \lim_{{h \to 0^+}} \frac{\frac{2}{1+h} - 2}{h} = -2 \).

Because the one-sided derivatives are different (2 and -2), the derivative doesn't exist at this point, and \( f(x) \) is not differentiable at \( x=1 \). Differentiability requires the same slope approach from both sides, which we clearly lack in this scenario.