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Use the definition of the derivative to show that the following functions are not differentiable at \(x=0\). [Hint for Exercises 5 and 6: Modify the calculations on pages \(164-165\) to apply to these functions.] $$f(x)=|3 x|$$

Short Answer

Expert verified
The function \( f(x) = |3x| \) is not differentiable at \( x = 0 \) because the one-sided limits are not equal.

Step by step solution

01

State the Definition of the Derivative

The definition of the derivative of a function \( f(x) \) at a point \( a \) is given by the limit \( f'(a) = \lim_{{h \to 0}} \frac{{f(a+h) - f(a)}}{h} \), provided this limit exists.
02

Specify the Function and Point of Interest

For the function \( f(x) = |3x| \), we are tasked with finding the derivative at \( x = 0 \). Thus, we look at \( f'(0) = \lim_{{h \to 0}} \frac{{|3(0+h)| - |3 \cdot 0|}}{h} \), which simplifies to \( f'(0) = \lim_{{h \to 0}} \frac{{|3h|}}{h} \).
03

Evaluate the One-Sided Limits

Evaluate the limit as \( h \to 0^+ \):\[ \lim_{{h \to 0^+}} \frac{{|3h|}}{h} = \lim_{{h \to 0^+}} \frac{{3h}}{h} = \lim_{{h \to 0^+}} 3 = 3. \]Evaluate the limit as \( h \to 0^- \):\[ \lim_{{h \to 0^-}} \frac{{|3h|}}{h} = \lim_{{h \to 0^-}} \frac{{-3h}}{h} = \lim_{{h \to 0^-}} -3 = -3. \]
04

Determine Differentiability Based on the Limits

The two one-sided limits from Step 3 do not agree, as \( \lim_{{h \to 0^+}} \frac{{|3h|}}{h} = 3 \) and \( \lim_{{h \to 0^-}} \frac{{|3h|}}{h} = -3 \). Since these limits are not equal, \( \lim_{{h \to 0}} \frac{{|3h|}}{h} \) does not exist, which means \( f'(0) \) is undefined.

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Key Concepts

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

Derivative
In calculus, the derivative is a core concept that measures how a function changes as its input changes. It's essentially the rate of change of the function with respect to its variable. For a given function \( f(x) \), the derivative at a specific point \( a \) is defined by the limit:\[f'(a) = \lim_{{h \to 0}} \frac{{f(a + h) - f(a)}}{h}\]If this limit exists, it gives us the slope of the tangent line to the function at \( x = a \).
  • Derivatives are widely used to find the rate of change in functions such as velocity and acceleration in physics.
  • They also help in finding maxima and minima points which are crucial in optimization problems.
For the absolute value function \( f(x) = |3x| \), the problem becomes evaluating this limit as \( h \to 0 \) to check if the function is differentiable at \( x = 0 \). Understanding the derivative through such definitions provides a fundamental basis for more complex mathematical analysis.
Differentiability
Differentiability is a property of a function that indicates whether a derivative exists at each point within its domain. A function is differentiable at a point if it has a defined derivative at that point. To check the differentiability of \( f(x) = |3x| \) at \( x = 0 \), we examine the one-sided limits of the derivative:
  • Right-hand limit as \( h \to 0^+ \): \( \lim_{{h \to 0^+}} \frac{{3h}}{h} = 3 \)
  • Left-hand limit as \( h \to 0^- \): \( \lim_{{h \to 0^-}} \frac{{-3h}}{h} = -3 \)
If these one-sided limits are equal, the derivative exists, and the function is differentiable at that point. However, since the one-sided limits for \( f(x) = |3x| \) at \( x = 0 \) don't match, the function is not differentiable there. This lack of equality in the limits reflects a corner or cusp in the graph at \( x = 0 \), showing the discontinuity of the derivative.
Absolute Value Function
The absolute value function, represented as \( f(x) = |x| \), outputs the non-negative value of \( x \). It has the peculiar property of changing behavior at \( x = 0 \), which affects its differentiability. For \( f(x) = |3x| \), it's useful to understand how the absolute value impacts derivation. As demonstrated through examination of limits:
  • For \( x > 0 \), \( |3x| = 3x \), behaving like a linear function.
  • For \( x < 0 \), \( |3x| = -3x \), flipping its direction.
This change in direction at \( x = 0 \) leads to different slopes on either side of zero, causing a sharp corner or cusp. Graphically, this manifests as a V-shape at the origin, indicating the abrupt change in direction and why the derivative doesn't exist at \( x = 0 \). Understanding the behavior of the absolute value function at this critical point reveals how it disrupts the continuity needed for a derivative to exist.

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