Problem 22 - The Calculus with Analytic Geometry

Question

Given $$ f(x)= \begin{cases}0 & \text { if } x \leq 0 \\ x^{n} & \text { if } x>0\end{cases} $$ where \(n\) is a positive integer. (a) For what values of \(n\) is \(f\) differentiable for all values of \(x ?\) (b) For what values of \(n\) is \(f^{\prime}\) continuous for all values of \(x ?\)

Step-by-Step Solution

Answer

(a) \(f(x)\) is differentiable for all \(x\) if \(n > 1\). (b) \(f'(x)\) is continuous for all \(x\) if \(n > 2\).

Step 1: Understand the Function

The function is piecewise defined: - For \(x \leq 0\), \(f(x) = 0\) - For \(x > 0\), \(f(x) = x^n\)

Step 2: Compute the Left-Hand and Right-Hand Derivatives

To find the derivative of \(f(x)\), consider the two cases separately and check the behavior at \(x = 0\). - For \(x \leq 0\), \(f(x) = 0\) implies \(f'(x) = 0\). - For \(x > 0\), \(f(x) = x^n\) implies \(f'(x) = nx^{n-1}\).

Step 3: Examine Differentiability at \(x = 0\)

To check differentiability at \(x = 0\), ensure that the left-hand and right-hand derivatives equal the same value: - Left-hand derivative at \(x = 0\) is \(0\). - Right-hand derivative as \(x \to 0^+\) is \(nx^{n-1}\). For these to be equal, the limit of \(nx^{n-1}\) as \(x\) approaches \(0\) must be \(0\). This happens if \(nx^{n-1} \to 0\), which requires \(n-1 > 0\) or \(n > 1\).

Step 4: Examine Continuity of \(f'(x)\)

To ensure that \(f'(x)\) is continuous for all values of \(x\), we need to check the continuity at \(x = 0\): - For \(x \leq 0\), \(f'(x) = 0\). - For \(x > 0\), \(f'(x) = nx^{n-1}\). The continuity of \(f'(x)\) at \(x = 0\) requires \[ \lim_{{x \to 0^+}} f'(x) = 0 \] which happens if \(n-1 > 1\) or \(n > 2\).

Step 5: Summarize the Conditions

(a) \(f(x)\) is differentiable for all values of \(x\) if \(n > 1\). (b) \(f'(x)\) is continuous for all values of \(x\) if \(n > 2\).

Key Concepts

Piecewise Functions

A piecewise function is defined by multiple sub-functions, each applying to a certain interval of the domain. These intervals are specified by conditions. For instance, in our exercise, the function \(f(x) = \begin{cases} 0 & \text{if } x \leq 0 \ x^n & \text{if } x > 0 \end{cases}\) is a piecewise function. It changes its definition at the point where \(x=0\).
Understanding how these functions behave at the points where their definitions change (known as breaking points) is crucial. It's important to analyze the function on both sides of these points to fully grasp its behavior.
In our function, for \(x \leq 0\), \(f(x) = 0\). For \(x > 0\), \(f(x) = x^n\). The point of interest is \(x = 0\) as this is where the definition changes.

Left-Hand and Right-Hand Derivatives

To determine the differentiability of a piecewise function at a point, we need to check both the left-hand and right-hand derivatives at that point. This means we look at the derivatives as we approach the point from both sides.
In our example, we need to compute the derivatives at \(x = 0\) from both sides:

• The left-hand derivative when \(x \leq 0\): Since \(f(x) = 0\), we have \(f'(x) = 0\).
• The right-hand derivative when \(x > 0\): Here, \(f(x) = x^n\), so \(f'(x) = nx^{n-1}\).

For \(f(x)\) to be differentiable at \(x = 0\), both derivatives must be equal at \(x = 0\).
We examine the limit of the right-hand derivative as \(x\) approaches \(0\): \(\lim_{{x \to 0^+}} nx^{n-1}\). This limit is \(0\) only if \(n-1 > 0\) or \(n > 1\). So, \(f(x)\) is differentiable at \(x=0\) if \(n > 1\).

Continuity of Derivatives

A function's derivative is continuous if there are no jumps or breaks in the derivative's value across its domain. For \(f'(x)\) to be continuous at \(x = 0\), \(f'(x)\) must have the same value as we approach \(x = 0\) from both the left and right.

• For \(x \leq 0\), \(f'(x) = 0\).
• For \(x > 0\), \(f'(x) = nx^{n-1}\).

We need the right-hand derivative's limit to be \(0\) at \(x=0\): \lim_{{x \to 0^+}} nx^{n-1}=0. For this to be the case, \(n-1\) must be strictly greater than \(1\), i.e., \(n > 2\).
Therefore, \(f'(x)\) is continuous for all \(x\) if \(n > 2\).