Solved on Feb 15, 2024

Determine if Rolle's Theorem applies to $f(x) = \sin 7x$ on $[\pi/7, 2\pi/7]$ . If so, find the point(s) guaranteed to exist. Select A and fill in the answer if Rolle's Theorem applies, otherwise select B.

STEP 1

Assumptions
1. The function given is $f(x) = \sin(7x)$ .
2. The interval given is $\left[\frac{\pi}{7}, \frac{2\pi}{7}\right]$ .
3. Rolle's Theorem states that if a function $f$ is continuous on the closed interval $[a, b]$ , differentiable on the open interval $(a, b)$ , and $f(a) = f(b)$ , then there exists at least one $c$ in $(a, b)$ such that $f'(c) = 0$ .

STEP 2

Verify if the function $f(x) = \sin(7x)$ is continuous on the closed interval $\left[\frac{\pi}{7}, \frac{2\pi}{7}\right]$ .
Since sine functions are continuous everywhere, $f(x)$ is continuous on $\left[\frac{\pi}{7}, \frac{2\pi}{7}\right]$ .

STEP 3

Verify if the function $f(x) = \sin(7x)$ is differentiable on the open interval $\left(\frac{\pi}{7}, \frac{2\pi}{7}\right)$ .
Since sine functions are differentiable everywhere, $f(x)$ is differentiable on $\left(\frac{\pi}{7}, \frac{2\pi}{7}\right)$ .

STEP 4

Check if $f\left(\frac{\pi}{7}\right) = f\left(\frac{2\pi}{7}\right)$ .
Calculate $f\left(\frac{\pi}{7}\right)$ :
$f\left(\frac{\pi}{7}\right) = \sin\left(7 \cdot \frac{\pi}{7}\right) = \sin(\pi) = 0$
Calculate $f\left(\frac{2\pi}{7}\right)$ :
$f\left(\frac{2\pi}{7}\right) = \sin\left(7 \cdot \frac{2\pi}{7}\right) = \sin(2\pi) = 0$
Since $f\left(\frac{\pi}{7}\right) = f\left(\frac{2\pi}{7}\right)$ , the function satisfies the conditions for Rolle's Theorem.

STEP 5

Find $f'(x)$ , the derivative of $f(x)$ .
$f'(x) = \frac{d}{dx}[\sin(7x)] = 7\cos(7x)$

STEP 6

Find the value(s) of $x$ in the open interval $\left(\frac{\pi}{7}, \frac{2\pi}{7}\right)$ such that $f'(x) = 0$ .
Set $f'(x)$ equal to zero and solve for $x$ :
$7\cos(7x) = 0$

STEP 7

Solve for $x$ when $\cos(7x) = 0$ .
The cosine function is zero at $\frac{\pi}{2} + k\pi$ , where $k$ is an integer.
$7x = \frac{\pi}{2} + k\pi$

STEP 8

Solve for $x$ :
$x = \frac{\pi}{14} + \frac{k\pi}{7}$

STEP 9

Find the value of $k$ such that $x$ is in the interval $\left(\frac{\pi}{7}, \frac{2\pi}{7}\right)$ .
Since $\frac{\pi}{14} < \frac{\pi}{7}$ , we need to find the smallest positive integer $k$ such that $x$ is greater than $\frac{\pi}{7}$ .

STEP 10

Test $k = 1$ :
$x = \frac{\pi}{14} + \frac{\pi}{7} = \frac{3\pi}{14}$
Since $\frac{3\pi}{14}$ is in the interval $\left(\frac{\pi}{7}, \frac{2\pi}{7}\right)$ , this is a valid solution.

STEP 11

Verify that there are no other values of $k$ that give a solution in the interval.
For $k = 2$ , $x = \frac{\pi}{14} + \frac{2\pi}{7} = \frac{5\pi}{14}$ , which is greater than $\frac{2\pi}{7}$ . Therefore, $k = 1$ is the only value that gives a solution in the interval.

STEP 12

Conclude that Rolle's Theorem applies and the point guaranteed to exist is $x = \frac{3\pi}{14}$ .
Rolle's Theorem applies and the point(s) guaranteed to exist is/are $x = \frac{3\pi}{14}$ .

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Determine if Rolle's Theorem applies to $f(x) = \sin 7x$ on $[\pi/7, 2\pi/7]$ . If so, find the point(s) guaranteed to exist. Select A and fill in the answer if Rolle's Theorem applies, otherwise select B.

STEP 1

STEP 2

Verify if the function $f(x) = \sin(7x)$ is continuous on the closed interval $\left[\frac{\pi}{7}, \frac{2\pi}{7}\right]$ .
Since sine functions are continuous everywhere, $f(x)$ is continuous on $\left[\frac{\pi}{7}, \frac{2\pi}{7}\right]$ .

STEP 3

Verify if the function $f(x) = \sin(7x)$ is differentiable on the open interval $\left(\frac{\pi}{7}, \frac{2\pi}{7}\right)$ .
Since sine functions are differentiable everywhere, $f(x)$ is differentiable on $\left(\frac{\pi}{7}, \frac{2\pi}{7}\right)$ .

STEP 4

STEP 5

Find $f'(x)$ , the derivative of $f(x)$ .
$f'(x) = \frac{d}{dx}[\sin(7x)] = 7\cos(7x)$

STEP 6

Find the value(s) of $x$ in the open interval $\left(\frac{\pi}{7}, \frac{2\pi}{7}\right)$ such that $f'(x) = 0$ .
Set $f'(x)$ equal to zero and solve for $x$ :
$7\cos(7x) = 0$

STEP 7

Solve for $x$ when $\cos(7x) = 0$ .
The cosine function is zero at $\frac{\pi}{2} + k\pi$ , where $k$ is an integer.
$7x = \frac{\pi}{2} + k\pi$

STEP 8

Solve for $x$ :
$x = \frac{\pi}{14} + \frac{k\pi}{7}$

STEP 9

Find the value of $k$ such that $x$ is in the interval $\left(\frac{\pi}{7}, \frac{2\pi}{7}\right)$ .
Since $\frac{\pi}{14} < \frac{\pi}{7}$ , we need to find the smallest positive integer $k$ such that $x$ is greater than $\frac{\pi}{7}$ .

STEP 10

Test $k = 1$ :
$x = \frac{\pi}{14} + \frac{\pi}{7} = \frac{3\pi}{14}$
Since $\frac{3\pi}{14}$ is in the interval $\left(\frac{\pi}{7}, \frac{2\pi}{7}\right)$ , this is a valid solution.

STEP 11

Verify that there are no other values of $k$ that give a solution in the interval.
For $k = 2$ , $x = \frac{\pi}{14} + \frac{2\pi}{7} = \frac{5\pi}{14}$ , which is greater than $\frac{2\pi}{7}$ . Therefore, $k = 1$ is the only value that gives a solution in the interval.

STEP 12

Conclude that Rolle's Theorem applies and the point guaranteed to exist is $x = \frac{3\pi}{14}$ .
Rolle's Theorem applies and the point(s) guaranteed to exist is/are $x = \frac{3\pi}{14}$ .

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Determine if Rolle's Theorem applies to f(x)=sin⁡7xf(x) = \sin 7xf(x)=sin7x on [π/7,2π/7][\pi/7, 2\pi/7][π/7,2π/7]. If so, find the point(s) guaranteed to exist. Select A and fill in the answer if Rolle's Theorem applies, otherwise select B.

STEP 1

STEP 2

STEP 3

STEP 4

STEP 5

Find f′(x) f'(x) f′(x), the derivative of f(x) f(x) f(x).f′(x)=ddx[sin⁡(7x)]=7cos⁡(7x) f'(x) = \frac{d}{dx}[\sin(7x)] = 7\cos(7x) f′(x)=dxd​[sin(7x)]=7cos(7x)

STEP 6

Find the value(s) of x x x in the open interval (π7,2π7) \left(\frac{\pi}{7}, \frac{2\pi}{7}\right) (7π​,72π​) such that f′(x)=0 f'(x) = 0 f′(x)=0.Set f′(x) f'(x) f′(x) equal to zero and solve for x x x:7cos⁡(7x)=0 7\cos(7x) = 0 7cos(7x)=0

STEP 7

Solve for x x x when cos⁡(7x)=0 \cos(7x) = 0 cos(7x)=0.The cosine function is zero at π2+kπ \frac{\pi}{2} + k\pi 2π​+kπ, where k k k is an integer.7x=π2+kπ 7x = \frac{\pi}{2} + k\pi 7x=2π​+kπ

STEP 8

Solve for x x x:x=π14+kπ7 x = \frac{\pi}{14} + \frac{k\pi}{7} x=14π​+7kπ​

STEP 9

STEP 10

Test k=1 k = 1 k=1:x=π14+π7=3π14 x = \frac{\pi}{14} + \frac{\pi}{7} = \frac{3\pi}{14} x=14π​+7π​=143π​Since 3π14 \frac{3\pi}{14} 143π​ is in the interval (π7,2π7) \left(\frac{\pi}{7}, \frac{2\pi}{7}\right) (7π​,72π​), this is a valid solution.

STEP 11

STEP 12

Conclude that Rolle's Theorem applies and the point guaranteed to exist is x=3π14 x = \frac{3\pi}{14} x=143π​.Rolle's Theorem applies and the point(s) guaranteed to exist is/are x=3π14 x = \frac{3\pi}{14} x=143π​.

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Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Determine if Rolle's Theorem applies to $f(x) = \sin 7x$ on $[\pi/7, 2\pi/7]$ . If so, find the point(s) guaranteed to exist. Select A and fill in the answer if Rolle's Theorem applies, otherwise select B.

Find $f'(x)$ , the derivative of $f(x)$ .
$f'(x) = \frac{d}{dx}[\sin(7x)] = 7\cos(7x)$

Find the value(s) of $x$ in the open interval $\left(\frac{\pi}{7}, \frac{2\pi}{7}\right)$ such that $f'(x) = 0$ .
Set $f'(x)$ equal to zero and solve for $x$ :
$7\cos(7x) = 0$

Solve for $x$ when $\cos(7x) = 0$ .
The cosine function is zero at $\frac{\pi}{2} + k\pi$ , where $k$ is an integer.
$7x = \frac{\pi}{2} + k\pi$

Solve for $x$ :
$x = \frac{\pi}{14} + \frac{k\pi}{7}$

Test $k = 1$ :
$x = \frac{\pi}{14} + \frac{\pi}{7} = \frac{3\pi}{14}$
Since $\frac{3\pi}{14}$ is in the interval $\left(\frac{\pi}{7}, \frac{2\pi}{7}\right)$ , this is a valid solution.

Conclude that Rolle's Theorem applies and the point guaranteed to exist is $x = \frac{3\pi}{14}$ .
Rolle's Theorem applies and the point(s) guaranteed to exist is/are $x = \frac{3\pi}{14}$ .