Problem 19 - Calculus

Question

In Problems \(1-20\), find the numbers \(c\) that make \(f(x)\) into (A) a continuous function and (B) a differentiable function. In one case \(f(x) \rightarrow f(a)\) at every point, in the other case \(\Delta f / \Delta x\) has a limit at every point. $$ f(x)=\left\\{\begin{array}{cc} (\sin x-x) / x^{e} & x \neq 0 \\ 0 & x=0 \end{array}\right. $$

Step-by-Step Solution

Answer

For continuity, the condition is \( c \geq 3 \); for differentiability, it remains \( c \geq 3 \).

Step 1: Understanding the Function

The function given is a piecewise function where \( f(x) = \frac{\sin x - x}{x^e} \) for \( x eq 0 \) and \( f(x) = 0 \) for \( x = 0 \). We need to examine the behavior of the function around \( x = 0 \).

Step 2: Continuity at x = 0

For continuity at \( x = 0 \), \( \lim_{x \to 0} f(x) = f(0) = 0 \). We need \( \lim_{x \to 0} \frac{\sin x - x}{x^e} = 0 \). Simplifying, this requires that the numerator's limit approaches zero faster than \( x^e \). Using L'Hôpital's Rule might be a valid approach, two times eventual.

Step 3: Using L'Hôpital's Rule

The indeterminate form \( \frac{0}{0} \) allows the use of L'Hôpital's Rule. First derivative gives \( \frac{\cos x - 1}{ex^{e-1}} \). Apply L'Hôpital's Rule again to get \( \frac{-\sin x}{e(e-1)x^{e-2}} \). Evaluate this limit as \( x \to 0 \), which approaches zero only if \( e \geq 3 \).

Step 4: Differentiability at x = 0

For differentiability, the derivative must exist and be continuous at \( x = 0 \). This means both one-sided derivatives as \( x \to 0 \) should be equal to \( f'(0) \). Calculate \( f'(x) \) at \( x eq 0 \) from the limit of \( \frac{h(x-h) - h}{h^e} \).

Step 5: Computing the Derivative for x ≠ 0

Using the derivative formula \( \lim_{h \to 0} \frac{f(h+h)-f(h)}{h} \), simplify and evaluate. The limit must evaluate to zero for \( e \geq 3 \) which ensures \( f'(0) = 0 \).

Step 6: Conclusion for Differentiability

Since \( e \geq 3 \) ensures both continuity and \( f'(x) = 0 \) at \( x = 0 \), \( c = e \) makes \( f(x) \) differentiable. Any greater value confirms all required properties.

Key Concepts

Piecewise Function

A piecewise function is a function that is defined by different expressions depending on the values of the input variable. In this exercise, the function \( f(x) \) is defined differently for two scenarios:

• \( f(x) = \frac{\sin x - x}{x^e} \) when \( x eq 0 \)
• \( f(x) = 0 \) when \( x = 0 \)

These set-ups allow the function to change dramatically at a particular point or to behave differently across its domain. In our problem, the focus is on how \( f(x) \) behaves around \( x = 0 \), which is called a 'point of discontinuity' or a 'knot'. If not handled properly, such points can make the function seem jagged or undefined, hence the need to ensure continuity and differentiability. Analyzing such functions often requires considering the limits of each piece and how they transition between each defined section.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful method for calculating limits that have an indeterminate form, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This technique involves taking the derivative of the numerator and the derivative of the denominator until the indeterminate form is resolved.
In the exercise, L'Hôpital's Rule is applied to evaluate the limit \( \lim_{x \to 0} \frac{\sin x - x}{x^e} \). Initially, this limit creates the indeterminate form \( \frac{0}{0} \). By applying L'Hôpital's Rule, we find:

• The first derivative \( \frac{\cos x - 1}{ex^{e-1}} \)
• The second derivative \( \frac{-\sin x}{e(e-1)x^{e-2}} \)

After applying L'Hôpital's Rule twice, the function's limit at \( x \to 0 \) approaches zero if \( e \geq 3 \). This result helps in establishing continuity at \( x = 0 \), which is crucial for the piecewise function to be well-behaved at that point.

Derivative Calculation

Derivative calculation involves determining the rate at which a function changes at any point on its curve. For the function \( f(x) = \frac{\sin x - x}{x^e} \), calculating its derivative at points where \( x eq 0 \) is essential to examine its differentiability.
We use the limit process to calculate the derivative, given by the formula:\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]In our situation, the challenge is to determine that this limit leads to a negligible derivative at \( x = 0 \), particularly because the function must be smooth (derivatives from either side should match) at this juncture.

• Calculate the limit to see \( f'(x) \), ensuring it evaluates to zero.
• If \( e \geq 3 \), the derivative existence, when taken around \( x =0 \), reflects a continuous behavior at this point.

The derivative calculation confirms that \( f(x) \) is both continuous and differentiable when \( e \geq 3 \), ensuring the function is consistently smooth around the critical point.