Problem 68 - Thomas Calculus in SI Units

Question

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?

Step-by-Step Solution

Answer

f'(0) does not exist; f'(3) does not exist; f'(-3) does not exist.

Step 1: Analyze f(x)

The function given is \(f(x) = |x^3 - 9x|\). This is a piecewise function based on the expression inside the absolute value.

Step 2: Determine Critical Points

The expression inside the absolute value is \(x^3 - 9x\). Set it equal to zero: \(x^3 - 9x = 0\). Factor it to find the critical points: \(x(x^2 - 9) = 0\), giving us \(x = 0, x = 3, x = -3\). These points are where the derivative might change behavior.

Step 3: Evaluate f'(0)

Evaluate the derivative \(f'(x)\) at \(x = 0\). At \(x = 0\), the function \(f(x)\) changes from \(x^3-9x\) for \(x > 0\) to \(-(x^3-9x)\) for \(x < 0\). Compute the right-hand and left-hand derivatives at \(x = 0\):- Right-hand derivative: \(f'_+(0) = \lim_{{h \to 0^+}} \frac{(h)^3 - 9(h)}{h} = -9\).- Left-hand derivative: \(f'_-(0) = \lim_{{h \to 0^-}} -\frac{(h)^3 - 9(h)}{h} = 9\).Since \(f'_+(0) eq f'_-(0)\), \(f'(0)\) does not exist.

Step 4: Evaluate f'(3)

Evaluate the derivative \(f'(x)\) at \(x = 3\). At \(x = 3\), note that the argument of the absolute value \(x^3 - 9x\) is zero. Compute the derivatives on both sides:- Right-hand derivative: \(f'_+(3) = \lim_{{h \to 0^+}} \frac{(27 + 27h + 9h^2 + h^3) - 27}{h} = 18\).- Left-hand derivative: \(f'_-(3) = \lim_{{h \to 0^-}} -\frac{(27 + 27h + 9h^2 + h^3) - 27}{h} = -18\).Since \(f'_+(3) eq f'_-(3)\), \(f'(3)\) does not exist.

Step 5: Evaluate f'(-3)

Evaluate the derivative \(f'(x)\) at \(x = -3\). At \(x = -3\), the argument of the absolute value \(x^3 - 9x\) is zero. Compute the derivatives on both sides:- Right-hand derivative: \(f'_+(-3) = \lim_{{h \to 0^+}} \frac{(-(27 + 27h + 9h^2 + h^3)) + 27}{h} = 18\).- Left-hand derivative: \(f'_(-3^-) = \lim_{{h \to 0^-}} \frac{(27 + 27h + 9h^2 + h^3) - 27}{h} = -18\).Since \(f'_+(-3) eq f'_-(3)\), \(f'(-3)\) does not exist.

Key Concepts

Piecewise Functions

A piecewise function is a function that is defined by multiple sub-functions, each of which applies to a specific interval in the domain. This means, instead of a single formula, several different formulas are used, depending on the value of the input variable. For our exercise, the function is defined as \( f(x) = |x^3 - 9x| \). Since it involves an absolute value, it implies that the function changes behavior, depending on whether the expression inside the absolute value is positive, negative, or zero.

Here's how this works for \( f(x) \):

• When \( x^3 - 9x \geq 0 \) (i.e., when \( x \) is outside the interval between the roots -3 and 3), \( f(x) = x^3 - 9x \).
• When \( x^3 - 9x < 0 \) (i.e., when \( x \) is between the roots -3 and 3), \( f(x) = -(x^3 - 9x) \).

By understanding piecewise functions, we can more easily determine where and why a function may not have a derivative at certain points.

Critical Points

Critical points of a function are values of \(x\) where the function changes its direction or behavior which can include turning points or points where the derivative does not exist. Critical points can be found by setting the derivative of the function to zero or finding where the derivative does not exist. For piecewise functions such as our \( f(x) = |x^3 - 9x| \), it’s useful to look at where the inside function \(x^3 - 9x\) equals zero, since these can also be points of change.

For \(x^3 - 9x = 0\), factor to get:

• \( x(x^2 - 9) = 0 \)
• Which results in the solutions: \( x = 0, 3, -3 \)

These are the critical points. They are potential locations where \( f(x) \) might not be smooth or might change direction. Evaluating the derivative at these points will help us confirm whether it exists or not.

Derivative Existence

The existence of a derivative at a given point tells us whether the function is smooth and well-behaved at that location. Derivatives mathematically measure how a function changes as its input changes. For the derivative to exist at a point, both the left-hand and right-hand derivatives must equal each other at that point.

In our example, we checked the derivative at \(x = 0\), \(x = 3\), and \(x = -3\).

For \(x = 0\):

• The right-hand derivative gives \(f'_+(0) = -9\),
• and the left-hand derivative gives \(f'_-(0) = 9\).

Since they are not equal, it shows \(f'(0)\) does not exist.

Likewise, for \(x = 3\) and \(x = -3\):

• For \(x = 3\), \(f'_+(3) = 18\) and \(f'_-(3) = -18\),
• For \(x = -3\), \(f'_+(-3) = 18\) and \(f'_-(3) = -18\).

In each case, the mismatched derivatives confirm there is no derivative at these points. Understanding when and why a derivative does not exist helps diagnose points of sharp turns or cusps within a function.