How do you verify that #y = x^3 + x  1# over [0,2] satisfies the hypotheses of the Mean Value Theorem?
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To verify that ( y = x^3 + x  1 ) over the interval ([0,2]) satisfies the hypotheses of the Mean Value Theorem, we need to check two conditions:

The function ( y = x^3 + x  1 ) must be continuous on the closed interval ([0,2]).

The function ( y = x^3 + x  1 ) must be differentiable on the open interval ((0,2)).

Continuity: To check continuity, evaluate the function at the endpoints of the interval ([0,2]), namely at (x = 0) and (x = 2).
(y(0) = (0)^3 + 0  1 = 1) (y(2) = (2)^3 + 2  1 = 9)
Since the function is a polynomial, it is continuous everywhere, including the interval ([0,2]).

Differentiability: To check differentiability, we need to verify that the derivative of the function exists for all points in the open interval ((0,2)).
(y'(x) = 3x^2 + 1)
The derivative exists for all real numbers, including the interval ((0,2)).
Since both conditions are satisfied, the function ( y = x^3 + x  1 ) over the interval ([0,2]) satisfies the hypotheses of the Mean Value Theorem.
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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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