Question

If x is real, find the maximum and minimum values of (x^2-x+1)/(x^2+x+1); also find the corresponding values of x?

Answer 1

Maximum value is `3` at `x=-1` and minimum value is `1/3` when `x=1`

Explanation:

Let `f(x)=(x^2-x+1)/(x^2+x+1)`

Observe that `lim_(x->+-oo)(x^2-x+1)/(x^2+x+1)=lim_(x->+-oo)(1-1/x+1/x^2)/(1+1/x+1/x^2)=1`

Now `(df)/(dx)=((x^2+x+1)(2x-1)-(x^2-x+1)(2x+1))/(x^2+x+1)^2`

= `(2x^3+x^2+x-1-(2x^3-x^2+x+1))/(x^2+x+1)^2`

= `(2x^2-2)/(x^2+x+1)^2`

= `(2(x+1)(x-1))/(x^2+x+1)^2`

Hence, we have extrema at `x=1` and `x=-1`

and `(d^2f)/(dx^2)=(4x(x^2+x+1)^2-2(x^2+x+1)(2x+1)*2(x^2-1))/(x^2+x+1)^4`

= `(4x^5+8x^4+12x^3+8x^2+4x-8x^5-12x^4-4x^3+8x^2+12x-4)/(x^2+x+1)^4`

= `(-4x^5-4x^4+8x^3+16x^2+16x-4)/(x^2+x+1)^4`

At `x=1`, we have `f(x)=1/3` and `(d^2f)/(dx^2)=28/81` and

at `x=-1`, we have `f(x)=3` and `(d^2f)/(dx^2)=-12/81` hence maxima.

graph{(x^2-x+1)/(x^2+x+1) [-5.625, 4.375, -0.92, 4.08]}