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

Question

Cameron Alexander · Accepted Answer

The function is decreasing.

Calculate the first derivative at the specified point; a positive value indicates an increasing function, and a negative value indicates a decreasing function. Thus, we possess that #f'(x)=\frac{(6x^2+4x+2)(x-3) - (2x^3+2x^2+2x+1)}{(x-3)^2}# #=\frac{4x^3-16x^2-12x-7}{(x-3)^2}# Evaluate #f'(2)#: #=\frac{4*2^3-16*2^2-12*2-7}{(2-3)^2} =4*8-16*4-24-7# #=32-64-24-7=-63# Thus, the function is getting smaller.

Samantha Adkison · Answer

undefined To determine if $ f(x) = \frac{{2x^3 + 2x^2 + 2x + 1}}{{x - 3}} $ is increasing or decreasing at $ x = 2 $, we need to evaluate the sign of its derivative at that point.

First, find the derivative of $ f(x) $ using the quotient rule:

$$ f'(x) = \frac{{(x - 3)(6x^2 + 4x + 2) - (2x^3 + 2x^2 + 2x + 1)(1)}}{{(x - 3)^2}} $$

Next, plug in $ x = 2 $ into $ f'(x) $ to find the sign:

$$ f'(2) = \frac{{(2 - 3)(6(2)^2 + 4(2) + 2) - (2(2)^3 + 2(2)^2 + 2(2) + 1)(1)}}{{(2 - 3)^2}} $$

$$ f'(2) = \frac{{(-1)(24 + 8 + 2) - (16 + 8 + 4 + 1)}}{{(-1)^2}} $$

$$ f'(2) = \frac{{-34 - 29}}{{1}} = -63 $$

Since $ f'(2) $ is negative, $ f(x) $ is decreasing at $ x = 2 $.