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

Question

Evelyn Agard · Accepted Answer

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.

Adam Adkins · Answer

undefined To determine if $ f(x) = \frac{{4x^3 + 2x^2 - 2x - 3}}{{x - 2}} $ is increasing or decreasing at $ x = -1 $, we can use the first derivative test.

First, find the derivative of $ f(x) $:
$$ f'(x) = \frac{{d}}{{dx}}\left( \frac{{4x^3 + 2x^2 - 2x - 3}}{{x - 2}} \right) $$
$$ f'(x) = \frac{{(12x^2 + 4x - 2)(x - 2) - (4x^3 + 2x^2 - 2x - 3)(1)}}{{(x - 2)^2}} $$
$$ f'(x) = \frac{{12x^3 - 4x^2 - 24x^2 + 8x - 2x + 4 - 4x^3 - 2x^2 + 2x + 3}}{{(x - 2)^2}} $$
$$ f'(x) = \frac{{-6x^2 - 16x + 7}}{{(x - 2)^2}} $$

Now, plug in $ x = -1 $ into $ f'(x) $ to determine the sign of the derivative at $ x = -1 $:
$$ f'(-1) = \frac{{-6(-1)^2 - 16(-1) + 7}}{{(-1 - 2)^2}} $$
$$ f'(-1) = \frac{{-6 + 16 + 7}}{{(-3)^2}} $$
$$ f'(-1) = \frac{{17}}{{9}} $$

Since $ f'(-1) > 0 $, $ f(x) $ is increasing at $ x = -1 $.