Question

How do you find the maximum and minimum values of the function `f(x)= x - ((64x)/(x+4))` on the interval [0,13]?

Answer 1

There is only a local minimum at `(12, -36)`

Explanation:

Calculate the first derivative of `f(x)`

The derivative of

`(x)'=1`

And by the quotient rule

`((64x)/(x+4))'=(64(x+4)-1(64x))/(x+4)^2`

`=256/(x+4)^2`

Therefore,

`f'(x)=1-256/(x+4)^2=(((x+4)^2)-64)/(x+4)^2`

`=(x^2+8x+16-256)/(x+4)^2`

`=(x^2+8x-240)/(x+4)^2`

`=((x+20)(x-12))/(x+4)^2`

The critical points are when

`f'(x)=0`

`=>`, `{(x=-20),(x=12):}`

Build a variation chart in the interval `[0,13]`

`color(white)(aaaa)` `"Interval"` `color(white)(aaaa)` `(0,12)` `color(white)(aaaa)` `(12,13)`

`color(white)(aaaa)` `"sign f'(x)"` `color(white)(aaaaa)` `-` `color(white)(aaaaaaa)` `+`

`color(white)(aaaa)` `" f(x)"` `color(white)(aaaaaaaa)` `↘` `color(white)(aaaaaaa)` `↗`

There is only a local minimum at `(12, -36)`

graph{x-((64x)/(x+4)) [-73.8, 113.7, -63, 30.8]}