Problem 54 - Thomas Calculus

Question

In Exercises 53 and \(54,\) give reasons for your answers. Let \(f(x)=\left|x^{3}-9 x\right|\) a. Does \(f^{\prime}(0)\) exist? b. Does \(f^{\prime}(3)\) exist? c. Does \(f^{\prime}(-3)\) exist? d. Determine all extrema of \(f\)

Step-by-Step Solution

Answer

f'(0), f'(3), and f'(-3) do not exist due to cusp points. Minima occur at x = -3, 0, 3.

Step 1: Understanding the Exercise

We are given the function \( f(x) = |x^3 - 9x| \), a piecewise function due to the absolute value. We need to find whether the derivative exists at specific points and determine the extrema of this function.

Step 1: Analyze the Absolute Value Function

The expression inside the absolute value \(x^3 - 9x\) can be factored to \(x(x^2 - 9) = x(x-3)(x+3)\). The zeros of this expression are \(x = 0, 3, -3\), where the function possibly exhibits non-smooth behavior, like cusps or corners.

Step 2: Investigate f'(0)

At \(x = 0\), \(f(x) = |x^3 - 9x| = |0| = 0\). For \(x < 0\), \(f(x) = -(x^3 - 9x)\). For \(x > 0\), \(f(x) = x^3 - 9x\). Differentiating these gives different forms at \(x = 0\), so \(f'(0)\) does not exist as the limits from left and right do not match.

Step 3: Check f'(3)

At \(x = 3\), \(f(x) = |0| = 0\). For \(x < 3\), \(f(x) = -(x^3 - 9x)\) because the expression is negative. For \(x > 3\), \(f(x) = x^3 - 9x\) because the expression is positive. The derivatives on either side of \(x = 3\) give different signs, indicating different slopes, so \(f'(3)\) does not exist.

Step 4: Evaluate f'(-3)

At \(x = -3\), \(f(x) = |0| = 0\). For \(x < -3\), \(f(x) = x^3 - 9x\) since \(x^3 - 9x\) is already negative. For \(x > -3\), \(f(x) = -(x^3 - 9x)\) because it's also negative within -3 to 0. Differentiation yields opposite signs, showing \(f'(-3)\) does not exist.

Step 5: Determine Extrema

Since \(f(x) = 0\) at \(x = -3, 0, 3\), these are candidate minima points. Since the differentiability fails at these points, check that all nearby points on \(f(x)\) are positive, confirming these are indeed minima. Check for other behavior changes to identify any potential extra extrema, but none are found beyond their associated analysis.

Key Concepts

Derivatives

In calculus, derivatives measure how a function changes as its input changes. It essentially tells us the rate at which the function's value is changing at any given point. To find a derivative, you usually perform a process called differentiation. This involves finding the limit of the average rate of change of the function as the interval approaches zero. However, for a derivative to exist at a certain point, the function must be smooth, or continuous, at that point.

When dealing with piecewise functions, like one involving an absolute value, the derivative can behave differently on either side of a point of interest. This is what happens with the function given in the exercise, particularly at points where the expression changes sign, such as at the zeros of the polynomial inside the absolute value.

• If a function has a sharp corner, cusp, or any discontinuity at a point, its derivative at that point does not exist.
• For instance, in the exercise with function \( f(x) = |x^3 - 9x| \), at \( x = 0, 3, -3 \), derivatives don't exist because of these characteristics of the function.

Absolute Value Function

The absolute value function, noted as \(|x|\), transforms any real number \(x\) into its non-negative counterpart. Hence, no matter if the input is negative or positive, the output will always be positive or zero. When analyzing the absolute value of another function or a polynomial, it turns the graph into a piecewise function.

The absolute value expression must be split into two separate functions to study its behavior properly:

• For the expression \(x^3 - 9x\), when it is negative (under the curve x-intercepts, \(x = -3, 0, 3\)), \( |x^3 - 9x| \) turns into \(- (x^3 - 9x)\).
• When it is positive (above those points), it remains \( (x^3 - 9x) \).

This causes the graph to have sharp edges or corners at the points where the expression changes from negative to positive, leading to potential issues in differentiability at those points.

Extrema

Extrema are the points at which a function takes its maximum or minimum value, either within a given range or across its entire domain. Knowing this helps identify critical points of a function, where it turns around or levels off. In the function \( f(x) = |x^3 - 9x| \), we look specifically at \( x = 0, 3, -3 \), as these are points where the derivative does not exist and create potential for minima.

These points are known as critical points and often help indicate which way the function is moving. Here's why minima are present at these points in our exercised function:

• \( f(x) = 0 \) at \( x = -3, 0, 3 \) since these zeros of the function transform from negative to positive values with the absolute value.
• In between or at these points, the function does not drop below zero, reinforcing them as minima.

The examination shows that no other extremum exists outside these specific points in our function given for analysis.

Calculus

Calculus is a branch of mathematics focused on change. It primarily deals with derivatives and integrals. The fundamental theorem of calculus links these two operations together. Derivatives measure the rate of change, while integrals compute accumulated change over time. These tools are essential for analyzing functions and understanding their properties like behavior, limits, and area under curves.

In practical applications, calculus helps us understand how functions behave under different conditions. For example, the given function \( f(x) = |x^3 - 9x| \) demonstrates key calculus principles: identifying non-differentiable points helps in understanding where the behavior of a function is irregular.

• Understanding the differentiation of polynomial expressions.
• Handling absolute value in calculus, which often results in non-smooth points.
• Determining extrema with the use of critical points.

The underlying ideas are vital to grasp the broader application and significance of calculus in various scientific and engineering fields.