Calculus of a Single Variable, Chapter 3, 3.2, Section 3.2, Problem 66

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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