Given that P(x) = x4 +ax3 + bx2 + cx + d
Hence, P’(x) = 4x3 + 3ax2 + 2bx + c
Now as P’(0) = 0 so the above equation gives c = 0.
Hence, P’(x) = x (4x2 + 3ax + 2b)
As x = 0 is teh only real root of P’(x) = 0, roots of 4x2 + 3ax + 2b must have imaginary roots.
Hence, 4x
2 + 3ax + 2b > 0
x ∈ R.
Thus P’(x) < 0 for x < 0
and P’(x) > 0 for x > 0
This gives that x = 0 is a point of local minima.
As P(-1) < P(1), we get P(1) is maximum but P(-1) is not minimum of P on [-1, 1].
Thanks & Regards
Latika Leekha
askIITians Faculty
