Key Concepts and Formulas

• Definition of Absolute Value: $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.
• Differentiability of Maximum Function: The function $h(x) = \max(g(x), s(x))$ is not differentiable at points where $g(x) = s(x)$ and the derivatives of $g(x)$ and $s(x)$ are different at those points.
• Derivative of $|x|$: $\frac{d}{dx}|x| = \frac{x}{|x|}$ for $x \ne 0$. This is equivalent to 1 for $x > 0$ and -1 for $x < 0$.
• Derivative of $\sqrt{a^2 - x^2}$: $\frac{d}{dx}\sqrt{a^2 - x^2} = \frac{-2x}{2\sqrt{a^2 - x^2}} = \frac{-x}{\sqrt{a^2 - x^2}}$ for $a^2 - x^2 > 0$.

Step-by-Step Solution

Step 1: Analyze the function and the domain. The function is given by $f(x) = \max\left\{ - \left| x \right|, - \sqrt {1 - {x^2}} \right\}$ for $x \in (-1, 1)$. We need to find the points in this interval where $f(x)$ is not differentiable.

Step 2: Simplify the terms inside the maximum function. We have two functions to compare: $g(x) = -|x|$ and $s(x) = -\sqrt{1 - x^2}$. Since the domain is $(-1, 1)$, we know that $1 - x^2 > 0$.

Step 3: Determine where the two functions are equal. The function $f(x)$ will be non-differentiable at points where $-|x| = -\sqrt{1 - x^2}$, which simplifies to $|x| = \sqrt{1 - x^2}$. Squaring both sides, we get $x^2 = 1 - x^2$. This leads to $2x^2 = 1$, so $x^2 = \frac{1}{2}$. Therefore, $x = \frac{1}{\sqrt{2}}$ and $x = -\frac{1}{\sqrt{2}}$. Both of these points lie within the domain $(-1, 1)$.

Step 4: Analyze the differentiability of each term.

• The function $g(x) = -|x|$ is not differentiable at $x=0$. However, $x=0$ is not a point where $g(x) = s(x)$.
• The function $s(x) = -\sqrt{1 - x^2}$ is differentiable for $x \in (-1, 1)$. Its derivative is $s'(x) = -\frac{-x}{\sqrt{1 - x^2}} = \frac{x}{\sqrt{1 - x^2}}$.

Step 5: Evaluate the derivatives of $g(x)$ and $s(x)$ at the points of equality. We found that the two functions are equal at $x = \frac{1}{\sqrt{2}}$ and $x = -\frac{1}{\sqrt{2}}$. Let's find the derivatives of $g(x) = -|x|$ and $s(x) = -\sqrt{1 - x^2}$ at these points.

For $x > 0$, $|x| = x$, so $g(x) = -x$. The derivative is $g'(x) = -1$. For $x < 0$, $|x| = -x$, so $g(x) = -(-x) = x$. The derivative is $g'(x) = 1$.

The derivative of $s(x)$ is $s'(x) = \frac{x}{\sqrt{1 - x^2}}$.

Case 1: $x = \frac{1}{\sqrt{2}}$ At this point, $x > 0$, so $g'(x) = -1$. The derivative of $s(x)$ is $s'\left(\frac{1}{\sqrt{2}}\right) = \frac{\frac{1}{\sqrt{2}}}{\sqrt{1 - \left(\frac{1}{\sqrt{2}}\right)^2}} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{1 - \frac{1}{2}}} = \frac{\frac{1}{\sqrt{2}}}{\sqrt{\frac{1}{2}}} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = 1$. Since $g'\left(\frac{1}{\sqrt{2}}\right) = -1$ and $s'\left(\frac{1}{\sqrt{2}}\right) = 1$, and $-1 \ne 1$, the function $f(x)$ is not differentiable at $x = \frac{1}{\sqrt{2}}$.

Case 2: $x = -\frac{1}{\sqrt{2}}$ At this point, $x < 0$, so $g'(x) = 1$. The derivative of $s(x)$ is $s'\left(-\frac{1}{\sqrt{2}}\right) = \frac{-\frac{1}{\sqrt{2}}}{\sqrt{1 - \left(-\frac{1}{\sqrt{2}}\right)^2}} = \frac{-\frac{1}{\sqrt{2}}}{\sqrt{1 - \frac{1}{2}}} = \frac{-\frac{1}{\sqrt{2}}}{\sqrt{\frac{1}{2}}} = \frac{-\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = -1$. Since $g'\left(-\frac{1}{\sqrt{2}}\right) = 1$ and $s'\left(-\frac{1}{\sqrt{2}}\right) = -1$, and $1 \ne -1$, the function $f(x)$ is not differentiable at $x = -\frac{1}{\sqrt{2}}$.

Step 6: Consider the point $x=0$. At $x=0$, $|x|=0$ and $\sqrt{1-x^2} = \sqrt{1-0} = 1$. So, $-|x| = 0$ and $-\sqrt{1-x^2} = -1$. Thus, $f(0) = \max\{0, -1\} = 0$. The function $g(x) = -|x|$ is not differentiable at $x=0$. However, the definition of $f(x)$ as the maximum of two functions means we only need to consider points where the two functions are equal and their derivatives differ. At $x=0$, the two functions are not equal. Let's check the differentiability of $f(x)$ around $x=0$. For $x$ close to 0 and $x>0$, $f(x) = \max\{-x, -\sqrt{1-x^2}\}$. Since $x$ is small and positive, $-x$ is small and negative. $-\sqrt{1-x^2}$ is close to $-1$. So $f(x) = -x$. For $x$ close to 0 and $x<0$, $f(x) = \max\{x, -\sqrt{1-x^2}\}$. Since $x$ is small and negative, $x$ is small and negative. $-\sqrt{1-x^2}$ is close to $-1$. So $f(x) = x$. Thus, for $x$ in a neighborhood of 0 (excluding 0 itself), $f(x) = -|x|$. The derivative of $-|x|$ is $-1$ for $x>0$ and $1$ for $x<0$. The left-hand derivative at $x=0$ is $1$ and the right-hand derivative is $-1$. Therefore, $f(x)$ is not differentiable at $x=0$.

Step 7: Consolidate the points of non-differentiability. The points where $f(x)$ is not differentiable are $x = \frac{1}{\sqrt{2}}$, $x = -\frac{1}{\sqrt{2}}$, and $x = 0$. All these points are within the domain $(-1, 1)$. Therefore, the set K of all points at which f is not differentiable has exactly three elements.

Common Mistakes & Tips

• Incorrectly handling the absolute value: Remember that $|x|$ has a point of non-differentiability at $x=0$. This needs to be considered even if it's not a point where the two functions inside the max are equal.
• Forgetting to check the derivatives at the points of equality: It's not enough for the two functions to be equal; their derivatives must also be different at those points for non-differentiability.
• Domain restrictions: Always ensure the points of non-differentiability lie within the given domain of the function.

Summary

The function $f(x) = \max\left\{ - \left| x \right|, - \sqrt {1 - {x^2}} \right\}$ is analyzed for differentiability in the interval $(-1, 1)$. The points of potential non-differentiability occur where the two arguments of the maximum function are equal, i.e., $-|x| = -\sqrt{1-x^2}$, which simplifies to $|x| = \sqrt{1-x^2}$. Squaring both sides leads to $x^2 = 1-x^2$, giving $x^2 = 1/2$, so $x = \pm \frac{1}{\sqrt{2}}$. At these points, we compared the derivatives of $-|x|$ and $-\sqrt{1-x^2}$ and found they were different, confirming non-differentiability. Additionally, the term $-|x|$ itself is not differentiable at $x=0$. We examined the behavior of $f(x)$ around $x=0$ and found that $f(x) = -|x|$ in a neighborhood of 0, making $f(x)$ non-differentiable at $x=0$. Thus, there are three points of non-differentiability: $0$, $\frac{1}{\sqrt{2}}$, and $-\frac{1}{\sqrt{2}}$.

The final answer is \boxed{B}.