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?

2 Answers
May 5, 2017

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

May 5, 2017

The function is increasing when x in (-oo,-1/sqrt6)uu(1/sqrt6,+oo)
The function is decreasing when x in (-1/sqrt6,0)uu(0,1/sqrt6)
See below

Explanation:

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)tcolor(white)(aaaa)-oocolor(white)(aaaa)-1/sqrt6color(white)(aaaaaaaa)0color(white)(aaaaaa)1/sqrt6color(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]}