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?
See below.
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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:
- ( f(x) ) must be continuous on ([1, 4]).
- ( 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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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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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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