How do you find the local extrema for #f(x) = 2x^3 - x^2 - 4x +3#?

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Answer 1 · 2016-12-29T12:50:12

#f(x)# has a local maximum for #x=-2/3# and a local minimum for #x=1#

You find the critical values for #f(x)# by identifying where the first derivative equals zero:

#f'(x) = 6x^2-2x-4 = 0#

#x= frac(1+-sqrt(1+24)) 6= frac (1+-5) 6#,

so: #x_1=-2/3# and #x_2=1#

Now, as #f(x)# is a second degree polynomial with leading positive coefficient, we know that:

#f'(x) > 0# for #x in (-oo,x_1)# and #x in (x_2,+oo)#
#f'(x)<0# for # x in (x_1,x_2)#

so #x_1# is a maximum and #x_2# is a minimum.

graph{2x^3-x^2-4x+3 [-10, 10, -5, 5]}