- #1

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I know we need to take the derivative, but from there I am lost.

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- Thread starter helpm3pl3ase
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- #1

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I know we need to take the derivative, but from there I am lost.

- #2

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If [tex] f(x) [/tex] is differentiable in the open interval [tex] (a,b) [/tex] and continuous on the closed interval [tex] [a,b] [/tex], then there is at least one point [tex] c [/tex] in [tex] (a,b) [/tex] such that:

[tex] f'(c) = \frac{f(b)-f(a)}{b-a} [/tex]

Assume that there are two real roots [tex] c_{1} [/tex] and [tex] c_{2} [/tex] where [tex] c_{1} < c_{2} [/tex].

Then [tex] f(c_{1}) = 0 = f(c_{2}) [/tex].

Thus [tex] 4x^{3} + 4 = 0 [/tex]

[tex] f'(c) = \frac{f(b)-f(a)}{b-a} [/tex]

Assume that there are two real roots [tex] c_{1} [/tex] and [tex] c_{2} [/tex] where [tex] c_{1} < c_{2} [/tex].

Then [tex] f(c_{1}) = 0 = f(c_{2}) [/tex].

Thus [tex] 4x^{3} + 4 = 0 [/tex]

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- #3

mathwonk

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- #4

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The function is a polynomial and is differentiable and continuous. Suppose a and b are distinct roots. There exists a c in which a<c<b such that 0 = f(b) - f(a). Since f'(x)= 4x^(2) + 4>0, f(a) != f(b). This is a contradiction; hence, a and b cannot both be roots.

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