Part 1 of 2 completed: 8 of 11 My score: $$ 7.5 / 11 $$ pts (68.18\%) 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. a. Choose the correct answer below. A. Yes; $$ f(x) $$ is not continuous on $$ \left[-\frac{\sqrt{2}}{2}, 0\right] $$ and not differentiable on $$ \left(-\frac{\sqrt{2}}{2}, 0\right) $$. B. No; $$ f(x) $$ is differentiable on $$ \left(-\frac{\sqrt{2}}{2}, 0\right) $$, but not continuous on $$ \left[-\frac{\sqrt{2}}{2}, 0\right] $$. C. No; $$ f(x) $$ is continuous on $$ \left[-\frac{\sqrt{2}}{2}, 0\right] $$, but not differentiable on $$ \left(-\frac{\sqrt{2}}{2}, 0\right) $$. D. Yes; $$ f(x) $$ is continuous on $$ \left[-\frac{\sqrt{2}}{2}, 0\right] $$ and differentiable on $$ \left(-\frac{\sqrt{2}}{2}, 0\right) $$.

Question

UpStudy AI · Accepted Answer

To determine whether the Mean Value Theorem (MVT) applies to the function $$ f(x) = \sin^{-1}(x) $$ on the interval $$ \left[-\frac{\sqrt{2}}{2}, 0\right] $$, we need to check two conditions:

1. **Continuity**: The function must be continuous on the closed interval $$ \left[-\frac{\sqrt{2}}{2}, 0\right] $$.
2. **Differentiability**: The function must be differentiable on the open interval $$ \left(-\frac{\sqrt{2}}{2}, 0\right) $$.

The function $$ f(x) = \sin^{-1}(x) $$ is continuous for all $$ x $$ in the interval $$ [-1, 1] $$ and differentiable for $$ x $$ in the interval $$ (-1, 1) $$. Since $$ -\frac{\sqrt{2}}{2} $$ and $$ 0 $$ are within this range, we can conclude that:

- $$ f(x) $$ is continuous on $$ \left[-\frac{\sqrt{2}}{2}, 0\right] $$.
- $$ f(x) $$ is differentiable on $$ \left(-\frac{\sqrt{2}}{2}, 0\right) $$.

Thus, the correct answer is D: Yes; $$ f(x) $$ is continuous on $$ \left[-\frac{\sqrt{2}}{2}, 0\right] $$ and differentiable on $$ \left(-\frac{\sqrt{2}}{2}, 0\right) $$.
Since the Mean Value Theorem applies, we can find the point(s) $$ c $$ in the interval $$ \left(-\frac{\sqrt{2}}{2}, 0\right) $$ such that:

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

where $$ a = -\frac{\sqrt{2}}{2} $$ and $$ b = 0 $$.

First, we calculate $$ f(a) $$ and $$ f(b) $$:

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

Now, we can find the average rate of change:

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

Next, we need to find $$ f'(x) $$:

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

Now we set $$ f'(c) = \frac{\pi \sqrt{2}}{4} $$:

$$
\frac{1}{\sqrt{1 - c^2}} = \frac{\pi \sqrt{2}}{4}
$$

Now we solve for $$ c $$:

$$
\sqrt{1 - c^2} = \frac{4}{\pi \sqrt{2}}
$$
$$
1 - c^2 = \left(\frac{4}{\pi \sqrt{2}}\right)^2
$$
$$
c^2 = 1 - \frac{16}{2\pi^2} = 1 - \frac{8}{\pi^2}
$$
$$
c = \pm \sqrt{1 - \frac{8}{\pi^2}}
$$

Since we are looking for $$ c $$ in the interval $$ \left(-\frac{\sqrt{2}}{2}, 0\right) $$, we take the negative root:

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

This value of $$ c $$ is guaranteed to exist by the Mean Value Theorem.
The correct answer for part a is D: Yes; $$ f(x) $$ is continuous on $$ \left[-\frac{\sqrt{2}}{2}, 0\right] $$ and differentiable on $$ \left(-\frac{\sqrt{2}}{2}, 0\right) $$. For part b, the point guaranteed by the Mean Value Theorem is $$ c = -\sqrt{1 - \frac{8}{\pi^2}} $$.

Upstudy AI Solution

Answer

Solution

Extra Insights

Related Questions

Latest Calculus Questions