# How do you determine if rolles theorem can be applied to #f(x) = 2x^2 − 5x + 1# on the interval [0,2] and if so how do you find all the values of c in the interval for which f'(c)=0?

The Rolles theorem says that if:

So:

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To determine if Rolle's Theorem can be applied to ( f(x) = 2x^2 - 5x + 1 ) on the interval ([0,2]), check if the function satisfies the conditions of Rolle's Theorem. The function must be continuous on the closed interval ([0,2]) and differentiable on the open interval ((0,2)), and ( f(0) = f(2) ).

Calculate ( f'(x) ), the derivative of ( f(x) ), and find the critical points in the interval ((0,2)) by setting ( f'(x) = 0 ) and solving for ( x ). Then, evaluate the critical points within the interval ([0,2]) to find the values of ( c ) for which ( f'(c) = 0 ).

First, find the derivative: [ f'(x) = 4x - 5 ]

Setting ( f'(x) ) equal to zero: [ 4x - 5 = 0 ] [ 4x = 5 ] [ x = \frac{5}{4} ]

This critical point falls within the interval ([0,2]), so Rolle's Theorem applies.

Therefore, ( c = \frac{5}{4} ) is the only value in the interval ([0,2]) for which ( f'(c) = 0 ).

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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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