For what values of x is #f(x)= x^2-x + 1/x-1/x^2 # concave or convex?

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Answer 1 · 2017-11-19T18:12:56

#x=-1, 3, (3+-sqrt(21))/2#

Find the second derivative of #f(x)#.
#f'(x)=2x-1-1/x^2+2/x^3#
#f''(x)=2+2/x^3-6/x^4#

Find the critical points of #f''(x)#.
#0=2+2/x^3-6/x^4#
#0=2(1+1/x^3-3/x^4)#
#-1=1/x^3-3/x^4#
#-1=x^3-x^4/3#
#-1=x(x^2-x^3/3)#

#x=-1#

#-1=x^2-x^3/3#
#x^3/3-1=x^2#
#(x^3-3)/3=x^2#
#x^3-3=3x^2#
#x^3-3x^2=3#
#x(x^2-3x)=3#

#x=3#

#x^2-3x=3#
#0=x^2-3x-3#

Use the quadratic formula.

#x=(-b+-sqrt(b^2-4ac))/(2a)=(3+-sqrt(21))/2#

#x=-1, 3, (3+-sqrt(21))/2#