Problem 60 - Thomas Calculus

Question

In Exercises 59 and \(60,\) give reasons for your answers. Let $$f(x)=\left|x^{3}-9 x\right|$$ $$\begin{array}{ll}{\text { a. Does } f^{\prime}(0) \text { exist? }} & {\text { b. Does } f^{\prime}(3) \text { exist? }} \\ {\text { c. Does } f^{\prime}(-3) \text { exist? }} & {\text { d. Determine all extrema of } f}\end{array}$$

Step-by-Step Solution

Answer

\( f'(0) \) exists and equals 0; \( f'(3) \) and \( f'(-3) \) do not exist; extrema at \((-3, -3)\), \((3, -3)\).

Step 1: Analyze Derivative at x = 0

To find if \( f'(0) \) exists, investigate the behavior of \( f(x) = |x^3 - 9x| \) at \( x = 0 \). The equation simplifies to \( f(x) = |x(x - 3)(x + 3)| \). At \( x = 0 \), \( f(x) = 0 \). Use the definition of the derivative:\[ f'(x) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} \]Since \( f(x) \) near \( x = 0 \) behaves as \( x^3 \), which is differentiable, \( f'(0) \) exists and equals 0.

Step 2: Analyze Derivative at x = 3

Examine \( f(x) = |x^3 - 9x| \) at \( x = 3 \). This becomes \( f(x) = |x - 3|(x^2 + 3x + 3) \). At \( x = 3 \), the expression inside the absolute value is zero, causing a sharp corner, indicating a non-differentiable point. For the derivative to exist both the left-hand derivative and right-hand derivative must be equal at \( x = 3 \), which they are not, thus \( f'(3) \) does not exist.

Step 3: Analyze Derivative at x = -3

Check \( f(x) = |x^3 - 9x| \) at \( x = -3 \). Similarly, this becomes \( f(x) = |x + 3|(x^2 - 3x + 3) \). At \( x = -3 \), the value again hits zero inside the absolute value function, indicating a point of non-differentiability. The different behaviors on either side of \( x = -3 \) confirm that \( f'( -3) \) does not exist.

Step 4: Determine Extrema Points

An extremum occurs when the derivative is zero or undefined. From earlier steps, we know the derivative does not exist at \( x = 3, -3 \), these are potential extrema. Examine function behavior over intervals:\- For \( x > 3 \), \( f(x) = x^3 - 9x \).- For \( -3 < x < 3 \), \( f(x) = 9x - x^3 \).- For \( x < -3 \), \( f(x) = x^3 - 9x \) again.Evaluate \( f(x) \) at critical points \( x = -3, 0, 3 \) and endpoints. We find \((-3, -3), (3, -3)\) are local minima, and the function is symmetric about the y-axis, confirming symmetry at extrema.

Key Concepts

Non-Differentiable Points

Understanding non-differentiable points is key to analyzing the behavior of functions. A point on a graph where the function does not have a derivative is known as a non-differentiable point. This typically happens at corners, cusps, or vertical tangents.
In our exercise, we have the function \( f(x) = |x^3 - 9x| \). The absolute value creates potential points of non-differentiability where the expression inside the |x| is zero.

• For \( x = 3 \), the expression becomes zero, leading to a sharp corner in the graph.
• Similarly, at \( x = -3 \), we also find a sharp corner indicating another non-differentiable point.

These points are important because they represent where the derivative does not exist, indicating potential extrema as seen in the exercise.

Extrema

Extrema are significant features of a function's graph. They represent the highest or lowest points, known as maxima and minima. There are local extrema (local maxima and minima), which occur at points within a specific interval, and global extrema that are absolute peaks or valleys over the entire domain.

In the context of derivatives, extrema occur where the derivative is zero (indicating a flat slope) or does not exist (potentially indicating a cusp or a sharp corner). From the given function \( f(x) = |x^3 - 9x| \), we identified:

• Non-differentiable points potentially acting as extrema.
• By evaluating at critical points \( x = -3, \) and \( x = 3, \) we find they are local minima.

The symmetry of the function across the y-axis further supports these findings, suggesting balance in its extremal points.

Absolute Value Function

The absolute value function plays a crucial role in determining the points where the function might be non-differentiable. The absolute value function, defined as \( |x| \), takes the magnitude of x, meaning it flips negative values to positive while leaving positive values unchanged.
This characteristic can introduce points in a function where the transition is not smooth, often creating corners or slopes that do not exist on the derivative.
In \( f(x) = |x^3 - 9x| \), the crucial moments occur where \( x^3 - 9x \) changes sign:

• At \( x = 3 \) and \( x = -3 \), the path of the function sharply changes, not allowing for a smooth slope and hence no derivative exists at these points.

This property is fundamental in analyzing the critical points discussed in derivative problems.

Critical Points

Critical points are key locations in a function's graph that are identified through its derivative. These are the points where the derivative is zero or undefined. Identifying these points helps understand the potential maximum or minimum values the function might take.
For \( f(x) = |x^3 - 9x| \), several analyses are conducted:

• Set the derivative \( f'(x) \) equal to zero (where applicable) or check where it does not exist, such as at \( x = 3 \) and \( x = -3 \).
• Evaluate the function at these critical points to analyze the behavior across the intervals.

Through this analysis, the exercise revealed a local minima at \( x = 3 \) and \( x = -3 \), as the derivative swaps from positive to negative across these points. The method of checking the derivative sign changes is crucial in pinpointing critical points for optimization analysis.