# Given the function #f(x)=abs(x-3)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [0,6] and find the c?

See below.

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To determine if ( f(x) = |x - 3| ) satisfies the hypotheses of the Mean Value Theorem (MVT) on the interval ([0, 6]), you need to check two conditions:

- The function must be continuous on the closed interval ([0, 6]).
- The function must be differentiable on the open interval ((0, 6)).

( f(x) = |x - 3| ) is continuous and differentiable everywhere except at ( x = 3 ). However, since ( x = 3 ) is not included in the interval ([0, 6]), we only need to check the continuity and differentiability of ( f(x) ) at the endpoints of the interval.

At ( x = 0 ) and ( x = 6 ), ( f(x) ) is continuous and differentiable. Therefore, ( f(x) = |x - 3| ) satisfies the hypotheses of the Mean Value Theorem on the interval ([0, 6]).

Now, to find ( c ), we use the formula of the Mean Value Theorem:

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

where ( a = 0 ) and ( b = 6 ).

( f'(x) = \frac{x - 3}{|x - 3|} ) on ( (0, 6) ), except at ( x = 3 ).

( f(6) - f(0) = |6 - 3| - |0 - 3| = 3 )

[ f'(c) = \frac{3}{6 - 0} = \frac{1}{2} ]

Solve for ( c ) in ( f'(c) = \frac{1}{2} ):

[ \frac{x - 3}{|x - 3|} = \frac{1}{2} ]

[ x - 3 = \pm\frac{1}{2}(x - 3) ]

This equation holds true when ( x = 1 ) or ( x = 5 ).

Therefore, ( c = 1 ) or ( c = 5 ).

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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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