Let's work through the two parts of this problem step-by-step.

Part a: Determine whether the Mean Value Theorem (MVT) applies to the function $f(x) = \sin^{-1}(x)$ on the interval $[ -\frac{\sqrt{2}}{2}, 0 ]$.

The Mean Value Theorem states that if a function $f$ is:

1. Continuous on the closed interval $[a, b]$,
2. Differentiable on the open interval $(a, b)$,

then there exists at least one point $c \in (a, b)$ such that:

$f'(c) = \frac{f(b) - f(a)}{b - a}.$

To determine if the Mean Value Theorem applies, we need to check whether:

• $f(x) = \sin^{-1}(x)$ is continuous on $[ -\frac{\sqrt{2}}{2}, 0 ]$,
• $f(x) = \sin^{-1}(x)$ is differentiable on $(-\frac{\sqrt{2}}{2}, 0)$.

1. Continuity: The function $f(x) = \sin^{-1}(x)$ is continuous on its entire domain, which is $[-1, 1]$. Since $-\frac{\sqrt{2}}{2} \approx -0.707$ and 0 are within this domain, $f(x)$ is continuous on the interval $[ -\frac{\sqrt{2}}{2}, 0 ]$.

2. Differentiability: The derivative of $f(x) = \sin^{-1}(x)$ is:

$f'(x) = \frac{1}{\sqrt{1 - x^2}}.$

This derivative exists and is defined as long as $x \in (-1, 1)$, and since $-\frac{\sqrt{2}}{2}$ and 0 are both within this domain, $f(x)$ is differentiable on the interval $(-\frac{\sqrt{2}}{2}, 0)$.

Since both conditions of the MVT are satisfied, the Mean Value Theorem applies to $f(x) = \sin^{-1}(x)$ on the interval $[ -\frac{\sqrt{2}}{2}, 0 ]$.

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Part b: Find or approximate the point(s) that are guaranteed to exist by the Mean Value Theorem.

By the Mean Value Theorem, there exists a point $c \in (-\frac{\sqrt{2}}{2}, 0)$ such that:

$f'(c) = \frac{f(0) - f\left(-\frac{\sqrt{2}}{2}\right)}{0 - \left(-\frac{\sqrt{2}}{2}\right)}.$

Let's calculate each part:

1. $f(0) = \sin^{-1}(0) = 0$,
2. $f\left(-\frac{\sqrt{2}}{2}\right) = \sin^{-1}\left(-\frac{\sqrt{2}}{2}\right) = -\frac{\pi}{4}$.

Thus, the equation becomes:

$f'(c) = \frac{0 - \left(-\frac{\pi}{4}\right)}{0 - \left(-\frac{\sqrt{2}}{2}\right)} = \frac{\frac{\pi}{4}}{\frac{\sqrt{2}}{2}}.$

Simplifying:

$f'(c) = \frac{\pi}{4} \times \frac{2}{\sqrt{2}} = \frac{\pi}{2\sqrt{2}}.$

Now, we set this equal to the derivative $f'(c) = \frac{1}{\sqrt{1 - c^2}}$:

$\frac{1}{\sqrt{1 - c^2}} = \frac{\pi}{2\sqrt{2}}.$

Squaring both sides:

$\frac{1}{1 - c^2} = \frac{\pi^2}{8}.$

Now, solve for $c^2$:

$1 - c^2 = \frac{8}{\pi^2},$ $c^2 = 1 - \frac{8}{\pi^2}.$

Taking the square root of both sides:

$c = \pm \sqrt{1 - \frac{8}{\pi^2}}.$

Since $c$ is within the interval $\left( -\frac{\sqrt{2}}{2}, 0 \right)$, we select the negative root:

$c = -\sqrt{1 - \frac{8}{\pi^2}}.$

This gives us the approximate value of $c$ where the Mean Value Theorem holds.

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Let me know if you'd like more details, clarifications, or further calculations.

Here are some related questions to explore:

1. What is the significance of the Mean Value Theorem in calculus?
2. How does the derivative of $\sin^{-1}(x)$ affect the behavior of the function?
3. What would happen if the interval were different for this function?
4. Can we generalize this result for other inverse trigonometric functions?
5. How can we apply the Mean Value Theorem to non-trigonometric functions?

Tip: When solving MVT problems, always check both the continuity and differentiability of the function before applying the theorem.