Given the function #f(x)=(x4)^21#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [3,0] and find the c?
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To determine whether ( f(x) = (x  4)^2  1 ) satisfies the hypotheses of the Mean Value Theorem (MVT) on the interval ([3, 4]), you need to check if the function is continuous on the closed interval and differentiable on the open interval.

Continuity: ( f(x) ) is a polynomial function, and polynomial functions are continuous everywhere. Therefore, ( f(x) ) is continuous on the closed interval ([3, 4]).

Differentiability: To check differentiability, compute ( f'(x) ). Then check if ( f'(x) ) is continuous on the open interval ((3, 4)).
[ f(x) = (x  4)^2  1 ]
[ f'(x) = 2(x  4) ]
[ f'(x) = 2x  8 ]
( f'(x) ) is a linear function and is continuous everywhere. So, it is continuous on the open interval ((3, 4)).
Since ( f(x) ) is both continuous on ([3, 4]) and differentiable on ((3, 4)), it satisfies the hypotheses of the Mean Value Theorem on the interval ([3, 4]).
To find the value of ( c ) that satisfies the conclusion of the Mean Value Theorem, use the formula:
[ f'(c) = \frac{f(b)  f(a)}{b  a} ]
Where ( a = 3 ) and ( b = 4 ).
[ f'(c) = \frac{f(4)  f(3)}{4  3} ]
[ f'(c) = \frac{(4  4)^2  1  ((3  4)^2  1)}{1} ]
[ f'(c) = \frac{1  (4)}{1} ]
[ f'(c) = \frac{3}{1} ]
[ f'(c) = 3 ]
Now, solve for ( c ):
[ 2c  8 = 3 ]
[ 2c = 11 ]
[ c = \frac{11}{2} ]
So, ( c = \frac{11}{2} ) satisfies the conclusion of the Mean Value Theorem on the interval ([3, 4]).
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