Given the function #f(x) = - 4 / x^2#, how do you determine whether f satisfies the hypotheses of the Mean Value Theorem on the interval [1,4] and find the c?

Answer 1

See below.

#f# is continuous at every real number except #0#. #0# is not in #[1,4]#. Therefore, #f# is continuous on #[1,4]#
#f'(x) = 8x^-3# is defined for all #x# other than #0#. #0# is not in #[1,4]#. Therefore, #f# is differentiable on #(1,4)#
To find #c#, solve #f'(x) = (f(4)-f(1))/(4-1)#. The solutions in #(1,4)# are values of #c#.
Solving, we get #x=root(3)(32/5)#
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Answer 2

To determine if ( f(x) = -\frac{4}{x^2} ) satisfies the hypotheses of the Mean Value Theorem on the interval ([1, 4]) and find ( c ), we need to check the following conditions:

  1. ( f(x) ) must be continuous on ([1, 4]).
  2. ( f(x) ) must be differentiable on ( (1, 4) ).

( f(x) ) is continuous and differentiable on ( (1, 4) ) since it's a rational function with no singularities or undefined points in that interval.

To find ( c ), we apply the Mean Value Theorem, which states that if ( f(x) ) satisfies the above conditions, then there exists a ( c ) in ( (1, 4) ) such that:

[ f'(c) = \frac{f(4) - f(1)}{4 - 1} ]

[ f'(c) = \frac{-\frac{4}{4^2} - (-\frac{4}{1^2})}{4 - 1} ]

[ f'(c) = \frac{-\frac{4}{16} + \frac{4}{1}}{3} ]

[ f'(c) = \frac{-\frac{1}{4} + 4}{3} ]

[ f'(c) = \frac{-\frac{1}{4} + \frac{16}{4}}{3} ]

[ f'(c) = \frac{\frac{15}{4}}{3} ]

[ f'(c) = \frac{5}{4} ]

Therefore, ( c = \frac{5}{4} ).

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Answer 3

To determine if the function ( f(x) = \frac{-4}{x^2} ) satisfies the hypotheses of the Mean Value Theorem (MVT) on the interval ([1, 4]), first, check if the function is continuous on the closed interval ([1, 4]) and differentiable on the open interval ((1, 4)).

The function ( f(x) = \frac{-4}{x^2} ) is continuous and differentiable on the interval ((1, 4)). It is also continuous on the closed interval ([1, 4]). Therefore, the function satisfies the hypotheses of the Mean Value Theorem on the interval ([1, 4]).

To find the value ( c ) guaranteed by the Mean Value Theorem, calculate the derivative of ( f(x) ), which is ( f'(x) = \frac{8}{x^3} ). Then, find the average rate of change of ( f(x) ) over the interval ([1, 4]) using the formula:

[ f'(c) = \frac{f(4) - f(1)}{4 - 1} ]

[ f'(c) = \frac{\frac{-4}{4^2} - \frac{-4}{1^2}}{4 - 1} ]

[ f'(c) = \frac{-1 - (-4)}{3} ]

[ f'(c) = \frac{3}{3} ]

[ f'(c) = 1 ]

Now, solve for ( c ):

[ \frac{8}{c^3} = 1 ]

[ 8 = c^3 ]

[ c = \sqrt[3]{8} ]

[ c = 2 ]

Therefore, ( c = 2 ) is the value guaranteed by the Mean Value Theorem on the interval ([1, 4]).

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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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