Question

What is the maximum value of the given function?

Answer 1

The answer is `(1)`

Explanation:

I would call this `y = x/(x^2 - 5x + 9)`.

Taking the derivative, we get

`y' = (1(x^2 - 5x + 9) - x(2x -5))/(x^2 -5x + 9)^2`

`y' = (x^2 - 5x + 9 - 2x^2 + 5x)/(x^2 - 5x+ 9)^2`

`y' = (-x^2 + 9)/(x^2 - 5x + 9)`

Maximums and minimums will occur when the derivative equals `0`.

`0 = 9 - x^2`

`0 = (3 + x)(3 -x)`

`0 =-3 or 3`

If we select test points, we see that points in `-3 < x < 3` when evaluated in the derivative give positive results. Thus, `x = 3` is a maximum. When evaluated this would be

`3/(3^2 - 15 + 9) =3/3 = 1`

Hopefully this helps!