Math Problem Statement

Let 𝑓𝑓(π‘₯π‘₯) = |π‘₯π‘₯ + 3|,If f differentiable at x = -3? Use the alternate definition of a derivative to show why or why not. (Hint:𝑓𝑓′ (𝑐𝑐) = 𝑓𝑓(π‘₯π‘₯)βˆ’π‘“π‘“(𝑐𝑐) π‘₯π‘₯βˆ’π‘π‘ )

Solution

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

fβ€²(c)=lim⁑xβ†’cf(x)βˆ’f(c)xβˆ’cf'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}

In this case, c=βˆ’3c = -3, so the expression becomes:

fβ€²(βˆ’3)=lim⁑xβ†’βˆ’3f(x)βˆ’f(βˆ’3)x+3f'(-3) = \lim_{x \to -3} \frac{f(x) - f(-3)}{x + 3}

Step 1: Evaluate f(βˆ’3)f(-3)

We know that f(x)=∣x+3∣f(x) = |x + 3|. At x=βˆ’3x = -3:

f(βˆ’3)=βˆ£βˆ’3+3∣=∣0∣=0f(-3) = |-3 + 3| = |0| = 0

Thus, f(βˆ’3)=0f(-3) = 0.

Step 2: Set up the difference quotient

We now need to analyze the difference quotient on both sides of x=βˆ’3x = -3:

fβ€²(βˆ’3)=lim⁑xβ†’βˆ’3∣x+3βˆ£βˆ’0x+3=lim⁑xβ†’βˆ’3∣x+3∣x+3f'( -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 (xβ†’βˆ’3βˆ’x \to -3^-)

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

Thus, the difference quotient becomes:

lim⁑xβ†’βˆ’3βˆ’βˆ’(x+3)x+3=lim⁑xβ†’βˆ’3βˆ’βˆ’1=βˆ’1\lim_{x \to -3^-} \frac{-(x + 3)}{x + 3} = \lim_{x \to -3^-} -1 = -1

Step 4: Right-hand limit (xβ†’βˆ’3+x \to -3^+)

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

Thus, the difference quotient becomes:

lim⁑xβ†’βˆ’3+x+3x+3=lim⁑xβ†’βˆ’3+1=1\lim_{x \to -3^+} \frac{x + 3}{x + 3} = \lim_{x \to -3^+} 1 = 1

Step 5: Conclusion

Since the left-hand limit (βˆ’1-1) and the right-hand limit (11) are not equal, the derivative does not exist at x=βˆ’3x = -3. Therefore, f(x)=∣x+3∣f(x) = |x + 3| is not differentiable at x=βˆ’3x = -3.

Key Reason:

The absolute value function ∣x+3∣|x + 3| introduces a sharp corner at x=βˆ’3x = -3, causing a discontinuity in the slope from either side, preventing the derivative from existing at that point.


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.

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Math Problem Analysis

Mathematical Concepts

Calculus
Differentiation
Absolute Value Functions
Limits

Formulas

f'(c) = lim (x β†’ c) [f(x) - f(c)] / (x - c)
f(x) = |x + 3|

Theorems

Non-differentiability at sharp corners
Left-hand limit and Right-hand limit

Suitable Grade Level

Grades 11-12 or introductory college-level calculus