Is #f(x)=(4x^3+2x^2-2x-3)/(x-2)# increasing or decreasing at #x=-1#?

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Answer 1 · 2018-01-05T14:35:58

Increasing at #x_0=-1#

#f(x)=(4x^3+2x^2-2x-3)/(x-2)#

#D_f=(-oo,2)uu(2,+oo)#

#f'(x)=((4x^3+2x^2-2x-3)'(x-2)-(4x^3+2x^2-2x-3)(x-2)')/(x-2)^2#

#=# #((12x^2+4x-2)(x-2)-(4x^3+2x^2-2x-3))/(x-2)^2# #=#

#(12x^3-24x^2+4x^2-8x-2x+4-4x^3-2x^2+2x+3)/(x-2)^2#

#=# #(8x^3-22x^2-8x+7)/(x-2)^2# #=#

#f'(-1)=(-8-22+8+7)/9=-15/9=-5/3<0#

so #f# is decreasing at #x_0=-1# as seen in the graph.

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