Purple Calculus 1

Probably not the intended method, but quick and easy:

$f(x) = x^3 -3x+1$ .

$f(2) = 8 - 6 +1 = 3$ and $f(1) = 1 - 3 +1 = -1$ . So there is a root, say $\alpha$ between 1 and 2. So $\lfloor\alpha\rfloor = 1$ .

$f(1) = 1 - 3 +1 = -1$ and $f(0) = 0 - 0 +1 = 1$ . So there is a root, say $\beta$ between 0 and 1. So $\lfloor\beta\rfloor = 0$ .

$f(-1) = -1 + 3 +1 = 3$ and $f(-2) = -8 + 6 +1 = -1$ . So there is a root, say $\gamma$ between -2 and -1. So $\lfloor\gamma\rfloor = -2$ .

So the answer is $\lfloor\alpha\rfloor + \lfloor\beta\rfloor + \lfloor\gamma\rfloor = 1 + 0 - 2 = -1$ .

To limit the range to look for the three roots, we can first find the local maximum and minimum by differentiating $f(x)$ .

$\begin{aligned} f(x) & = x^3-3x+1 \\ f'(x) & = 3x^2-3 & \small \color{#3D99F6} \text{Equating it to 0} \\ \implies 3x^2 & = 3 \\ x & = \pm 1 & \small \color{#3D99F6} \text{Differentiating again.} \\ f''(x) & = 6x \\ f''(-1) & = -6 < 0 & \color{#3D99F6} f(-1) = 3 \small \text{ is a local maximum} \\ f''(1) & = 6 < 0 & \color{#3D99F6} f(1) = -1 \small \text{ is a local minimum} \end{aligned}$

Therefore, there is a root for $x < -1$ , a root for $-1 < x < 1$ and a root for $x > 1$ .

Since $\begin{cases} f(-2) = -1, & f(-1) = 3 & \implies \lfloor \alpha \rfloor = -2 \\ f(0) = 1, & f(1) = -1 & \implies \lfloor \beta \rfloor = 0 \\ f(1) = -1, & f(2) = 3 & \implies \lfloor \gamma \rfloor = 1 \end{cases}$

Therefore, $\lfloor \alpha \rfloor + \lfloor \beta \rfloor + \lfloor \gamma \rfloor = -2+0+1 = \boxed{-1}$