Question

find the value of n for which f(x) = (x^2-4)^n ( x^2-x+1) n E N assumes a local minima at x = 2

Answer 1

Given, f(x) = (x^2 - 4)^n ( x^2-x+1) assumes local minima at x = 2.

then, f(2) < f( 2- h) and f(2) < f(2+h)   {h>0 and f(2)=0}

f(2-h) > 0 and f(2+h) > 0 , ∀ h> 0

= (-h)^n (4-h)^n . (h^2-3h+1) > 0

= h^n ( 4+h)^n (h^2 + 5h + 1) > 0

Hence, (-h)^n > 0

Note: (4 - h) > 0, h^2 - 3h + 1 > 0 , 4+h > 0, h^2+5h+1>0, ∀ h > 0.