f(x)=3x4−4x3−12x2+8 Domain: (−∞,∞)
f′(x)=(3x4−4x3−12x2+8)′
=12x3−12x2−24x
=12x(x2−x−2) Find the critical number(s)
f′(x)=0=>12x(x2−x−2)=0
12x(x+1)(x−2)=0 Critical numbers: −1,0,2.
If x<−1,f′(x)<0,f(x) decreases.
If −1<x<0,f′(x)>0,f(x) increases.
If 0<x<2,f′(x)<0,f(x) decreases.
If x>2,f′(x)>0,f(x) increases.
f(−1)=3
f(0)=8
f(2)=−24
The function f(x) has a local maximum with value of 8 at x=0.
The function f(x) has a local minimum with value of 3 at x=−1.
The function f(x) has a local minimum with value of −24 at x=2.
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