Showing that $f(x)=x^3-3x+1$ has at least two zeros in the interval $[0,2]$

Question

I was given this task by my professor:

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $f(x)=x^3-3x+1$ Show that $f$ has at least two zeros in the interval $I := [0,2]$.

My answer is:

Since $f:\mathbb{R}\rightarrow\mathbb{R}$ and $f$ is an polynomial it follows that f is continuous

$$f(0) = 1; f(0,5)=-0,375$$

by the intermediate value theorem since f(0,5) < 0 < f(0) $\Rightarrow \exists x_1\in[0;0,5]: f(x_1) = 0$ and since $[0;0,5] \subset [0,2]$ it follows that $f$ has at least one zero in the intervall $I:= [0,2]$

$$f(1) = -1; f(2) = 3$$

by the intermediate value theorem since f(1) < 0 < f(2) $\Rightarrow \exists x_2\in[1,2]: f(x_2) = 0$ and since $[1,2] \subset [0,2]$ it follows that $f$ has at least tow zeros in the intervall $I:= [0,2]$

q.e.d

Is this formally and content correct?

Answer 1

Yes, your proof is correct, good work !

Answer 2

It looks correct...but pretty messy. Why not simply:

$$\begin{cases}f(0)=1>0\\{}\\ f(1)=-1<0\end{cases}\implies\,\text{there's a root in}\,(0,1)\;,\begin{cases}f(1)=-1<0\\{}\\f(2)=3>0\end{cases}\implies\,\text{there's a root in}\;(1,2)$$

...and voila !