What are the points of inflection of #f(x)= (x^2 - 8)/(x+3) #?

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Answer 1 · 2017-08-11T07:01:32

There are no points of inflections

We need

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

Here,

#f(x)=(x^2-8)/(x+3)#

#u(x)=x^2-8#, #=>#, #u'(x)=2x#

#v(x)=x+3#, #=>#, #v'(x)=1#

Therefore, the first derivative is

#f'(x)=(2x*(x+3)-(x^2-8)*1)/(x+3)^2#

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

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

And the second derivative is

#u(x)=x^2+6x+8#, #=>#, #u'(x)=2x+6#

#v(x)=(x+3)^2#, #=>#, #v'(x)=2(x+3)#

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

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

#=((2x^2+6x+6x+18)-(2x^2+12x+16))/(x+3)^3#

#=2/(x+3)^3#

The inflection points are when #f''(x)=0#

#2/(x+3)^3=0#, #=>#, #S=O/#

graph{(x^2-8)/(x+3) [-58.53, 58.55, -29.24, 29.23]}