Let f: R (0, ) be a twice differentiable function such that f(3)=18, f^ (3)=0 and f^ (3)=4 . Then _ x 1 ( _ e ( f(2+x) f(3) )^ 18 (x-1)² ) is equal to :

Step-by-step solution

Let T = _ x 1 ( f(x+2) f(3) )^ 18 (x-1)^2 ; 1^ form T = e^ _ x 1 18 (x-1)^2 ( f(x+2) - f(3) f(3) ) T = e^ _ x 1 18 (x-1)^2 ( f(x+2) - f(3) 18 ) T = e^ _ x 1 ( f(x+2) - f(3) (x-1)^2 ) 0 0 form, apply L'pital T = e^ _ x 1 f'(x+2) 2(x-1) ; 0 0 form, apply L'pital T = e^ _ x 1 f''(x+2) 2 = e^ 4 2 = e^2 _e(T) = 2