What are the critical values, if any, of # f(x)= (x-2)/(x^2-4)+2x#?

Question

2821 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-07T21:29:10

The critical points are:

#x= -2+- 1/sqrt(2)#

We can simplify the function and remove one discontinuity noting that:

#(x^2-4) = (x-2)(x+2)#

hence:

#f(x) = (x-2)/(x^2-4)+2x = (x-2)/((x-2)(x+2))+2x =1/(x+2) +2x#

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

So we can find the critical points as roots of the equation:

#-1/((x+2)^2)+2 = 0#

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

#(x+2)^2 = 1/2#

#x= -2+- 1/sqrt(2)#

As:

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

we have:

#f''(-2- 1/sqrt(2)) = 2/((-2- 1/sqrt(2)+2)^3)=-4sqrt(2) <0#

#f''(-2+ 1/sqrt(2)) = 2/((-2+ 1/sqrt(2)+2)^3)=4sqrt(2) >0#

so #x=-2- 1/sqrt(2)# is a local maximum and #x=-2+ 1/sqrt(2)# is a local minimum.

graph{(x-2)/(x^2-4)+2x [-23.13, 16.87, -13.6, 6.4]}