a. Determine whether the Mean Value Theorem applies to the function $$ f(x)=\sin ^{-1} x $$ on the interval $$ \left[-\frac{\sqrt{2}}{2}, 0\right] $$ b. If so, find or approximate the point(s) that are guaranteed to exist by the Mean Value Theorem. B. No; $$ f(x) $$ is differentiable on $$ \left\lfloor-\frac{\sqrt{L}}{2}, 0\right) $$, but not continuous on $$ \left\lfloor-\frac{\sqrt{L}}{2}, 0\right] $$. C. No; $$ f(x) $$ is continuous on $$ \left[-\frac{\sqrt{2}}{2}, 0\right] $$, but not differentiable on $$ \left(-\frac{\sqrt{2}}{2}, 0\right) $$. Yes; $$ f(x) $$ is continuous on $$ \left[-\frac{\sqrt{2}}{2}, 0\right] $$ and differentiable on $$ \left(-\frac{\sqrt{2}}{2}, 0\right) $$. A. Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The point(s) is/are $$ x=\square $$. (Type an exact answer, using $$ \pi $$ as needed. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) B. The Men Value Theorem does not apply in this case.

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We first check the conditions for the Mean Value Theorem (MVT). The MVT requires that the function is continuous on the closed interval and differentiable on the open interval.

1. Since $$ f(x)=\sin^{-1}x $$ is continuous on $$[-1,1]$$ and $$\left[-\frac{\sqrt{2}}{2},0\right] \subset [-1,1]$$, it is continuous on $$\left[-\frac{\sqrt{2}}{2},0\right]$$.

2. The derivative of $$ f(x)=\sin^{-1}x $$ is
   $$
   f'(x)=\frac{1}{\sqrt{1-x^2}},
   $$
   which is defined for $$ x $$ in the open interval $$\left(-\frac{\sqrt{2}}{2},0\right)$$ because $$1-x^2>0$$ for these $$ x $$. Hence, $$ f $$ is differentiable on $$\left(-\frac{\sqrt{2}}{2},0\right)$$.

Since both conditions are met, the MVT applies.

Next, we find the point $$ c $$ that satisfies the MVT.

The MVT tells us that there exists $$ c $$ in $$\left(-\frac{\sqrt{2}}{2},0\right)$$ such that
$$
f'(c)=\frac{f(0)-f\left(-\frac{\sqrt{2}}{2}\right)}{0-\left(-\frac{\sqrt{2}}{2}\right)}.
$$

We compute the function values:
- $$ f(0)=\sin^{-1}(0)=0 $$.
- $$ f\left(-\frac{\sqrt{2}}{2}\right)=\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)=-\frac{\pi}{4} $$.

Thus, the average rate of change is:
$$
\frac{0- \left(-\frac{\pi}{4}\right)}{\frac{\sqrt{2}}{2}}=\frac{\frac{\pi}{4}}{\frac{\sqrt{2}}{2}}=\frac{\pi}{4}\cdot \frac{2}{\sqrt{2}}=\frac{\pi}{2\sqrt{2}}.
$$

We now set the derivative equal to the average rate:
$$
\frac{1}{\sqrt{1-c^2}}=\frac{\pi}{2\sqrt{2}}.
$$

Solve for $$ c $$:
$$
\sqrt{1-c^2}=\frac{2\sqrt{2}}{\pi}.
$$
Square both sides:
$$
1-c^2=\frac{8}{\pi^2}.
$$
Then,
$$
c^2=1-\frac{8}{\pi^2}.
$$

Since $$ c $$ lies in $$\left(-\frac{\sqrt{2}}{2},0\right)$$ (i.e. $$ c $$ is negative), we have:
$$
c=-\sqrt{1-\frac{8}{\pi^2}}.
$$

Thus, the Mean Value Theorem applies and the point $$ c $$ is
$$
\boxed{c=-\sqrt{1-\frac{8}{\pi^2}}}.
$$
The Mean Value Theorem applies to the function $$ f(x) = \sin^{-1}x $$ on the interval $$\left[-\frac{\sqrt{2}}{2}, 0\right]$$. There exists a point $$ c = -\sqrt{1 - \frac{8}{\pi^2}} $$ within the open interval $$\left(-\frac{\sqrt{2}}{2}, 0\right)$$ where the derivative equals the average rate of change over the interval.

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