Critical point ,con cavity and graph ?

Question

f(x)=x⁴-6x³

Answer 1

f(x)=x⁴-6x³

1.D(f)=R(f)=R

f'(x)=4x³-18x²

Critical points: f'(x)=0,  x=0, x=9/2.

f''(x)=12x²-36x

f''(x)=0 -->x=0 and x=3 are inflection points.

f''(3)=81>0 --> x=9/2 is minimum. (f(9/2)=-2187/16)

f''(x)>0 for x<0 and x>3. So f(x)in concave in (-∞, 0)∪(0, 3).

f''(x)<0 for 0<x<3, f(x) is convex.