We have x4 – 4x3 + 12x2 + x – 1 = 0
Now, f(– 1) = 1 + 4 + 12 – 1 – 1 = 15 > 0
f(0) = – 1 < 0
and f(1) = 1 – 4 + 12 + 1 – 1 = 9 > 0
So f(x) = 0 has a root in (–1, 0) and a root in (0, 1).
Thus, f(x) = 0 has at least two distinct real roots.
Also, f'(x) = 4x3 – 12x2 + 24x + 1
and f"(x) = 12x2 – 24x + 24
= 12(x2 – 2x + 1) + 1
= 12(x – 1)2 + 1 > 0, " x ∈ R
⇒ f'(x) increases on R.
⇒ y = f'(x) intersects the x-axis exactly once.
⇒ y = f(x) has exactly three real roots.
