`x^5 + x^3 + x + 1 = 0` Use the Intermediate Value Theorem and Rolle’s Theorem to prove that the equation has exactly one real solution.

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Chapter 3, 3.2 - Problem 65 - Calculus of a Single Variable (10th Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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Consider x=-1 and x=0. For x=-1 f(x)<0 and for x=0 f(x)>0. By Intermediate Value Theorem there is at least one root of f at the interval (-1, 0).

Let's suppose that there are two roots. Then by Rolle's Theorem there is a point where f'(x)=0. But f'(x)=5x^4+3x^2+1 which is always >0. This contradiction proves that there is only one root.

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