Question

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

Answer 1

`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]}