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

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Answer 1 · 2015-12-14T12:26:51

The function is decreasing.

If #f'(0)<0#, then #f(x)# is decreasing when #x=0#.
If #f'(0)>0#, then #f(x)# is increasing when #x=0#.

Find #f'(x)# using the quotient rule.

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

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

#f'(x)=(4x^2+4x-8x-8-2x^2-4x+4)/(4(x-2)^2#

#f'(x)=(2x^2-8x-4)/(4(x-2)^2#

#f'(x)=(x^2-4x-2)/(2(x-2)^2#

#f'(0)=(-2)/8=-1/4#

Since #-1/4<0#, #f(x)# is decreasing when #x=0#.