Let 𝑓𝑓(𝑥𝑥) = |𝑥𝑥 + 3|,If f differentiable at x = -3? Use the alternate definition of a derivative to show why or why not. (Hint:𝑓𝑓′ (𝑐𝑐) = 𝑓𝑓(𝑥𝑥)−𝑓𝑓(𝑐𝑐) 𝑥𝑥−𝑐𝑐 )

To determine if \( f(x) = |x + 3| \) is differentiable at \( x = -3 \), we use the **alternate definition of a derivative**:

\[
f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}
\]

In this case, \( c = -3 \), so the expression becomes:

\[
f'(-3) = \lim_{x \to -3} \frac{f(x) - f(-3)}{x + 3}
\]

### Step 1: Evaluate \( f(-3) \)

We know that \( f(x) = |x + 3| \). At \( x = -3 \):

\[
f(-3) = |-3 + 3| = |0| = 0
\]

Thus, \( f(-3) = 0 \).

### Step 2: Set up the difference quotient

We now need to analyze the difference quotient on both sides of \( x = -3 \):

\[
f'( -3 ) = \lim_{x \to -3} \frac{|x + 3| - 0}{x + 3} = \lim_{x \to -3} \frac{|x + 3|}{x + 3}
\]

### Step 3: Left-hand limit (\( x \to -3^- \))

When \( x \to -3^- \) (approaching from the left), \( x + 3 \) is negative. Therefore, \( |x + 3| = -(x + 3) \).

Thus, the difference quotient becomes:

\[
\lim_{x \to -3^-} \frac{-(x + 3)}{x + 3} = \lim_{x \to -3^-} -1 = -1
\]

### Step 4: Right-hand limit (\( x \to -3^+ \))

When \( x \to -3^+ \) (approaching from the right), \( x + 3 \) is positive. Therefore, \( |x + 3| = (x + 3) \).

Thus, the difference quotient becomes:

\[
\lim_{x \to -3^+} \frac{x + 3}{x + 3} = \lim_{x \to -3^+} 1 = 1
\]

### Step 5: Conclusion

Since the left-hand limit (\( -1 \)) and the right-hand limit (\( 1 \)) are **not equal**, the derivative does not exist at \( x = -3 \). Therefore, \( f(x) = |x + 3| \) is **not differentiable at \( x = -3 \)**.

### Key Reason:
The absolute value function \( |x + 3| \) introduces a sharp corner at \( x = -3 \), causing a discontinuity in the slope from either side, preventing the derivative from existing at that point.

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Would you like further clarification, or do you have any questions?

Here are 5 related questions to consider:
1. How does the absolute value function affect differentiability?
2. What are the conditions for differentiability at a point?
3. Can a function be continuous but not differentiable at a point? Why?
4. What is the geometric interpretation of non-differentiability at sharp corners?
5. How can the piecewise definition of absolute value functions help in calculating derivatives?

**Tip:** Always check both left-hand and right-hand limits when verifying the existence of a derivative at a point, especially for piecewise or absolute value functions.

Let f(x) = |x + 3|. Is f differentiable at x = -3? Use the alternate definition of a derivative to show why or why not.

Explore whether the function f(x) = |x + 3| is differentiable at x = -3. Learn how to apply the alternate definition of a derivative, analyze limits, and understand why absolute value functions introduce sharp corners where differentiability fails. Suitable for Grades 11-12 or early college-level calculus.

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