How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for `f(t) = 6t + 1/t`?

`f (t)= 6t+1/t`
`f'(t)= 6-1/(t^2)`
`f'(t)>0` for `t>1/(sqrt6) or t <-1/(sqrt6)`
`:. f (t)` increases for `t>1/(sqrt6) or t <-1/(sqrt6)`
And `f (t)` decreases for `-1/ sqrt6<``t``<``1/sqrt6`
Now as the derivative changes from positive to negative at `t=-1/sqrt6` therefore it is a point of Maximum
Also as derivative changes from negative to positive at `t= 1/ sqrt6` therefore it is a point of minimum

Let's calculate the first derivative

`f(t)=6t+1/t`

The domain of `f(t)` is `D_(f(t))=RR-{0}`

`f'(t)=6-1/t^2`

The critical points are when `f'(t)=0`, that is

`6-1/t^2=0`

`6=1/t^2`

`t^2=1/6`

`t=+-1/sqrt6`

Let's build a chart

`color(white)(aaaa)``t``color(white)(aaaa)``-oo``color(white)(aaaa)``-1/sqrt6``color(white)(aaaaaaaa)``0``color(white)(aaaaaa)``1/sqrt6``color(white)(aaaa)``+oo`

`color(white)(aaaa)``f'(t)``color(white)(aaaaaa)``+``color(white)(aaaaaa)``-``color(white)(aaaa)``||``color(white)(aaaa)``-``color(white)(aaaa)``+`

`color(white)(aaaa)``f(t)``color(white)(aaaaaa)``↗``color(white)(aaaaaa)``↘``color(white)(aaaa)``||``color(white)(aaaa)``↘``color(white)(aaaa)``↗`

Let's calculate the second derivative

`f''(t)=2/t^3`

There is no point of inflection.

`f''(-1/sqrt6)=-29.39`, as `f''(-1/sqrt6)<0`, this a local maximum at `(-0.408,-4.899)`

`f''(1/sqrt6)=29.39`, as `f''(1/sqrt6)>0`, this a local minimum at `(0.408,4.899)`

graph{6x+1/x [-41.1, 41.08, -20.54, 20.57]}