Is #f(x)=(-2x^3+9x^2-5x+6)/(x-2)# increasing or decreasing at #x=0#?

Question

1292 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-24T04:53:41

#f# is #uarr# at #x=0#.

We know that, #f# is #uarr or darr at x=0# according as f'(0) >, or, < 0"#.

So, we need to check #f'(0)#.

Now, #f(x)=(-2x^3+9x^2-5x+6)/(x-2)#

#={ul(-2x^3+4x^2)+ul(5x^2-10x)+ul(5x-10)+16}/(x-2)#

#={-2x^2(x-2)+5x(x-2)+5(x-2)+16}/(x-2)#

#={(cancel((x-2))(-2x^2+5x+5))/cancel((x-2))+16/(x-2)}#

#:. f(x)=-2x^2+5x+5+16/(x-2)................(1)#

Knowing that, #d/dt(1/t)=-1/t^2, "we have, by" (1)#,

#f'(x)=-4x+5-16/(x-2)^2#

#rArr f'(0)=5-16/(-2)^2=5-4=1>0#

Hence, #f# is #uarr# at #x=0#.

Enjoy Maths.!