How do you find the number c that satisfies the conclusion of the Mean Value Theorem for the function #f(x)=x-2sinx# on the interval [0, pi]?

Answer 1

Solve the equation #f'(x) = (f(pi)-f(0))/(pi-0)# on the interval #(0,pi)#.

#f'(x) = 1-2cosx#
#(f(pi)-f(0))/(pi-0) = ((pi-2sinpi)-(0+2sin0))/pi = 1#
#1-2cosx = 1# at #x=pi/2 + pik# for integer #k#.
The only solution in #(0,pi)# is #pi/2# so
#c = pi/2# is the only #c# that satisfies the conclusion of the mean value theorem for this function on this interval.
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Answer 2

To find the number ( c ) that satisfies the conclusion of the Mean Value Theorem for the function ( f(x) = x - 2\sin(x) ) on the interval ([0, \pi]), we first need to check if the conditions of the Mean Value Theorem are met. The conditions are:

  1. ( f(x) ) must be continuous on the closed interval ([a, b]).
  2. ( f(x) ) must be differentiable on the open interval ((a, b)).

Given that ( f(x) = x - 2\sin(x) ), it's continuous and differentiable everywhere, including the interval ([0, \pi]).

Now, to find the number ( c ), we apply the conclusion of the Mean Value Theorem, which states:

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

Substituting the values into the equation:

[ f'(c) = \frac{f(\pi) - f(0)}{\pi - 0} ]

[ f'(c) = \frac{\pi - 2\sin(\pi) - (0 - 2\sin(0))}{\pi} ]

[ f'(c) = \frac{\pi - 0 - 0}{\pi} ]

[ f'(c) = \frac{\pi}{\pi} ]

[ f'(c) = 1 ]

Now, we need to find the derivative of ( f(x) ), which is:

[ f'(x) = 1 - 2\cos(x) ]

Setting ( f'(c) = 1 ), we have:

[ 1 - 2\cos(c) = 1 ]

[ -2\cos(c) = 0 ]

[ \cos(c) = 0 ]

The solutions to ( \cos(c) = 0 ) on the interval ([0, \pi]) are ( c = \frac{\pi}{2} ) and ( c = \frac{3\pi}{2} ). However, only ( c = \frac{\pi}{2} ) falls within the interval ([0, \pi]).

Therefore, the number ( c ) that satisfies the conclusion of the Mean Value Theorem for the function ( f(x) = x - 2\sin(x) ) on the interval ([0, \pi]) is ( c = \frac{\pi}{2} ).

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Answer from HIX Tutor

When evaluating a one-sided 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 one-sided 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 one-sided 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 one-sided 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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