How do use the first derivative test to determine the local extrema #f(x) = x^3 - x^2 - x + 3#?

Question

1330 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-26T01:31:57

See the explanation.

#f'=3x^2-2x-1#
#f'=0 <=> 3x^2-2x-1=0#
#3x^2-3x+x-1=0#
#3x(x-1)+x-1=0#
#(x-1)(3x+1)=0#
#x=1 vv x=-1/3#

#a=3 >0# function is concave.

graph{3x^2-2x-1 [-10, 10, -5, 5]}

#AAx in (-oo,-1/3)uu(1,+oo) f'(x)>0# function is increasing

#AAx in (-1/3,1) f'(x)<0# function is decreasing

#f'(x)# changes sign in #x=-1/3# and #f(x)# has a maximim value #f_max=f(-1/3)=86/27#

#f'(x)# changes sign in #x=1# and #f(x)# has a minimum value #f_min=f(1)=2#