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

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Answer 1 · 2016-06-16T20:03:25

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)<f(0)#
So, the function is decreasing.