How do you find points of inflection and determine the intervals of concavity given #y=(3x+2)/(x-2)#?

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Answer 1 · 2017-11-17T07:52:19

There are no points of inflection.
The interval of concavity is #(-oo,2)# and the interval of convexity is #(2,+oo)#

Calculate the first derivative

#y=(3x+2)/(x-2)#

This is a quotient #u/v# and the derivative is

#(u/v)'=(u'v-uv')/(v^2)#

Here,

#u=3x+2#, #=>#, #u'=3#

#v=x-2#, #=>#, #v'=1#

Therefore,

#dy/dx=(3(x-2)-1(3x+2))/(x-2)^2=(3x-6-3x-2)/(x-2)^2#

#=-8/(x-2)^2#

And the second derivative is

#(d^2y)/dx^2=(-8(x-2)^-2)'=-8*-2/(x-2)^3=16/(x-2)^3#

The points of inflection are when #(d^2y)/dx^2=0#

Here, the second derivative is #!=0#

So, there are no points of inflection

Build a sign chart for the concavity

#color(white)(aaaa)##Interval##color(white)(aaaa)##(-oo,2)##color(white)(aaaa)##(2,+oo)#

#color(white)(aaaa)##Sign (d^2y)/dx^2##color(white)(aaaaaaa)##-##color(white)(aaaaaaaa)##+#

#color(white)(aaaaaaa)##y##color(white)(aaaaaaaaaaa)##nn##color(white)(aaaaaaaa)##uu#

The interval of concavity is #(-oo,2)# and the interval of convexity is #(2,+oo)#

graph{(3x+2)/(x-2) [-41.1, 41.08, -20.56, 20.56]}