How do you determine whether the function satisfies the hypotheses of the Mean Value Theorem for #f(x)=x^(1/3)# on the interval [5,4]?
It doesn't. But it does satisfy the conclusion. See below.
There are two hypotheses for MVT.
This function does not satisfy the hypotheses of the Mean Value Theorem on this interval.
Bonus material
graph{(yx^(1/3)) (y4^{1/3)((4^(1/3)+5^(1/3))/9)(x4))= 0 [6.06, 5.04, 2.864, 2.685]}
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To determine if the function satisfies the hypotheses of the Mean Value Theorem on the interval [5, 4], you need to check two conditions:
 The function must be continuous on the closed interval [5, 4].
 The function must be differentiable on the open interval (5, 4).
For ( f(x) = x^{1/3} ), it is continuous for all real numbers. Also, it is differentiable for all ( x \neq 0 ). However, at ( x = 0 ), the derivative does not exist since the function has a sharp corner at that point.
Since the interval given is [5, 4], which does not include ( x = 0 ), the function is continuous and differentiable on the interval (5, 4). Therefore, it satisfies the hypotheses of the Mean Value Theorem on the interval [5, 4].
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To determine if the function ( f(x) = x^{1/3} ) satisfies the hypotheses of the Mean Value Theorem on the interval ([5, 4]), follow these steps:

Continuity: Check if ( f(x) ) is continuous on the closed interval ([5, 4]).

Differentiability: Determine if ( f(x) ) is differentiable on the open interval ((5, 4)).

Conclusion: If both continuity and differentiability are satisfied on the given interval, then the function satisfies the hypotheses of the Mean Value Theorem for that interval.
Let's evaluate each step:

Continuity: The function ( f(x) = x^{1/3} ) is continuous for all real numbers, including the interval ([5, 4]). Hence, it is continuous on the closed interval.

Differentiability: To determine differentiability, check if the derivative of ( f(x) ) exists for all ( x ) in the open interval ((5, 4)). The derivative of ( f(x) ) is ( f'(x) = \frac{1}{3}x^{2/3} ). This derivative exists for all ( x \neq 0 ). Since the open interval ((5, 4)) does not contain ( x = 0 ), the function is differentiable on ((5, 4)).

Conclusion: Since the function is both continuous on ([5, 4]) and differentiable on ((5, 4)), it satisfies the hypotheses of the Mean Value Theorem for the interval ([5, 4]).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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