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For the Function f(x)=2x312x2+2f ( x ) = 2 x ^ { 3 } - 12 x ^ { 2 } + 2

Question 71

Multiple Choice

For the function f(x) =2x312x2+2f ( x ) = 2 x ^ { 3 } - 12 x ^ { 2 } + 2 : (a) Find the critical numbers of f (if any) ;
(b) Find the open intervals where the function is increasing or decreasing; and
(c) Apply the First Derivative Test to identify all relative extrema.
Then use a graphing utility to confirm your results.


A) (a) x=0,4x = 0,4
(b) increasing: (,0) (4,) ( - \infty , 0 ) \cup ( 4 , \infty ) ; decreasing: (0,4) ( 0,4 )
(c) relative max: f(0) =2f ( 0 ) = 2 ; relative min: f(4) =62f ( 4 ) = - 62
B) (a) x=0,4x = 0,4
(b) decreasing: (,0) (4,) ( - \infty , 0 ) \cup ( 4 , \infty ) ; increasing: (0,4) ( 0,4 )
(c) relative min: f(0) =2f ( 0 ) = 2 ; relative max: f(4) =62f ( 4 ) = - 62
C) (a) x=0,1x = 0,1
(b) increasing: (,0) (1,) ( - \infty , 0 ) \cup ( 1 , \infty ) ; decreasing: (0,1) ( 0,1 )
(c) relative max: f(0) =2f ( 0 ) = 2 ; relative min: f(1) =8f ( 1 ) = - 8
D) (a) x=0,1x = 0,1
(b) decreasing: (,0) (1,) ( - \infty , 0 ) \cup ( 1 , \infty ) ; increasing: (0,1) ( 0,1 )
(c) relative min: f(0) =2f ( 0 ) = 2 ; relative max: f(1) =8f ( 1 ) = - 8
E) (a) x=0,1x = 0,1
(b) increasing: (,0) (1,) ( - \infty , 0 ) \cup ( 1 , \infty ) ; decreasing: (0,1) ( 0,1 )
(c) relative max:f(0) =2\max : f ( 0 ) = 2 ; no relative min.

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