Number of Non Differentiable Points of Piecewise Defined Function

To determine the total number of points at which $h(x)$ fails to be differentiable, we first write explicit algebraic expressions for $h(x)$ across different intervals of $x$.

The given functions are:

$$f(x) = |x| + 1 = \begin{cases} 1 - x & \text{if } x \le 0, \\ 1 + x & \text{if } x > 0, \end{cases}$$

$g(x) = x^2 + 1.$

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Step 1: Definition of $h(x)$ for $x \le 0$

For $x \le 0$, we have $h(x) = \max\{f(x), g(x)\} = \max\{1 - x, x^2 + 1\}$.

Comparing $f(x)$ and $g(x)$: $f(x) - g(x) = (1 - x) - (x^2 + 1) = -x(x + 1).$

• Interval $x < -1$: Here, $x < 0$ and $x + 1 < 0$, so $-x(x + 1) < 0 \implies f(x) < g(x)$. $\implies h(x) = g(x) = x^2 + 1.$

• Interval $-1 \le x \le 0$: Here, $x \le 0$ and $x + 1 \ge 0$, so $-x(x + 1) \ge 0 \implies f(x) \ge g(x)$. $\implies h(x) = f(x) = 1 - x.$

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Step 2: Definition of $h(x)$ for $x > 0$

For $x > 0$, we have $h(x) = \min\{f(x), g(x)\} = \min\{1 + x, x^2 + 1\}$.

Comparing $f(x)$ and $g(x)$: $f(x) - g(x) = (1 + x) - (x^2 + 1) = x(1 - x).$

• Interval $0 < x < 1$: Here, $x > 0$ and $1 - x > 0$, so $x(1 - x) > 0 \implies f(x) > g(x)$. $\implies h(x) = g(x) = x^2 + 1.$

• Interval $x \ge 1$: Here, $x > 0$ and $1 - x \le 0$, so $x(1 - x) \le 0 \implies f(x) \le g(x)$. $\implies h(x) = f(x) = 1 + x.$

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Step 3: Piecewise Definition of $h(x)$

Combining the sub-intervals, we get:

$$h(x) = \begin{cases} x^2 + 1 & \text{if } x < -1, \\ 1 - x & \text{if } -1 \le x \le 0, \\ x^2 + 1 & \text{if } 0 < x < 1, \\ x + 1 & \text{if } x \ge 1. \end{cases}$$

Since polynomials are smooth (infinitely differentiable) on open intervals, $h(x)$ is differentiable on $\mathbb{R} \setminus \{-1, 0, 1\}$. We now analyze differentiability at the boundary points $x = -1$, $x = 0$, and $x = 1$.

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Step 4: Differentiability Analysis at Boundary Points

1. At $x = -1$:

   • Continuity check: $\lim_{x \to -1^-} h(x) = (-1)^2 + 1 = 2, \quad \lim_{x \to -1^+} h(x) = 1 - (-1) = 2 = h(-1).$ Thus, $h(x)$ is continuous at $x = -1$.
   • Left-hand derivative: $h'(-1^-) = \left. \frac{d}{dx}(x^2 + 1) \right|_{x = -1} = 2(-1) = -2.$
   • Right-hand derivative: $h'(-1^+) = \left. \frac{d}{dx}(1 - x) \right|_{x = -1} = -1.$
   • Since $h'(-1^-) \neq h'(-1^+)$, $h(x)$ is not differentiable at $x = -1$.

2. At $x = 0$:

   • Continuity check: $\lim_{x \to 0^-} h(x) = 1 - 0 = 1, \quad \lim_{x \to 0^+} h(x) = 0^2 + 1 = 1 = h(0).$ Thus, $h(x)$ is continuous at $x = 0$.
   • Left-hand derivative: $h'(0^-) = \left. \frac{d}{dx}(1 - x) \right|_{x = 0} = -1.$
   • Right-hand derivative: $h'(0^+) = \left. \frac{d}{dx}(x^2 + 1) \right|_{x = 0} = 2(0) = 0.$
   • Since $h'(0^-) \neq h'(0^+)$, $h(x)$ is not differentiable at $x = 0$.

3. At $x = 1$:

   • Continuity check: $\lim_{x \to 1^-} h(x) = 1^2 + 1 = 2, \quad \lim_{x \to 1^+} h(x) = 1 + 1 = 2 = h(1).$ Thus, $h(x)$ is continuous at $x = 1$.
   • Left-hand derivative: $h'(1^-) = \left. \frac{d}{dx}(x^2 + 1) \right|_{x = 1} = 2(1) = 2.$
   • Right-hand derivative: $h'(1^+) = \left. \frac{d}{dx}(1 + x) \right|_{x = 1} = 1.$
   • Since $h'(1^-) \neq h'(1^+)$, $h(x)$ is not differentiable at $x = 1$.

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Conclusion

The function $h(x)$ fails to be differentiable at $3$ points: $x = -1$, $x = 0$, and $x = 1$.