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

1 Answer
Jan 22, 2016

#f(x)# is convex on #(-oo,-1)uu(0,+oo)# and is concave on #(-1,0)#.

Explanation:

Convexity and concavity are determined by the sign of the second derivative:

  • If #f''(a)<0#, then the function #f(x)# is concave at #x=a#.
  • If #f''(a)>0#, then the function #f(x)# is convex at #x=a#.

To find the second derivative, rewrite the final term with a negative exponent. From there, differentiation is a simple application of the power rule.

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

Working with the second derivative will be simpler if we put it into fractional form. To do this, multiply it by #x^3/x^3#.

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

We now must find the intervals on which #f''(x)# is positive and negative. To do this, we must find the times when the sign of the function could change, which is when the second derivative is equal to #0# or does not exist.

To find the time(s) when the function is equal to #0#, set the numerator of #f''(x)# equal to #0#.

#2x^3+2=0#
#x^3=-1#
#x=-1#

The second derivative is undefined when the denominator is equal to #0#.

#x^3=0#
#x=0#

We now have found the two points at which the derivative could change signs, #x=-1# and #x=0#. Thus, to determine the convexity and concavity of the function, we should test each interval where the convexity/concavity could be different. The intervals are #(-oo,-1),(-1,0),# and #(0,+oo)#.

#mathbf((-oo,-1)#

Test point of #x=-3#:

#f''(-3)=(2(-3)^3+2)/(-3)^3=(2(-27)+2)/(-27)=52/27#

Since this is #>0#, we know this entire interval will have a second derivative value #>0#, meaning the function is convex on #(-oo,-1)#.

#mathbf((-1,0)#

Test point of #x=-1/2#:

#f''(-1/2)=(2(-1/2)^3+2)/(-1/2)^3=(-1/4+2)/(-1/8)=-14#

Since this is #<0#, the function is concave on the interval #(-1,0)#.

#mathbf((0,+oo)#

Test point of #x=1#:

#f''(1)=(2(1^3)+2)/1^3=(2+2)/1=4#

Since this is #>0#, the function is convex on the interval #(0,+oo)#.

Thus, #f(x)# is convex on #(-oo,-1)uu(0,+oo)# and is concave on #(-1,0)#.

We can check a graph of #f(x)#:

graph{x^2-x + 1/x [-6, 6, -17, 17]}