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

##### 1 Answer

#### 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

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

We now must find the intervals on which *could* change, which is when the second derivative is equal to

To find the time(s) when the function is equal to

#2x^3+2=0#

#x^3=-1#

#x=-1#

The second derivative is undefined when the denominator is equal to

#x^3=0#

#x=0#

We now have found the two points at which the derivative could change signs,

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)# .

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)# .

Test point of

#x=1# :

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

Since this is

Thus,

We can check a graph of

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