Problem 4 (6 points ): Let f(x) = |x ^ 3 + x ^ 2 - x - 1| . Analyze the differentiability of f(x) and determine at which points, if any, f(x) is differentiable. Justify your answer graphically (using Desmos or a Graphing Calculator


The function \( f(x) = |x^3 + x^2 - x - 1| \) involves the absolute value of a cubic polynomial. To analyze its differentiability, we need to focus on two aspects:

1. **Differentiability of the underlying cubic polynomial**: The polynomial \( g(x) = x^3 + x^2 - x - 1 \) is differentiable everywhere because polynomials are smooth and differentiable on \( \mathbb{R} \).

2. **Effect of the absolute value on differentiability**: The absolute value function can cause non-differentiability where the argument inside the absolute value changes sign (i.e., where \( g(x) = 0 \)).

### Steps to Analyze Differentiability:

#### Step 1: Find where the polynomial \( g(x) = x^3 + x^2 - x - 1 \) is zero
We need to solve:
\[
x^3 + x^2 - x - 1 = 0
\]
By trial or synthetic division, you can find that \( x = -1 \) is a root. So we factor the cubic as:
\[
x^3 + x^2 - x - 1 = (x + 1)(x^2 - 1)
\]
Further factoring gives:
\[
(x + 1)(x - 1)(x + 1) = (x + 1)^2(x - 1)
\]
Thus, the zeros of the cubic are \( x = -1 \) (a double root) and \( x = 1 \) (a single root).

#### Step 2: Analyze differentiability at \( x = -1 \) and \( x = 1 \)
- At \( x = -1 \): Since \( g(x) = 0 \) and \( g'(x) \) is also 0 here (due to the double root), there is no cusp or corner at this point. Hence, \( f(x) \) should be differentiable at \( x = -1 \).
- At \( x = 1 \): \( g(x) = 0 \), but \( g'(x) \neq 0 \). This suggests that there will be a cusp or corner at \( x = 1 \), making \( f(x) \) non-differentiable at this point.

#### Step 3: Verify using the derivative
Let’s compute the derivative of \( f(x) \) piecewise:
- For \( g(x) \geq 0 \) (i.e., \( x \leq -1 \) or \( x \geq 1 \)), \( f(x) = g(x) \), so \( f'(x) = g'(x) \).
- For \( g(x) < 0 \) (i.e., \( -1 < x < 1 \)), \( f(x) = -g(x) \), so \( f'(x) = -g'(x) \).

At \( x = 1 \), the derivative will not match from the left and right, confirming a cusp.

#### Step 4: Graphical Justification
Graphing \( f(x) \) on a tool like Desmos would show a smooth transition through \( x = -1 \), but a sharp corner (non-differentiable point) at \( x = 1 \).

### Conclusion:
- \( f(x) \) is differentiable everywhere except at \( x = 1 \), where it has a cusp.
  
Would you like more details or have any questions?

Here are five follow-up questions to deepen understanding:
1. Why is differentiability lost at cusps or corners?
2. How can the double root at \( x = -1 \) affect smoothness and differentiability?
3. What is the graphical behavior of \( g(x) = x^3 + x^2 - x - 1 \) near the roots?
4. How do you interpret the role of \( f'(x) \) at points of non-differentiability?
5. Can you extend this analysis to higher-degree polynomials within absolute values?

**Tip**: Always check for points where the argument inside the absolute value is zero when determining differentiability.

Problem 4 (6 points ): Let f(x) = |x ^ 3 + x ^ 2 - x - 1| . Analyze the differentiability of f(x) and determine at which points, if any, f(x) is differentiable. Justify your answer graphically (using Desmos or a Graphing Calculator).

Explore the differentiability of f(x) = |x^3 + x^2 - x - 1|, focusing on how the absolute value impacts smoothness at certain points. Learn step-by-step how to find where f(x) is differentiable, analyze graphical behavior, and identify the cusp at x=1. Suitable for students in Grades 11-12.

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