Answer:
There is exactly one real root.
Step-by-step explanation:
There are two parts to arrive at the solution.
(i) The polynomial has at least one real root.
(ii) The polynomial has exactly one real root.
We prove (i) using Intermediary Value Theorem.
f(x) = x³ + x - 1 = 0 is a polynomial. So, it is continuous.
At x = 1, f(1) = 1
At x = o, f(0) = -1
Since, there is a change of sign it should have crossed through zero.
Now, to prove there is exactly one real root we use Rolle's theorem.
Let us assume there are two real roots to the polynomial, say 'a' and 'b'.
Then f(a) = 0 and f(b) = 0.
⇒ f(a) = f(b)
To use Rolle's theorem we need the function to be continuous, differentiable and for any two points a,b f(a) = f(b) there should exist a 'c' such that f'(c) = 0.
Now, f'(x) = 3x² + 1
Note that f'(x) is always greater than equal to 1.
It can never be zero for any c. This contradicts Rolle's Theorem. o, our assumption that two real roots exist must be wrong.
Hence, we conclude that there is exactly one real root to the polynomial.
