In: Math
Let p(x) be a real polynomial of least degree which has a local maximum at x = 1 and a local minimum at x = 3. If p(1) = 6 and p(3) = 2, then p’(0) is
Let p’(x) = k(x-1)(x-3)
= k(x² – 4x + 3)
p(x) = k(x³/3 – 2x² + 3x) + c (Integrating)
Given p(1) = 6.
=> k(⅓ – 2 + 3) + c = 6
=> 4k/3 + c = 6 …(i)
Also given p(3) = 2
=> k(9 – 18 + 9) + c = 2
=> c = 2
Put c in (i), we get
4k/3 + 2 = 6
=> 4k/3 = 4
=> k = 3
So p’(x) = 3(x-1)(x-3)
p’(0) = 3(0-1)(0-3)
= 9
Value of p’(0) is 9.