Set f(x)=2x^3-3x^2-1 show that the equation f(x)=0 has a root between 1 and 2

You may evaluate the function at `x = 1` and `x = 2` and
then you need to check if the product `f(1)*f(2)<0` , such
that:

`f(1) = 2*1^3 - 3*1^2 - 1 => f(1) = 2 - 3 - 1
= -2 < 0`

`f(2) = 2*2^3 - 3*2^2 - 1 => f(2)
= 16 - 12 - 1 = 3 > 0`

You may evaluate the product
`f(1)*f(2)` such that:

`f(1)*f(2) = -2*3 = -6 < 0`

Since the given function is continuous, you
may notice that the values of the function at `x = 1` and `x = 2 ` proves that the
function passes through 0 for `x in (1,2)` , hence, the equation `2x^3 - 3x^2 - 1 = 0`
has a root `x in (1,2).`