Question

Нужно найти интервала вогнутости и выпуклости: f(x)=x^4-2x^3+36x^2-x+7

Answer 1

f'=4x^3-6x^2+72x-1

f''=12x^2-12x+72

f''=0

x^2-x-6=0

x1=3

x2=-2

(-2;3) f''<0 кривая выпукла

x<-2 U x>3 f''>0  кривая вогнута

Answer 2

f '[x] = 6x^2-6x-36 = 0 
x^2 -x-6 =0
(x -3)(x +2) =0
x=3, -2
f[3] = 2 * 3^3 -3 * 3^2 -36 * 3 = 2 *27 -27- 108 = -81
f ''[x] = 12 x -6
f ''[3] = 36-6 = 30 > {3, -81} 
f[-2] = 2 * (-2)^3 -3 (-2)^2 +72 = -16-12+72 = 44 
f ''[-2] = 12*-2-6 =-30 <0  {-2,44} 
f''[x] = 12x-6=0 -> x=1/2 
f[1/2] = 2 * (1/2)^3-3 (1/2)^2-18 = 1/4-3/4 -18 = -37/2

{1/2 ,-37/2}