How do you determine all values of c that satisfy the mean value theorem on the interval [0, 2] for #y = x^3 + x  1#?
See below.
The conclusion of the mean value theorem says:
In this case solve
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To determine all values of ( c ) that satisfy the Mean Value Theorem on the interval ([0, 2]) for ( y = x^3 + x  1 ), we first find the average rate of change of the function over the interval. Then, we find the derivative of the function and evaluate it at some ( c ) within the interval.

Average rate of change over ([0, 2]): [ \text{Average rate of change} = \frac{f(2)  f(0)}{2  0} ]

Find the derivative of the function: [ f'(x) = 3x^2 + 1 ]

Set up the Mean Value Theorem equation: [ f'(c) = \frac{f(2)  f(0)}{2  0} ]

Solve for ( c ): [ 3c^2 + 1 = \frac{(2)^3 + 2  1  ((0)^3 + 0  1)}{2} ]
[ 3c^2 + 1 = \frac{8 + 2  1}{2} ]
[ 3c^2 + 1 = \frac{9}{2} ]
[ 3c^2 = \frac{9}{2}  1 ]
[ 3c^2 = \frac{7}{2} ]
[ c^2 = \frac{7}{6} ]
[ c = \pm \sqrt{\frac{7}{6}} ]
So, the values of ( c ) that satisfy the Mean Value Theorem on the interval ([0, 2]) for ( y = x^3 + x  1 ) are ( c = \sqrt{\frac{7}{6}} ) and ( c = \sqrt{\frac{7}{6}} ).
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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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