Is #f(x)=(3x^3+x^2-2x-14)/(x-1)# increasing or decreasing at #x=2#?

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Answer 1 · 2016-10-13T09:50:09

At #x=2#, #f(x)=(3x^3+x^2-2x-14)/(x-1)# is increasing.

For any function, if #(df)/(dx)# is positive, it is increasing, if #(df)/(dx)# is negative, it is decreasing and if if #(df)/(dx)=0#, it is flat.

in the case of #f(x)=(3x^3+x^2-2x-14)/(x-1)#

#(df)/(dx)=((x-1)(9x^2+2x-2)-(3x^3+x^2-2x-14)xx1)/(x-1)^2#

= #(9x^3+2x^2-2x-9x^2-2x+2-3x^3-x^2+2x+14)/(x-1)^2#

= #(6x^3-8x^2-2x+16)/(x-1)^2#

And at #x=2#,

#(df)/(dx)=(6*2^3-8*2^2-2*2+16)/(2-1)^2#

= #(48-32-4+16)/1=28#

Hence, at #x=2#, #(3x^3+x^2-2x-14)/(x-1)# is increasing.