Given the function #f(x) = 1 / (x1)#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [2,5] and find the c?
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To determine if the function ( f(x) = \frac{1}{x1} ) satisfies the hypotheses of the Mean Value Theorem on the interval ([2,5]), we need to check if it is continuous on ([2,5]) and differentiable on ((2,5)).

Continuity: The function ( f(x) = \frac{1}{x1} ) is continuous on ([2,5]) except at (x = 1), as the denominator becomes zero there.

Differentiability: The function ( f(x) = \frac{1}{x1} ) is differentiable on ((2,5)) because it is the quotient of two differentiable functions, (1) and (x1), and the denominator (x1) is not zero on ((2,5)).
Since ( f(x) ) is continuous on ([2,5]) and differentiable on ((2,5)), it satisfies the hypotheses of the Mean Value Theorem.
Now, to find the value of ( c ) guaranteed by the Mean Value Theorem, we use the formula:
[ f'(c) = \frac{f(b)  f(a)}{b  a} ]
where ( a = 2 ) and ( b = 5 ).

Calculate ( f(b) ) and ( f(a) ): [ f(5) = \frac{1}{51} = \frac{1}{4} ] [ f(2) = \frac{1}{21} = 1 ]

Calculate ( f'(c) ): [ f'(c) = \frac{\frac{1}{4}  1}{5  2} = \frac{3/4}{3} = \frac{1}{4} ]
To find ( c ), solve the equation ( f'(c) = \frac{1}{4} ): [ \frac{1}{4} = \frac{1}{(c1)^2} ] [ (c1)^2 = 4 ] [ c  1 = \pm 2i ] [ c = 1 \pm 2i ]
Since ( c ) must be a real number within the interval ((2,5)), the only valid solution is ( c = 1 ). Therefore, ( c = 1 ) is the value guaranteed by the Mean Value Theorem.
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