Key Concepts and Formulas

• Composition of Functions (fog)(x): $(f \circ g)(x) = f(g(x))$. To find the composite function, substitute the inner function $g(x)$ into the outer function $f(x)$.
• Differentiability of Piecewise Functions: A piecewise function is differentiable at a point where the definition changes if:
  • It is continuous at that point.
  • The left-hand derivative (LHD) equals the right-hand derivative (RHD) at that point.
• Continuity Check at a Point 'a': A function $h(x)$ is continuous at $x=a$ if $\lim_{x \to a^-} h(x) = \lim_{x \to a^+} h(x) = h(a)$.
• Differentiability Check at a Point 'a': A function $h(x)$ is differentiable at $x=a$ if the limit $\lim_{x \to a} \frac{h(x) - h(a)}{x-a}$ exists. This is checked by comparing the LHD and RHD.
  • LHD: $\lim_{h \to 0^-} \frac{h(a+h) - h(a)}{h}$ or $\lim_{h \to 0^+} \frac{h(a-h) - h(a)}{-h}$
  • RHD: $\lim_{h \to 0^+} \frac{h(a+h) - h(a)}{h}$

Step-by-Step Solution

Step 1: Define the composite function (fog)(x). We are given $f(x) = \begin{cases} x+2, & x < 0 \\ x^2, & x \ge 0 \end{cases}$ and $g(x) = \begin{cases} x^3, & x < 1 \\ 3x-2, & x \ge 1 \end{cases}$. To find $(f \circ g)(x) = f(g(x))$, we need to consider the conditions on $g(x)$ and then apply them to $f$.

• Case 1: $g(x) < 0$. This occurs when $x^3 < 0$, which implies $x < 0$. In this case, $f(g(x)) = g(x) + 2 = x^3 + 2$. So, $(f \circ g)(x) = x^3 + 2$ for $x < 0$.

• Case 2: $g(x) \ge 0$. This requires us to consider the two sub-cases for $g(x)$.

  • Sub-case 2a: $x < 1$ and $g(x) \ge 0$. Here $g(x) = x^3$. So, $x^3 \ge 0$, which implies $x \ge 0$. Combining with $x < 1$, this sub-case is for $0 \le x < 1$. In this range, $f(g(x)) = f(x^3) = (x^3)^2 = x^6$. So, $(f \circ g)(x) = x^6$ for $0 \le x < 1$.

  • Sub-case 2b: $x \ge 1$ and $g(x) \ge 0$. Here $g(x) = 3x-2$. So, $3x-2 \ge 0$, which implies $x \ge 2/3$. Combining with $x \ge 1$, this sub-case is for $x \ge 1$. In this range, $f(g(x)) = f(3x-2) = (3x-2)^2$. So, $(f \circ g)(x) = (3x-2)^2$ for $x \ge 1$.

Combining all cases, the composite function $(f \circ g)(x)$ is: $(f \circ g)(x) = \begin{cases} x^3 + 2, & x < 0 \\ x^6, & 0 \le x < 1 \\ (3x-2)^2, & x \ge 1 \end{cases}$ Note: At $x=0$, $g(0) = 0^3 = 0$, so $f(g(0)) = f(0) = 0^2 = 0$. The first case gives $0^3+2 = 2$. The second case gives $0^6=0$. So, we need to be careful with the intervals. Let's re-evaluate the intervals.

Let $h(x) = (f \circ g)(x)$.

• If $x < 0$, then $g(x) = x^3 < 0$. Thus $f(g(x)) = g(x) + 2 = x^3 + 2$.
• If $0 \le x < 1$, then $g(x) = x^3$. Since $0 \le x < 1$, we have $0 \le x^3 < 1$. So $g(x) \ge 0$. Thus $f(g(x)) = (g(x))^2 = (x^3)^2 = x^6$.
• If $x \ge 1$, then $g(x) = 3x-2$. Since $x \ge 1$, $3x \ge 3$, so $3x-2 \ge 1$. Thus $g(x) \ge 0$. Thus $f(g(x)) = (g(x))^2 = (3x-2)^2$.

The composite function is: $h(x) = (f \circ g)(x) = \begin{cases} x^3 + 2, & x < 0 \\ x^6, & 0 \le x < 1 \\ (3x-2)^2, & x \ge 1 \end{cases}$ The points where differentiability might be an issue are $x=0$ and $x=1$, where the definition of the function changes.

Step 2: Check for continuity and differentiability at x = 0.

• Continuity at x = 0:

  • Left-hand limit: $\lim_{x \to 0^-} h(x) = \lim_{x \to 0^-} (x^3 + 2) = 0^3 + 2 = 2$.
  • Right-hand limit: $\lim_{x \to 0^+} h(x) = \lim_{x \to 0^+} x^6 = 0^6 = 0$.
  • Function value: $h(0) = 0^6 = 0$.

  Since $\lim_{x \to 0^-} h(x) \neq \lim_{x \to 0^+} h(x)$, the function $h(x)$ is discontinuous at $x=0$. A function that is discontinuous at a point cannot be differentiable at that point. Therefore, $h(x)$ is not differentiable at $x=0$.

Step 3: Check for continuity and differentiability at x = 1.

• Continuity at x = 1:

  • Left-hand limit: $\lim_{x \to 1^-} h(x) = \lim_{x \to 1^-} x^6 = 1^6 = 1$.
  • Right-hand limit: $\lim_{x \to 1^+} h(x) = \lim_{x \to 1^+} (3x-2)^2 = (3(1)-2)^2 = (1)^2 = 1$.
  • Function value: $h(1) = (3(1)-2)^2 = (1)^2 = 1$.

  Since $\lim_{x \to 1^-} h(x) = \lim_{x \to 1^+} h(x) = h(1) = 1$, the function $h(x)$ is continuous at $x=1$.

• Differentiability at x = 1: We need to calculate the Left-Hand Derivative (LHD) and the Right-Hand Derivative (RHD) at $x=1$.

  • RHD at x = 1: We use the definition $h'(1^+) = \lim_{h \to 0^+} \frac{h(1+h) - h(1)}{h}$. For $x \ge 1$, $h(x) = (3x-2)^2$. So $h(1) = (3(1)-2)^2 = 1$. $h(1+h) = (3(1+h)-2)^2 = (3+3h-2)^2 = (1+3h)^2$. RHD $= \lim_{h \to 0^+} \frac{(1+3h)^2 - 1}{h} = \lim_{h \to 0^+} \frac{1 + 6h + 9h^2 - 1}{h}$ RHD $= \lim_{h \to 0^+} \frac{6h + 9h^2}{h} = \lim_{h \to 0^+} (6 + 9h) = 6$.

  • LHD at x = 1: We use the definition $h'(1^-) = \lim_{h \to 0^+} \frac{h(1-h) - h(1)}{-h}$. For $0 \le x < 1$, $h(x) = x^6$. So $h(1) = 1$. $h(1-h) = (1-h)^6$. LHD $= \lim_{h \to 0^+} \frac{(1-h)^6 - 1}{-h}$. We can use the binomial expansion of $(1-h)^6 = 1 - 6h + \binom{6}{2}h^2 - \dots$. LHD $= \lim_{h \to 0^+} \frac{(1 - 6h + O(h^2)) - 1}{-h} = \lim_{h \to 0^+} \frac{-6h + O(h^2)}{-h}$ LHD $= \lim_{h \to 0^+} (6 + O(h)) = 6$.

  Alternatively, for LHD, we can differentiate the function $x^6$ and evaluate at $x=1$. The derivative of $x^6$ is $6x^5$. At $x=1$, the derivative is $6(1)^5 = 6$. This corresponds to the LHD.

  Since LHD = RHD = 6, the function $h(x)$ is differentiable at $x=1$.

Step 4: Determine the total number of points of non-differentiability. From Step 2, we found that $h(x)$ is not differentiable at $x=0$ due to discontinuity. From Step 3, we found that $h(x)$ is differentiable at $x=1$. The function $h(x)$ is composed of polynomials ($x^3+2$, $x^6$, $(3x-2)^2$) in the open intervals $(-\infty, 0)$, $(0, 1)$, and $(1, \infty)$. Polynomials are differentiable everywhere. Therefore, the only points where non-differentiability can occur are the points where the definition changes, i.e., $x=0$ and $x=1$.

We have identified only one point of non-differentiability, which is $x=0$.

Common Mistakes & Tips

• Incorrectly defining the composite function intervals: Carefully check the conditions on $g(x)$ and how they map to the conditions of $f(x)$. Ensure that the domain intervals for the composite function are correctly derived.
• Assuming differentiability after continuity: Even if a function is continuous at a point, it may not be differentiable there. Always check the LHD and RHD.
• Errors in calculating limits for LHD/RHD: Pay close attention to the signs and algebraic manipulations, especially when dealing with powers or binomial expansions in the limit calculation.

Summary

We first constructed the composite function $(f \circ g)(x)$ by substituting $g(x)$ into $f(x)$ and determining the correct intervals for each piece. We then analyzed the differentiability of this piecewise function at the points where its definition changes, namely $x=0$ and $x=1$. At $x=0$, we found that the function is discontinuous, which implies it is not differentiable. At $x=1$, we verified that the function is continuous and that its left-hand and right-hand derivatives are equal, meaning it is differentiable at $x=1$. Therefore, there is only one point where $(f \circ g)(x)$ is not differentiable.

The final answer is \boxed{0}.