Question

What are the extrema of `f(x)=x/(x^2+9)` on the interval [0,5]?

Answer 1

Find the critical values of `f(x)` on the interval `[0,5]`.

`f'(x)=((x^2+9)d/dx[x]-xd/dx[x^2+9])/(x^2+9)^2`

`f'(x)=(x^2+9-2x^2)/(x^2+9)^2`

`f'(x)=-(x^2-9)/(x^2+9)^2`

`f'(x)=0` when `x=+-3`.
`f'(x)` is never undefined.

To find the extrema, plug in the endpoints of the interval and any critical numbers inside the interval into `f(x)`, which, in this case, is only `3`.

`f(0)=0larr"absolute minimum"`

`f(3)=1/6larr"absolute maximum"`

`f(5)=5/36`

Check a graph:

graph{x/(x^2+9) [-0.02, 5, -0.02, 0.2]}