`2x^5 + 7x - 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 66 - Calculus of a Single Variable (10th Edition, Ron Larson).
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gsarora17 | (Level 2) Associate Educator

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`f(x)=2x^5+7x-1`

`f(0)=2*0^5+7*0-1=-1`

`f(1)=2*1^5+7*1-1=8`

So,f(0) is negative and f(1) is positive. Since f(x) is continuous, by the Intermediate value theorem there is a number c between 0 and 1 such that f(c)=0. Thus the given equation has a root.

Assuming contrary to the equation that `2x^5+7x-1=0` , has at least two roots a and b that is f(a)=0 and f(b)=0.

Thus by Rolle's theorem there is a number c between a and b such that f'(c)=0, which is impossible as,

`f'(x)=10x^4+7>0` for any point x in `(-oo,oo)` 

Thus it is a contradiction to our assumption. So the equation has only one real root.

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