Solved on Feb 15, 2024

Determine if Rolle's Theorem applies to f(x)=sin7xf(x) = \sin 7x on [π/7,2π/7][\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) f(x) = \sin(7x) .
2. The interval given is [π7,2π7] \left[\frac{\pi}{7}, \frac{2\pi}{7}\right] .
3. Rolle's Theorem states that if a function f f is continuous on the closed interval [a,b] [a, b] , differentiable on the open interval (a,b) (a, b) , and f(a)=f(b) f(a) = f(b) , then there exists at least one c c in (a,b) (a, b) such that f(c)=0 f'(c) = 0 .

STEP 2

Verify if the function f(x)=sin(7x) f(x) = \sin(7x) is continuous on the closed interval [π7,2π7] \left[\frac{\pi}{7}, \frac{2\pi}{7}\right] .
Since sine functions are continuous everywhere, f(x) f(x) is continuous on [π7,2π7] \left[\frac{\pi}{7}, \frac{2\pi}{7}\right] .

STEP 3

Verify if the function f(x)=sin(7x) f(x) = \sin(7x) is differentiable on the open interval (π7,2π7) \left(\frac{\pi}{7}, \frac{2\pi}{7}\right) .
Since sine functions are differentiable everywhere, f(x) f(x) is differentiable on (π7,2π7) \left(\frac{\pi}{7}, \frac{2\pi}{7}\right) .

STEP 4

Check if f(π7)=f(2π7) f\left(\frac{\pi}{7}\right) = f\left(\frac{2\pi}{7}\right) .
Calculate f(π7) f\left(\frac{\pi}{7}\right) :
f(π7)=sin(7π7)=sin(π)=0 f\left(\frac{\pi}{7}\right) = \sin\left(7 \cdot \frac{\pi}{7}\right) = \sin(\pi) = 0
Calculate f(2π7) f\left(\frac{2\pi}{7}\right) :
f(2π7)=sin(72π7)=sin(2π)=0 f\left(\frac{2\pi}{7}\right) = \sin\left(7 \cdot \frac{2\pi}{7}\right) = \sin(2\pi) = 0
Since f(π7)=f(2π7) 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) f'(x) , the derivative of f(x) f(x) .
f(x)=ddx[sin(7x)]=7cos(7x) f'(x) = \frac{d}{dx}[\sin(7x)] = 7\cos(7x)

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

Find the value of k k such that x x is in the interval (π7,2π7) \left(\frac{\pi}{7}, \frac{2\pi}{7}\right) .
Since π14<π7 \frac{\pi}{14} < \frac{\pi}{7} , we need to find the smallest positive integer k k such that x x is greater than π7 \frac{\pi}{7} .

STEP 10

Test k=1 k = 1 :
x=π14+π7=3π14 x = \frac{\pi}{14} + \frac{\pi}{7} = \frac{3\pi}{14}
Since 3π14 \frac{3\pi}{14} is in the interval (π7,2π7) \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 k that give a solution in the interval.
For k=2 k = 2 , x=π14+2π7=5π14 x = \frac{\pi}{14} + \frac{2\pi}{7} = \frac{5\pi}{14} , which is greater than 2π7 \frac{2\pi}{7} . Therefore, k=1 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=3π14 x = \frac{3\pi}{14} .
Rolle's Theorem applies and the point(s) guaranteed to exist is/are x=3π14 x = \frac{3\pi}{14} .

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