Key Concepts and Formulas

• Definition of Differentiability: A function $g(x)$ is differentiable at a point $c$ if the limit of the difference quotient exists: $g'(c) = \lim_{x \to c} \frac{g(x) - g(c)}{x - c}$
• Properties of Absolute Value Function: For any real number $a$, $|a| \ge 0$. Also, $|a| = a$ if $a \ge 0$ and $|a| = -a$ if $a < 0$.
• Differentiability of $|f(x)|$ at a point where $f(c)=0$: If $f(x)$ is differentiable at $c$ and $f(c)=0$, then $|f(x)|$ is differentiable at $c$ if and only if $f'(c)=0$.

Step-by-Step Solution

Step 1: Understand the definition of the function g(x). We are given that $g(x) = |f(x)|$, and we know that $f$ is differentiable at $c$ and $f(c) = 0$. We need to determine the differentiability of $g(x)$ at $x=c$.

Step 2: Apply the definition of the derivative to g(x) at x = c. The derivative of $g(x)$ at $c$, denoted by $g'(c)$, is defined as: $g'(c) = \lim_{x \to c} \frac{g(x) - g(c)}{x - c}$

Step 3: Substitute the given information into the definition of the derivative. We are given $g(x) = |f(x)|$ and $f(c) = 0$. Since $g(c) = |f(c)|$, we have $g(c) = |0| = 0$. Substituting these into the limit expression: $g'(c) = \lim_{x \to c} \frac{|f(x)| - 0}{x - c}$ $g'(c) = \lim_{x \to c} \frac{|f(x)|}{x - c}$

Step 4: Analyze the limit by considering the sign of f(x) near c. Since $f$ is differentiable at $c$, it is also continuous at $c$. As $x \to c$, $f(x) \to f(c) = 0$. We need to consider two cases for the sign of $f(x)$ as $x$ approaches $c$:

• Case 1: f(x) > 0 as x approaches c. In this case, $|f(x)| = f(x)$. The limit becomes: $g'(c) = \lim_{x \to c} \frac{f(x)}{x - c}$ This is the definition of $f'(c)$. So, if $f(x) > 0$ near $c$, then $g'(c) = f'(c)$.

• Case 2: f(x) < 0 as x approaches c. In this case, $|f(x)| = -f(x)$. The limit becomes: $g'(c) = \lim_{x \to c} \frac{-f(x)}{x - c} = - \lim_{x \to c} \frac{f(x)}{x - c}$ This is equal to $-f'(c)$. So, if $f(x) < 0$ near $c$, then $g'(c) = -f'(c)$.

Step 5: Consider the condition for differentiability of g(x) at c. For $g'(c)$ to exist, the limit must be the same regardless of whether $f(x)$ is positive or negative near $c$. This means the value of the limit from Case 1 must equal the value of the limit from Case 2. Therefore, we must have: $f'(c) = -f'(c)$

Step 6: Solve for f'(c). Adding $f'(c)$ to both sides of the equation $f'(c) = -f'(c)$: $f'(c) + f'(c) = 0$ $2f'(c) = 0$ $f'(c) = 0$

Step 7: Interpret the result in relation to the given options. Our analysis shows that for $g(x) = |f(x)|$ to be differentiable at $x=c$ (given $f(c)=0$ and $f$ is differentiable at $c$), it is necessary that $f'(c) = 0$. Conversely, if $f'(c) = 0$, then $f'(c) = -f'(c)$, which implies the left-hand and right-hand limits for $g'(c)$ are equal. Let's confirm this. If $f'(c)=0$, then $\lim_{x \to c} \frac{f(x)}{x-c} = 0$. We need to check if $\lim_{x \to c} \frac{|f(x)|}{x - c}$ exists.

Consider the one-sided limits:

• Right-hand derivative: $\lim_{x \to c^+} \frac{|f(x)|}{x - c}$
• Left-hand derivative: $\lim_{x \to c^-} \frac{|f(x)|}{x - c}$

If $f'(c) = 0$, then $f(x) \approx f(c) + f'(c)(x-c) = 0 + 0(x-c) = 0$ for $x$ close to $c$. More precisely, $f(x)$ will have the same sign as $f'(c)(x-c)$ for $x$ sufficiently close to $c$ (if $f''(c)$ exists and is non-zero, or if $f'(c)$ is the first non-zero derivative). However, we don't assume $f''(c)$ exists.

Let's use the property that if $f'(c) = 0$, then $\lim_{x \to c} \frac{f(x)}{x-c} = 0$. We want to evaluate $\lim_{x \to c} \frac{|f(x)|}{x - c}$.

If $f(x)$ is identically zero in a neighborhood of $c$, then $g(x)=0$ and $g'(c)=0$. If $f(x)$ is not identically zero, then $f'(c)=0$ means that the tangent line to $f(x)$ at $c$ is horizontal. This implies that $f(x)$ touches the x-axis at $c$ without crossing it, or it crosses it.

Let's consider the definition of $f'(c)=0$: $\lim_{x \to c} \frac{f(x) - f(c)}{x - c} = 0 \implies \lim_{x \to c} \frac{f(x)}{x - c} = 0$.

This implies that for any $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x-c| < \delta$, then $|\frac{f(x)}{x-c}| < \epsilon$. This means $|f(x)| < \epsilon |x-c|$.

Now consider $\lim_{x \to c} \frac{|f(x)|}{x - c}$. We have $|\frac{|f(x)|}{x - c}| = \frac{|f(x)|}{|x - c|} = |\frac{f(x)}{x - c}|$. Since $\lim_{x \to c} |\frac{f(x)}{x - c}| = |\lim_{x \to c} \frac{f(x)}{x - c}| = |0| = 0$, by the Squeeze Theorem (or the definition of the limit of absolute value), we have: $\lim_{x \to c} \frac{|f(x)|}{x - c} = 0$.

This shows that if $f'(c) = 0$, then $g'(c) = 0$. Thus, $g(x)$ is differentiable at $x=c$.

Step 8: Evaluate the given options.

• (A) differentiable if $f'(c) = 0$: This matches our conclusion.
• (B) differentiable if $f'(c) \ne 0$: If $f'(c) \ne 0$, then $f'(c) \ne -f'(c)$, so $g(x)$ is not differentiable.
• (C) not differentiable: This is not always true; it depends on $f'(c)$.
• (D) not differentiable if $f'(c) = 0$: This is the opposite of our conclusion.

Our derivation confirms that $g(x)$ is differentiable at $x=c$ if and only if $f'(c)=0$.

Common Mistakes & Tips

• Confusing differentiability with continuity: While differentiability implies continuity, the converse is not true. The function $|x|$ is continuous at $x=0$ but not differentiable.
• Not considering the absolute value: Forgetting to handle the absolute value by considering cases for the sign of $f(x)$ will lead to incorrect conclusions.
• Assuming f(x) has a specific sign: When $f(c)=0$ and $f'(c) \ne 0$, $f(x)$ will change sign at $c$. If $f'(c)=0$, $f(x)$ might not change sign, or it might. The key is that for $|f(x)|$ to be differentiable, the "rate of change" of $f(x)$ must be zero at $c$.

Summary

We examined the differentiability of $g(x) = |f(x)|$ at $x=c$, given that $f$ is differentiable at $c$ and $f(c)=0$. By applying the definition of the derivative and considering the behavior of $|f(x)|$ near $c$, we found that the derivative of $g(x)$ at $c$ exists if and only if $f'(c) = 0$. This is because for the limit of the difference quotient of $|f(x)|$ to exist, the left-hand and right-hand derivatives must be equal, which leads to the condition $f'(c) = -f'(c)$, implying $f'(c)=0$.

The final answer is \boxed{A}.