
Bobosharif S. answered 12/01/21
PhD in Math, MS's in Calulus
f(x)= 3x^4−12x^3,,
f'(x)= 12x^3−36x^2,,
12x^3−36x^2=0
x=0, x=3
From this follows that x=0, x=3 are critical points
f''(x)=36x^2-72x (x=0 and x=2 are inclination points)
f''(x)=f''(2)=0,
f''(x)<0 at (0,2) and f''(x)>0 for x<0 and x>2,
Now can you see where the function is increasing or decreasing and where it is upward/downward?