`2x^5 + 7x - 1 = 0` Use the Intermediate Value Theorem and Rolle’s Theorem to prove that the equation has exactly one real solution.
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.