Question:

Step 1: Define $f(x)$ and $g(x)$ on the interval $[-1, 1]$. The given definitions for $f(x)$ and $g(x)$ contain ambiguous intervals ($1 \leq x \leq 0$ and $1 \geq x \geq 0$). Assuming these are intended to be $0 \leq x \leq 1$, the functions are: $$f(x) = \begin{cases} \cos^{-1} x, & -1 \leq x < 0 \\ \sin^{-1} x, & 0 \leq x \leq 1 \end{cases}$$ $$g(x) = \begin{cases} \sin^{-1} x, & -1 \leq x < 0 \\ \cos^{-1} x, & 0 \leq x \leq 1 \end{cases}$$ Step 2: Determine $h(x) = \min\{f(x), g(x)\}$. For $x \in [-1, 0)$: $f(x) = \cos^{-1}x$. The range of $\cos^{-1}x$ for $x \in [-1, 0)$ is $(\pi/2, \pi]$. $g(x) = \sin^{-1}x$. The range of $\sin^{-1}x$ for $x \in [-1, 0)$ is $[-\pi/2, 0)$. Since $g(x)$ is negative and $f(x)$ is positive on this interval, $h(x) = \sin^{-1}x$. For $x \in [0, 1]$: $f(x) = \sin^{-1}x$. $g(x) = \cos^{-1}x$. To find $\min\{\sin^{-1}x, \cos^{-1}x\}$, we compare the two functions. They are equal when $\sin^{-1}x = \cos^{-1}x$, which implies $x = 1/\sqrt{2}$. For $x \in [0, 1/\sqrt{2}]$, $\sin^{-1}x \leq \cos^{-1}x$, so $h(x) = \sin^{-1}x$. For $x \in (1/\sqrt{2}, 1]$, $\sin^{-1}x > \cos^{-1}x$, so $h(x) = \cos^{-1}x$. Combining these results, the function $h(x)$ is defined as: $$h(x) = \begin{cases} \sin^{-1} x, & -1 \leq x \leq 1/\sqrt{2} \\ \cos^{-1} x, & 1/\sqrt{2} < x \leq 1 \end{cases}$$ Step 3: Analyze the continuity of $h(x)$ on $[-1, 1]$. The functions $\sin^{-1}x$ and $\cos^{-1}x$ are continuous on their respective domains. We examine continuity at the transition points $x=0$ and $x=1/\sqrt{2}$. At $x=0$: $\lim_{x \to 0^-} h(x) = \lim_{x \to 0^-} \sin^{-1}x = \sin^{-1}(0) = 0$. $h(0) = \sin^{-1}(0) = 0$. $\lim_{x \to 0^+} h(x) = \lim_{x \to 0^+} \sin^{-1}x = \sin^{-1}(0) = 0$. Since $\lim_{x \to 0^-} h(x) = h(0) = \lim_{x \to 0^+} h(x)$, $h(x)$ is continuous at $x=0$. At $x=1/\sqrt{2}$: $\lim_{x \to (1/\sqrt{2})^-} h(x) = \lim_{x \to (1/\sqrt{2})^-} \sin^{-1}x = \sin^{-1}(1/\sqrt{2}) = \pi/4$. $h(1/\sqrt{2}) = \sin^{-1}(1/\sqrt{2}) = \pi/4$. $\lim_{x \to (1/\sqrt{2})^+} h(x) = \lim_{x \to (1/\sqrt{2})^+} \cos^{-1}x = \cos^{-1}(1/\sqrt{2}) = \pi/4$. Since $\lim_{x \to (1/\sqrt{2})^-} h(x) = h(1/\sqrt{2}) = \lim_{x \to (1/\sqrt{2})^+} h(x)$, $h(x)$ is continuous at $x=1/\sqrt{2}$. Therefore, $h(x)$ is continuous on the entire interval $[-1, 1]$. Step 4: Analyze the differentiability of $h(x)$ on $(-1, 1)$. The derivatives of $\sin^{-1}x$ and $\cos^{-1}x$ are $\frac{1}{\sqrt{1-x^2}}$ and $-\frac{1}{\sqrt{1-x^2}}$, respectively. We check differentiability at the transition points $x=0$ and $x=1/\sqrt{2}$. At $x=0$: The left-hand derivative is $h'(0^-) = \frac{d}{dx}(\sin^{-1}x)\Big|_{x=0} = \frac{1}{\sqrt{1-0^2}} = 1$. The right-hand derivative is $h'(0^+) = \frac{d}{dx}(\sin^{-1}x)\Big|_{x=0} = \frac{1}{\sqrt{1-0^2}} = 1$. Since $h'(0^-) = h'(0^+)$, $h(x)$ is differentiable at $x=0$. At $x=1/\sqrt{2}$: The left-hand derivative is $h'((1/\sqrt{2})^-) = \frac{d}{dx}(\sin^{-1}x)\Big|_{x=1/\sqrt{2}} = \frac{1}{\sqrt{1-(1/\sqrt{2})^2}} = \frac{1}{\sqrt{1-1/2}} = \frac{1}{\sqrt{1/2}} = \sqrt{2}$. The right-hand derivative is $h'((1/\sqrt{2})^+)= \frac{d}{dx}(\cos^{-1}x)\Big|_{x=1/\sqrt{2}} = -\frac{1}{\sqrt{1-(1/\sqrt{2})^2}} = -\frac{1}{\sqrt{1-1/2}} = -\frac{1}{\sqrt{1/2}} = -\sqrt{2}$. Since $h'((1/\sqrt{2})^-) \neq h'((1/\sqrt{2})^+)$, $h(x)$ is not differentiable at $x=1/\sqrt{2}$. Thus, $h(x)$ is non-derivable at exactly one point in $x \in (-1, 1)$, which is $x=1/\sqrt{2}$. Step 5: Determine the minimum value of $h(x)$. The function $h(x)$ is defined as: $$h(x) = \begin{cases} \sin^{-1} x, & -1 \leq x \leq 1/\sqrt{2} \\ \cos^{-1} x, & 1/\sqrt{2} < x \leq 1 \end{cases}$$ On the interval $[-1, 1/\sqrt{2}]$, $h(x) = \sin^{-1}x$. This function is strictly increasing, so its minimum value on this interval occurs at $x=-1$, which is $h(-1) = \sin^{-1}(-1) = -\pi/2$. On the interval $(1/\sqrt{2}, 1]$, $h(x) = \cos^{-1}x$. This function is strictly decreasing, so its minimum value on this interval occurs at $x=1$, which is $h(1) = \cos^{-1}(1) = 0$. Comparing the minimum values from both intervals, the overall minimum value of $h(x)$ on $[-1, 1]$ is $-\pi/2$. Step 6: Determine the maximum value of $h(x)$. On the interval $[-1, 1/\sqrt{2}]$, $h(x) = \sin^{-1}x$. This function is strictly increasing, so its maximum value on this interval occurs at $x=1/\sqrt{2}$, which is $h(1/\sqrt{2}) = \sin^{-1}(1/\sqrt{2}) = \pi/4$. On the interval $(1/\sqrt{2}, 1]$, $h(x) = \cos^{-1}x$. This function is strictly decreasing, so its maximum value on this interval approaches $\cos^{-1}(1/\sqrt{2}) = \pi/4$ as $x \to (1/\sqrt{2})^+$. Comparing the maximum values from both intervals, the overall maximum value of $h(x)$ on $[-1, 1]$ is $\pi/4$.