Key Concepts and Formulas

• Differentiability of Absolute Value Functions: A function of the form $|g(x)|$ is not differentiable at points where $g(x) = 0$, unless $g'(x) = 0$ at those points as well. More generally, points where the argument of an absolute value function changes sign are critical points where differentiability needs to be checked.
• Piecewise Functions and Differentiability: For a piecewise function defined over different intervals, differentiability at the points where the definition changes (boundary points) must be checked by comparing the left-hand and right-hand derivatives.
• Finding Critical Points: To determine potential points of non-differentiability for a function involving absolute values, we identify the values of $x$ that make the arguments of the absolute value functions equal to zero.

Step-by-Step Solution

Step 1: Identify the critical points. The function is given by $f(x) = |2x + 1| - 3|x + 2| + |x^2 + x - 2|$. The critical points where the arguments of the absolute value functions can change sign are found by setting each argument to zero:

1. $2x + 1 = 0 \implies x = -\frac{1}{2}$
2. $x + 2 = 0 \implies x = -2$
3. $x^2 + x - 2 = 0 \implies (x + 2)(x - 1) = 0 \implies x = 1$ or $x = -2$.

The distinct critical points are $x = -2$, $x = -\frac{1}{2}$, and $x = 1$. These points divide the real number line into intervals.

Step 2: Determine the signs of the arguments in each interval. We need to analyze the sign of $2x+1$, $x+2$, and $x^2+x-2$ in the intervals defined by the critical points: $(-\infty, -2)$, $(-2, -\frac{1}{2})$, $(-\frac{1}{2}, 1)$, and $(1, \infty)$.

• Interval $(-\infty, -2)$:

  • $2x + 1 < 0$ (e.g., $x=-3 \implies -5$)
  • $x + 2 < 0$ (e.g., $x=-3 \implies -1$)
  • $x^2 + x - 2 = (x+2)(x-1) > 0$ (e.g., $x=-3 \implies (-1)(-4) = 4$) In this interval, $f(x) = -(2x+1) - 3(-(x+2)) + (x^2+x-2) = -2x-1 + 3x+6 + x^2+x-2 = x^2 + 2x + 3$.

• Interval $(-2, -\frac{1}{2})$:

  • $2x + 1 < 0$ (e.g., $x=-1 \implies -1$)
  • $x + 2 > 0$ (e.g., $x=-1 \implies 1$)
  • $x^2 + x - 2 = (x+2)(x-1) < 0$ (e.g., $x=-1 \implies (1)(-2) = -2$) In this interval, $f(x) = -(2x+1) - 3(x+2) + -(x^2+x-2) = -2x-1 - 3x-6 - x^2-x+2 = -x^2 - 6x - 5$.

• Interval $(-\frac{1}{2}, 1)$:

  • $2x + 1 > 0$ (e.g., $x=0 \implies 1$)
  • $x + 2 > 0$ (e.g., $x=0 \implies 2$)
  • $x^2 + x - 2 = (x+2)(x-1) < 0$ (e.g., $x=0 \implies (2)(-1) = -2$) In this interval, $f(x) = (2x+1) - 3(x+2) + -(x^2+x-2) = 2x+1 - 3x-6 - x^2-x+2 = -x^2 - 2x - 3$.

• Interval $(1, \infty)$:

  • $2x + 1 > 0$ (e.g., $x=2 \implies 5$)
  • $x + 2 > 0$ (e.g., $x=2 \implies 4$)
  • $x^2 + x - 2 = (x+2)(x-1) > 0$ (e.g., $x=2 \implies (4)(1) = 4$) In this interval, $f(x) = (2x+1) - 3(x+2) + (x^2+x-2) = 2x+1 - 3x-6 + x^2+x-2 = x^2 - 7$.

So, the piecewise definition of $f(x)$ is: f(x) = \left\{ {\matrix{ {{x^2} + 2x + 3;} & {x < - 2} \cr { - {x^2} - 6x - 5;} & { - 2 < x < {{ - 1} \over 2}} \cr { - {x^2} - 2x - 3;} & { - {1 \over 2} < x < 1} \cr {{x^2} - 7;} & {x > 1} \cr } } \right.

Step 3: Calculate the derivative in each open interval. We find the derivative of each piece:

• For $x < -2$, $f'(x) = \frac{d}{dx}(x^2 + 2x + 3) = 2x + 2$.
• For $-2 < x < -\frac{1}{2}$, $f'(x) = \frac{d}{dx}(-x^2 - 6x - 5) = -2x - 6$.
• For $-\frac{1}{2} < x < 1$, $f'(x) = \frac{d}{dx}(-x^2 - 2x - 3) = -2x - 2$.
• For $x > 1$, $f'(x) = \frac{d}{dx}(x^2 - 7) = 2x$.

The piecewise definition of $f'(x)$ for the open intervals is: f'(x) = \left\{ {\matrix{ {2x + 2;} & {x < - 2} \cr { - 2x - 6;} & { - 2 < x < {{ - 1} \over 2}} \cr { - 2x - 2;} & { - {1 \over 2} < x < 1} \cr {2x;} & {x > 1} \cr } } \right.

Step 4: Check differentiability at the critical points. We need to check if the left-hand derivative equals the right-hand derivative at $x = -2$, $x = -\frac{1}{2}$, and $x = 1$.

• At $x = -2$:

  • Left-hand derivative: $\lim_{x \to -2^-} f'(x) = \lim_{x \to -2^-} (2x + 2) = 2(-2) + 2 = -4 + 2 = -2$.
  • Right-hand derivative: $\lim_{x \to -2^+} f'(x) = \lim_{x \to -2^+} (-2x - 6) = -2(-2) - 6 = 4 - 6 = -2$. Since the left-hand derivative equals the right-hand derivative, the function is differentiable at $x = -2$.

• At $x = -\frac{1}{2}$:

  • Left-hand derivative: $\lim_{x \to (-\frac{1}{2})^-} f'(x) = \lim_{x \to (-\frac{1}{2})^-} (-2x - 6) = -2(-\frac{1}{2}) - 6 = 1 - 6 = -5$.
  • Right-hand derivative: $\lim_{x \to (-\frac{1}{2})^+} f'(x) = \lim_{x \to (-\frac{1}{2})^+} (-2x - 2) = -2(-\frac{1}{2}) - 2 = 1 - 2 = -1$. Since the left-hand derivative ($-5$) does not equal the right-hand derivative ($-1$), the function is not differentiable at $x = -\frac{1}{2}$.

• At $x = 1$:

  • Left-hand derivative: $\lim_{x \to 1^-} f'(x) = \lim_{x \to 1^-} (-2x - 2) = -2(1) - 2 = -2 - 2 = -4$.
  • Right-hand derivative: $\lim_{x \to 1^+} f'(x) = \lim_{x \to 1^+} (2x) = 2(1) = 2$. Since the left-hand derivative ($-4$) does not equal the right-hand derivative ($2$), the function is not differentiable at $x = 1$.

Step 5: Count the points of non-differentiability. The function is not differentiable at $x = -\frac{1}{2}$ and $x = 1$. Therefore, there are 2 points where the function is not differentiable.

Common Mistakes & Tips

• Incorrectly simplifying absolute values: Ensure that the sign of the expression inside the absolute value is correctly determined for each interval. A common mistake is to assume the sign is constant across an interval where it might change.
• Forgetting to check continuity: Before checking differentiability at a point, ensure the function is continuous at that point. If a function is not continuous, it cannot be differentiable. In this case, the piecewise functions were constructed to be continuous at the boundaries.
• Algebraic errors in differentiation or limit calculation: Double-check all algebraic manipulations, especially when dealing with negative signs and fractions.

Summary

To find the points of non-differentiability of the given function, we first identified the critical points where the arguments of the absolute value functions are zero. These points divided the real line into intervals. We then determined the sign of each argument in each interval to rewrite the function as a piecewise polynomial. Subsequently, we computed the derivative for each open interval. Finally, we checked the differentiability at the critical points by comparing the left-hand and right-hand derivatives. The function was found to be non-differentiable at $x = -\frac{1}{2}$ and $x = 1$.

The final answer is $\boxed{2}$.