Is #f(x)=(x^3-2x^2+5x-4)/(x-2)# increasing or decreasing at #x=0#?

Question

1263 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-21T18:07:26

#f'(0)# is negative, therefore #f(x)# is decreasing at #x=0#.

This is the same as asking if #f'(x)# is positive or negative at #x=0#.
We have to first find #f'(x)#, then plug in 0 to get #f'(0)#.

#f(x)=[x^3-2x^2+5x-4]/[x-2]#

#f'(x)={(x-2)(3x^2-4x+5)-(x^3-2x^2+5x-4)(1)}/{(x-2)^2}#

#f'(x)={(3x^3-4x^2+5x-6x^2+8x-10)-(x^3-2x^2+5x-4)}/{(x-2)^2}#

#f'(x)={(3x^3-10x^2+13x-10)-(x^3-2x^2+5x-4)}/{(x-2)^2}#

#f'(x)={2x^3-8x^2+8x-6}/{(x-2)^2}#

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

#f'(0)=[(2)(-3)]/[(-2)^2]#
#=(-6)/(4)#
#=-3/2#

#f'(0)# is negative, therefore #f(x)# is decreasing at #x=0#.