Cubic polynomials give third degree burns to my brain

It is given that $f(x)$ is a polynomial of degree three. Hence, the degree of $f'(x)$ and $f''(x)$ will be $2 \text{ and } 1$ respectively. Given that $f'(x)$ has a maximum at $x=1$ implying $f''(1)=0$ and as $f''(x)$ is linear, therefore
$\Rightarrow f''(x) = k(x-1)$ , where $k$ is some constant. Integrating, we get,
$\displaystyle \Rightarrow f'(x) = k \times (\frac{x^2}{2} - x) + C$ , where $C$ is the constant of integration. To evaluate $C$ , we can use the information that $f(x)$ has a minimum at $x=0$ implying $f'(0) = 0$ . Since, $f'(0) = k(0) + C = C$ , therefore, $C=0$ .
Also, clearly $f'(x) = 0$ for, $x=0, x=2$ . Hence $f(x)$ has extremums at these two values.
Integrating $f'(x)$ , we get,
$\displaystyle f(x) = k \times ( \frac{x^3}{6} - \frac{x^2}{2} ) + C_{1}$ , where $C_{1}$ is constant of integration. Using the given values of $f(3) \text{ and } f(-1)$ , we can get the values of $k \text{ and } C_{1}$ .
Hence, $f(x) = -x^3 + 3x^2 + 5$ . It has a local minimum at $x=0$ and a local maximum at $x=2$ .

$\displaystyle \Rightarrow f(0) = 5 \text{and} f(2) = 9$ . Thus the required distance is the distance between the points $(0,5) \text{and} (2,9)$ which equals $2\sqrt{5} = 4.4721$

By using given conditions we get the function f(x)= -x^3+3x^2+5 For this local min and local max obtained at (0,5) and (2,9) distance between these two points is √20=4.4721