# By the Mean Value Theorem, we know there exists a c in the open interval (2,4) such that f′(c) is equal to this mean slope, how do you find the value of c in the interval which works for #f(x)=−3x^3−4x^2−3x+3#?

So, you need to solve:

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To find the value of ( c ) in the interval ( (2,4) ) which works for ( f(x) = -3x^3 - 4x^2 - 3x + 3 ) using the Mean Value Theorem, first find the derivative of ( f(x) ), then calculate the mean slope between the endpoints of the interval ( (2,4) ) using the formula:

[ \text{Mean slope} = \frac{f(4) - f(2)}{4 - 2} ]

Next, set ( f'(c) ) equal to the mean slope and solve for ( c ). This gives the value of ( c ) within the interval ( (2,4) ) for which the Mean Value Theorem holds.

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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.

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