Question

What are the absolute extrema of `f(x)=x / (x^2 -6) in[3,7]`?

Answer 1

The absolute extrema can either occur on the boundaries, on local extrema, or undefined points.

Let us find the values of `f(x)` on the boundaries `x=3` and `x=7`. This gives us `f(3)=1` and `f(7)=7/43`.

Then, find the local extrema by the derivative. The derivative of `f(x)=x/(x^2-6)` can be found using the quotient rule: `d/dx(u/v)=((du)/dxv-u(dv)/dx)/v^2` where `u=x` and `v=x^2-6`.

Thus, `f'(x)=-(x^2+6)/(x^2-6)^2`. The local extrema occurs when `f'(x)=0`, but nowhere in `x in [3,7]` is `f'(x)=0`.

Then, find any undefined points. However, for all `x in [3,7]`, `f(x)` is defined.

Therefore, it means that the absolute maximum is `(3,2)` and the absolute minimum is `(7,7/43)`.