Understanding the Derivative of a Piecewise Function

The question asks us to find the correct expression for the derivative, \(f'(x)\), of the given piecewise function defined over the interval \((0, \infty)\). A piecewise function has different definitions for different intervals of its domain.

The function is given as:

\( f(x) = \begin{cases} 1-x^2 & for& , 0 < x \leq 1 \quad \\ In \ x & for &, 1 < x \leq 2 \\ In \ 2 - 1 + 0.5x & for &, 2 < x < \infty \end{cases} \)

To find the derivative \(f'(x)\), we need to find the derivative of each piece of the function with respect to \(x\) for the corresponding open intervals. At the points where the definition changes (\(x=1\) and \(x=2\)), the derivative may or may not exist. The derivative \(f'(x)\) is generally defined for open intervals.

Calculating the Derivative for Each Interval

Let's calculate the derivative for each defined interval:

1. For the interval \(0 < x \leq 1\), the function is \(f(x) = 1 - x^2\).
   • The derivative for the open interval \(0 < x < 1\) is:
   • \( f'(x) = \frac{d}{dx}(1 - x^2) = \frac{d}{dx}(1) - \frac{d}{dx}(x^2) = 0 - 2x = -2x \)
   • So, for \(0 < x < 1\), \(f'(x) = -2x\).
2. For the interval \(1 < x \leq 2\), the function is \(f(x) = \ln x\).
   • The derivative for the open interval \(1 < x < 2\) is:
   • \( f'(x) = \frac{d}{dx}(\ln x) = \frac{1}{x} \)
   • So, for \(1 < x < 2\), \(f'(x) = \frac{1}{x}\).
3. For the interval \(2 < x < \infty\), the function is \(f(x) = \ln 2 - 1 + 0.5x\).
   • The derivative for the open interval \(2 < x < \infty\) is:
   • \( f'(x) = \frac{d}{dx}(\ln 2 - 1 + 0.5x) = \frac{d}{dx}(\ln 2) - \frac{d}{dx}(1) + \frac{d}{dx}(0.5x) = 0 - 0 + 0.5 = 0.5 \)
   • So, for \(2 < x < \infty\), \(f'(x) = 0.5\).

Checking Differentiability at Endpoints

The derivative \(f'(x)\) is defined for open intervals. At points where the function definition changes, such as \(x=1\) and \(x=2\), we need to check if the function is differentiable. Differentiability at a point requires the left-hand derivative and the right-hand derivative to exist and be equal at that point. Also, the function must be continuous at that point for it to be differentiable there.

Check at x = 1:

First, let's check continuity at \(x=1\):

• Left limit: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (1 - x^2) = 1 - 1^2 = 0 \)
• Right limit: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (\ln x) = \ln 1 = 0 \)
• Function value: \( f(1) = 1 - 1^2 = 0 \) (from the first definition)

Since the left limit, right limit, and function value are all equal (\(0\)), the function is continuous at \(x=1\).

Now, let's check differentiability at \(x=1\) by comparing the derivatives from the left and right sides:

• Left derivative: For \(x < 1\), \(f'(x) = -2x\). As \(x \to 1^-\), the left-hand derivative \(f'(1^-) = \lim_{x \to 1^-} (-2x) = -2(1) = -2\).
• Right derivative: For \(x > 1\), \(f'(x) = \frac{1}{x}\). As \(x \to 1^+\), the right-hand derivative \(f'(1^+) = \lim_{x \to 1^+} (\frac{1}{x}) = \frac{1}{1} = 1\).

Since the left-hand derivative (\(-2\)) and the right-hand derivative (\(1\)) are not equal at \(x=1\), the function \(f(x)\) is not differentiable at \(x=1\).

Check at x = 2:

First, let's check continuity at \(x=2\):

• Left limit: \( \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (\ln x) = \ln 2 \)
• Right limit: \( \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (\ln 2 - 1 + 0.5x) = \ln 2 - 1 + 0.5(2) = \ln 2 - 1 + 1 = \ln 2 \)
• Function value: \( f(2) = \ln 2 \) (from the second definition)

Since the left limit, right limit, and function value are all equal (\(\ln 2\)), the function is continuous at \(x=2\).

Now, let's check differentiability at \(x=2\):

• Left derivative: For \(x < 2\), \(f'(x) = \frac{1}{x}\). As \(x \to 2^-\), the left-hand derivative \(f'(2^-) = \lim_{x \to 2^-} (\frac{1}{x}) = \frac{1}{2} = 0.5\).
• Right derivative: For \(x > 2\), \(f'(x) = 0.5\). As \(x \to 2^+\), the right-hand derivative \(f'(2^+) = \lim_{x \to 2^+} (0.5) = 0.5\).

Since the left-hand derivative (\(0.5\)) and the right-hand derivative (\(0.5\)) are equal at \(x=2\), the function \(f(x)\) is differentiable at \(x=2\), and \(f'(2) = 0.5\).

Summarizing the Derivative f'(x)

Based on our calculations, the derivative \(f'(x)\) is:

• \(f'(x) = -2x\) for \(0 < x < 1\)
• \(f'(x) = \frac{1}{x}\) for \(1 < x < 2\)
• \(f'(x) = 0.5\) for \(x = 2\)
• \(f'(x) = 0.5\) for \(2 < x < \infty\)

Combining the last two, we have \(f'(x) = 0.5\) for \(2 \leq x < \infty\). So, the derivative function is:

\( f'(x) = \begin{cases} -2x & for& , 0 < x < 1 \quad \\ \frac{1}{x} & for &, 1 < x < 2 \\ 0.5 & for &, 2 \leq x < \infty \end{cases} \)

Note that \(f'(1)\) is undefined because the function is not differentiable at \(x=1\).

Evaluating the Options

Let's look at the provided options:

1. f’(x) = 2x for 0 < x ≤ 1: This is incorrect. The derivative is \(-2x\) for \(0 < x < 1\).
2. f’(x) = -2x for 0 < x ≤ 1: This is incorrect because the function is not differentiable at \(x=1\), so \(f'(x)\) is not equal to \(-2x\) (or any value) for the *closed* interval \(0 < x \le 1\). It's only \(-2x\) for the *open* interval \(0 < x < 1\).
3. f’(x) = -2x for 0 < x < 1: This matches our calculated derivative for the open interval \(0 < x < 1\).
4. f’(x) = 0 for 0 < x < ∞: This is incorrect. The derivative is not 0 over the entire domain.

Based on our analysis, the statement "f’(x) = -2x for 0 < x < 1" accurately describes the derivative of the function over that specific open interval.

Conclusion on Derivative of Piecewise Function

The correct statement about the derivative of the function \(f(x)\) is that \(f'(x) = -2x\) for the open interval \(0 < x < 1\). The function is not differentiable at \(x=1\), so the derivative is not defined at that point, and thus the statement cannot hold for \(0 < x \le 1\).

Revision Table: Key Concepts

Additional Information: Piecewise Differentiability

When dealing with the derivative of a piecewise function, it is essential to compute the derivative for each definition over its corresponding open interval. The points where the function definition changes are called "transition points" or "split points."

To determine if the derivative exists at a transition point \(c\), you must perform two checks:

1. Continuity Check: Verify if \(f(x)\) is continuous at \(x=c\). This means \( \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c) \). If the function is not continuous at \(c\), it cannot be differentiable at \(c\).
2. Differentiability Check: If the function is continuous at \(c\), check if the left-hand derivative equals the right-hand derivative at \(c\). This is often done by checking if \( \lim_{x \to c^-} f'(x) = \lim_{x \to c^+} f'(x) \), provided \(f'(x)\) is continuous on either side near \(c\). More formally, it involves evaluating \( \lim_{h \to 0^-} \frac{f(c+h) - f(c)}{h} \) and \( \lim_{h \to 0^+} \frac{f(c+h) - f(c)}{h} \) and seeing if they are equal.

If both checks pass, the function is differentiable at \(c\), and the derivative \(f'(c)\) exists. If either check fails, the function is not differentiable at \(c\), and \(f'(c)\) is undefined.

In this problem, we found that the function is not differentiable at \(x=1\) because the left and right derivatives are not equal, even though the function is continuous there. At \(x=2\), the function is continuous and the left and right derivatives are equal, so it is differentiable at \(x=2\).