Application of Derivatives Part 1

We have P(x)=x4+ax3+bx2+cx+d
⇒P′(x)=4x3+3ax2+2bx+c
But P′(0)=0⇒c=0
∴P(x)=x4+ax3+bx2+d
As given that P(−1)<P(a)
⇒1−a+b+d<1+a+b+d⇒a>0
Now P′(x)=4x3+3ax2+2bx=x(4x2+3ax+2b)
As P′(x)=0, there is only one solution x=0,
therefore 4x2+3ax+2b=0 should not have any real roots i.e. D<0
⇒9a2−32b<0
⇒b>‌

9a2

32

>0
Hence a,b>0⇒P′(x)=4x3+3ax2+2bx>0
∀x>0
∴P(x) is an increasing function on (0,1)
∴P(0)<P(a)
Similarly we can prove P(x) is decreasing on (−1,0)
∴P(−1)>P(0)
So we can conclude that
Max‌P(x)=P(1)‌ and ‌Min‌P(x)=P(0)
⇒P(−1) is not minimum but P(1) is the maximum of P.