Daniel B. answered 12/13/20
A retired computer professional to teach math, physics
By differentiating f(x) we get
f'(x) = x6 + 3x5 - 5x4 -15x3 + 4x2 +12x
At this point you could simply evaluate f' at the values
-3.1, -1.5, 0.4, 1.5, 3.1 and compare them.
However, I am pretty sure that your teacher wants you to convert f'(x) to a factored form.
Perhaps because sums are less numerically stable than products?
Did you talk about that?
In any case here is how to convert f'(x) to factored form:
f'(x) = x(x5 + 3x4 - 5x3 -15x2 + 4x +12)
= x(x4(x+3) - 5x2(x+3) + 4(x+3))
= x(x+3)(x4 - 5x2 + 4)
= x(x+3)(x2 - 4)(x2 - 1)
= x(x+3)(x-2)(x+2)(x-1)(x+1)
From the factored form we see that the roots of f'(x) are
-3, -2, -1, 0, 1, -2
If we evaluate f'(x) at -3.1, -1.5, 0.4, 1.5, 3.1
we see that none of the four identities are true.
I am sorry, I do not see what was intended by this exercise,
except for the practice of factoring.