Rolles Theorem, IVT, Mean Value theorem

Question

How to prove function $g ~\colon (0, 2) \to \mathbb{R}$, given by $g(x)=x^3+x-3$ cannot have two zeros.

Deducing that the function has exactly one zero.

Edit: Note that domain is in $\mathbb{R}$, questions is related to real roots.

Answer 1

If a function $f(x)$ is an increasing ($f'(x)>0$)/decreasing $(f'(x)<0)$ in a domain $x \in D$, it can have at most one real zero. So it cannot have two real zeros.

For exactly one zero in a domain $D$: $f'(x)>0$ or $f'(x)<0$ and $f(p)f(q) <0$ for some $p,q \in D.$

Answer 2

Here you can use Descartes' rule of signs.
Let $\ g(x)= x^3+x-3$
Then $\ g(-x)=-x^3-x-3$

The signs in the sequence of coefficients of $\ g(x)$ are $\ +\,+\,-$
There is only one variation of signs and therefore $\ g(x)=0$ has exactly one positive root.

The signs in the sequence of coefficients of $\ g(-x)$ are $\ -\,-\,-$
There is no variation of signs and therefore $\ g(x)=0$ has no negative root.

Thus the number of real roots of $\ g(x)=0$ is $\ 1$.

Answer 3

Now you can try with Sturm's method.

Let $\ g(x)=x^3+x-3$
So $\ g_1(x)=3x^2+1$
Now $$ \require{enclose} \begin{array}{r} x-9 \\[-3pt] 3x^2+1 \enclose{longdiv}{x^3+x-3} \\[-3pt] \underline{\times{3}\phantom{8900000}} \\[-3pt] 3x^3+3x-9 \\[-3pt] \underline{3x^3+x}\phantom{-911}\\[-3pt] 2x-9\end{array} $$
So $\ g_2(x)=-2x+9$
Again $$ \require{enclose} \begin{array}{r} -3x-27 \\[-3pt] -2x+9 \enclose{longdiv}{3x^2+1} \\[-3pt] \underline{\times{2}\phantom{3x^2}} \\[-3pt] 6x^2+2 \\[-3pt] \underline{6x^2-27x}\\[-3pt] 27x+2 \\[-3pt] \underline{\times{2}\phantom{3x^2}} \\[-3pt] 54x+4 \\[-3pt] \underline{54x-243}\\[-3pt] -247 \\[-3pt]\end{array} $$
So $\ g_3(x)=247$
$$ \begin{matrix} &g(x) & g_1(x) & g_2(x) &g_3(x)& \text{changes of sign} \\ -\infty & - &- &+ &+ &1 \\ 0 & - & + &+ &+ &1\\ \infty &+ &+&-&+ &2\\ \end{matrix} $$
So $\ g(x)=0$ has one positive root. Hence $\ g(x)$ cannot have two zeros.