Understanding Rolle's theorem.

Question

I think this is a very basic question but somehow I am unable to understand the answer.

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Find the number of zeroes of $f(x) = x^3 + x + 1$.

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Answer in the book :

$f^\prime (x) = 3x^2 + 1$ since $3x^2 + 1 \ge 0$ for all $x \in \mathbb{R}$ therefore $f^\prime (x) \ge 1 $ for all $x \in \mathbb{R}$

Therefore by Rolle's theorem $f(x)$ has at most one zero in its domain.

Now $f(-1) = -1$ and $f(0) = 1$ therefore by Intermediate value theorem $f(x)$ has at least one zero in the interval $[-1, 0]$.

Thus $f(x)$ has one zero.

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Rolle's theorem : A function which is continuous $[a,b]$ and differentiable on $(a,b)$ such that $f(a) = f(b) = 0$, then there exist at least a point $c \in [a,b]$ for which $f^\prime(c) = 0$

In the first part of the proof I am unable to understand how Rolle's theorem is applied because we don't know any $a,b$ where $f(a) = f(b) = 0$ .

How can conclude that $f(x)$ has at most one zero ?

Answer 1

Suppose the function has two zeros, then by Rolle, $f'(x)$ must be zero for some real number $c$ strictly between the two roots. This is impossible, since $f'(x) > 0$ for all real $x.$ So, the function has at most one zero.To show it has a zero, note that $f(-1) < 0$ and $f(0)>0,$ and appeal to the intermediate value theorem.

Answer 2

The derivative is always positive, so the function is monotonically increasing, hence it will have one root at most if the minimum value is less than 0.