Is #f(x)=(-2x^3-6x^2-3x+2)/(2x-1)# increasing or decreasing at #x=0#?

Question

Elijah Abram · Accepted Answer

Decreasing

We think about evaluating #f(1)# and #f(0)# If #f(1)-f(0)<0# then it is decreasing if#f(1)-f(0)>0# then it is increasing Let's calculate #f(1)# and #f(0)# #f(1)=(-2(1)^3-6(1)^2-3(1)+2)/(2(1)-1)# #f(1)=(-2-6-3+2)/(2-1)# #f(1)=-9# #f(0)=(-2(0)^3-6(0)^2-3(0)+2)/(2(0)-1)# #f(0)=-2# We have: #f(1)-f(0)=-9-(-2)=-9+2=-7# So,#f(1)-f(0)<0# #f(1)

William Abdalla · Answer

undefined To determine if $ f(x) = \frac{-2x^3 - 6x^2 - 3x + 2}{2x - 1} $ is increasing or decreasing at $ x = 0 $, we can analyze the sign of the derivative of $ f(x) $ evaluated at $ x = 0 $.

The derivative of $ f(x) $ with respect to $ x $ can be found using the quotient rule:

$$ f'(x) = \frac{(2x - 1)(-6x^2 - 12x - 3) - (-2x^3 - 6x^2 - 3x + 2)(2)}{(2x - 1)^2} $$

Evaluating $ f'(0) $, we get:

$$ f'(0) = \frac{(0 - 1)(0 - 0 - 3) - (2)(2)}{(0 - 1)^2} = \frac{3 - 4}{1} = -1 $$

Since $ f'(0) = -1 < 0 $, the function $ f(x) $ is decreasing at $ x = 0 $.