# How do you determine all values of c that satisfy the mean value theorem on the interval [0,1] for # sqrt (x(1-x))#?

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To find all values of ( c ) that satisfy the Mean Value Theorem (MVT) for the function ( f(x) = \sqrt{x(1-x)} ) on the interval ([0,1]), follow these steps:

- Calculate the derivative of ( f(x) ): ( f'(x) = \frac{1}{2\sqrt{x(1-x)}} - \frac{x - 1}{2\sqrt{x(1-x)}} ).
- Simplify the derivative: ( f'(x) = \frac{1 - 2x}{2\sqrt{x(1-x)}} ).
- Apply the MVT: There exists ( c ) in ((0,1)) such that ( f'(c) = \frac{f(1) - f(0)}{1 - 0} ).
- Find ( f(1) ) and ( f(0) ): ( f(1) = \sqrt{1(1-1)} = 0 ) and ( f(0) = \sqrt{0(1-0)} = 0 ).
- Substitute into the MVT equation: ( f'(c) = \frac{0 - 0}{1 - 0} = 0 ).
- Set the derivative equal to 0 and solve for ( c ): ( \frac{1 - 2c}{2\sqrt{c(1-c)}} = 0 ).
- Solve for ( c ): ( 1 - 2c = 0 \Rightarrow 2c = 1 \Rightarrow c = \frac{1}{2} ).

Therefore, the only value of ( c ) in the interval ([0,1]) that satisfies the Mean Value Theorem for ( f(x) = \sqrt{x(1-x)} ) is ( c = \frac{1}{2} ).

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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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