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

1 Answer
Nov 26, 2017

There is no point of inflexion of #f(x)# but there is a minimum point #(0,-8/3)#.

Explanation:

Some basic rules of differentiation are as follows, where #u# and #v# are functions of #x#:

  1. Addition / subtraction rule
    If #y=u+-v#, #dy/dx=d/dx(u)+-d/dx(v)#

  2. Chain rule
    #dy/dx=dy/(du)xx(du)/dx#

  3. Product rule
    If #y=uv#, #dy/dx=u(dv)/dx+v(du)/dx#

  4. Quotient rule
    If #y=u/v#, #dy/dx=(v(du)/dx-u(dv)/dx)/v^2#

Let's get started,

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

First, differentiate #f(x)#,

#f'(x)=([d/dx(x^2-8)] (x^2+3)-[d/dx(x^2+3)] (x^2-8))/(x^2+3)^2# ( Quotient rule )
#color(white)(f'(x))=((2x) (x^2+3)-(2x) (x^2-8))/(x^2+3)^2#
#color(white)(f'(x))=((2x) (cancel(x^2)+3cancel(-x^2)+8))/(x^2+3)^2#
#color(white)(f'(x))=((2x)(11))/(x^2+3)^2#
#color(white)(f'(x))=(22x)/(x^2+3)^2#

Since, it is an stationary point, #d/dx[f(x)]=0#,

#(22x)/(x^2+3)^2=0#
#color(white)(xx/xxx.)22x=0#
#color(white)(xxxxxx)x=0#

Find the #y#-coordinate of the stationary point by substituting #x=0# into #f(x)#,

#y= ((0)^2 - 8)/((0)^2+3)#
#y=-8/3#

Hence, the co-ordinate of the stationary point is #(0,-8/3)#.

To find the nature of the stationary point, differentiate #f'(x)#,

#f''(x)=([d/dx(22x)] (x^2+3)^2-[d/dx(x^2+3)^2] (22x))/(x^2+3)^4# ( Quotient rule )
#color(white)(f''(x))=((22)(x^2+3)^2-(4x)(x^2+3)(22x))/(x^2+3)^4# ( Product rule )
#color(white)(f''(x))=((22)(x^2+3)[(x^2+3)-(4x)(x)])/(x^2+3)^4#
#color(white)(f''(x))=((22)(3-3x))/(x^2+3)^3#
#color(white)(f''(x))=(66-66x^2)/(x^2+3)^3#

Substitute #x=0# into #f''(x)#,

#f''(x)=(66-66(0)^2)/((0)^2+3)^3#
#color(white)(f''(x))=66/27 > 0# ( minimum point )

Hence, the stationary point #(0,-8/3)# is, in fact, a minimum point and not an inflexion point.

Check:
graph{(x^2 - 8)/(x^2+3) [-10, 10, -5, 5]}