How do you verify that the function #f(x)= sin(22pix)# satisfies the three hypotheses of Rolle's Theorem on the given interval [1/11, 1/11] and then find all numbers c that satisfy the conclusion of Rolle's Theorem?
In this context, "Verify" mean "Prove" or "Show". Think of this as a writing assignment.
(Then do some algebra to prove to your grader (1) I know what the conclusion says and (2) I can do algebra to solve an equation.)
Go through the hypotheses in order and tell you reader why they should agree that the hypothesis is true for this function on this interval:
H3: The function has equal values at the endpoints or the interval
From trigonometry we know that the solutions are:
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To verify that the function (f(x) = \sin(22\pi x)) satisfies the three hypotheses of Rolle's Theorem on the interval ([1/11, 1/11]), and then find all numbers (c) that satisfy the conclusion of Rolle's Theorem, follow these steps:

Hypothesis 1: The function (f(x)) must be continuous on the closed interval ([1/11, 1/11]).
 The sine function is continuous everywhere, so (f(x)) is continuous on the interval ([1/11, 1/11]).

Hypothesis 2: The function (f(x)) must be differentiable on the open interval ((1/11, 1/11)).
 The derivative of (f(x) = \sin(22\pi x)) is (f'(x) = 22\pi \cos(22\pi x)).
 The cosine function is also continuous everywhere, so (f'(x)) exists and is continuous on ([1/11, 1/11]), making (f(x)) differentiable on ((1/11, 1/11)).

Hypothesis 3: (f(1/11) = f(1/11)).
 (f(1/11) = \sin\left(22\pi \cdot \frac{1}{11}\right) = \sin\left(2\pi\right) = 0)
 (f(1/11) = \sin\left(22\pi \cdot \frac{1}{11}\right) = \sin\left(2\pi\right) = 0)
 (f(1/11) = f(1/11)), fulfilling the third hypothesis.
Since all three hypotheses are satisfied, we can conclude that there exists at least one number (c) in the open interval ((1/11, 1/11)) such that (f'(c) = 0).
To find (c), we need to find where the derivative (f'(x) = 22\pi \cos(22\pi x)) equals zero on the interval ((1/11, 1/11)).
[ 22\pi \cos(22\pi x) = 0 ]
For cosine function to be zero, (22\pi x) should be (\frac{\pi}{2}), ( \frac{3\pi}{2}), ( \frac{5\pi}{2}), and so on.
So, we have: [ 22\pi x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots ] Solving for (x) gives: [ x = \frac{1}{44}, \frac{3}{44}, \frac{5}{44}, \ldots ]
Since these values of (x) are all within the interval ((1/11, 1/11)), they are all valid values of (c).
Therefore, all numbers (c) that satisfy the conclusion of Rolle's Theorem are (c = \frac{1}{44}, \frac{3}{44}, \frac{5}{44}, \ldots).
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To verify that the function ( f(x) = \sin(22\pi x) ) satisfies the three hypotheses of Rolle's Theorem on the given interval ( [\frac{1}{11}, \frac{1}{11}] ), we need to check the following:
 Continuity: The function ( f(x) ) is continuous on the closed interval ( [\frac{1}{11}, \frac{1}{11}] ).
 Differentiability: The function ( f(x) ) is differentiable on the open interval ( (\frac{1}{11}, \frac{1}{11}) ).
 Equal Function Values at Endpoints: The function ( f(x) ) has equal values at the endpoints of the interval, i.e., ( f\left(\frac{1}{11}\right) = f\left(\frac{1}{11}\right) ).
After verifying these hypotheses, we can apply Rolle's Theorem, which states that if a function satisfies these three conditions, then there exists at least one ( c ) in the open interval such that ( f'(c) = 0 ).
To find all numbers ( c ) that satisfy the conclusion of Rolle's Theorem, we need to find the derivative of ( f(x) ), which is ( f'(x) = 22\pi \cos(22\pi x) ). Then, we solve the equation ( f'(c) = 0 ) to find the critical points ( c ).
Since ( f'(x) = 22\pi \cos(22\pi x) ), we set ( f'(c) = 0 ) and solve for ( c ): [ 22\pi \cos(22\pi c) = 0 ]
Since ( \cos(22\pi c) = 0 ) when ( 22\pi c = \frac{\pi}{2} + k\pi ) for ( k ) being an integer, we have: [ 22\pi c = \frac{\pi}{2} + k\pi ] [ c = \frac{1}{44} + \frac{k}{22} ]
Given that ( c ) must lie in the interval ( (\frac{1}{11}, \frac{1}{11}) ), we find that the only ( c ) that satisfies this condition is ( c = 0 ). Therefore, ( c = 0 ) is the only number that satisfies the conclusion of Rolle's Theorem on the interval ( [\frac{1}{11}, \frac{1}{11}] ).
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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