Posted by Mary on .
Find any critical numbers of the function. (Enter your answers as a commaseparated list.)
g(t) = t sqrt6 − t , t < 13/3
I don't even know how to start out!

Calculas 
Reiny,
I will assume you meant
g(t) = t √(6t) = t(6t)^(1/2)
g'(t) = t (1/2)(6t)^(1/2) (1) + (6t)^(1/2) using the product rule
= 3)t4)/(2√(6t) )
=0 for any max/min values of the function
t/(2(6t)^(1/2) = (6t)^(1/2)
t = 2(6t)
t = 12  2t
3t = 12
t = 4
so g(4) = 4√(64) = 4√2
to see if there are any points of inflection
differentiate 3(t4)/(2√(6t) ) , set that result equal to zero and solve for t
(hint: there are no points of intersection)
You should get a second derivative of 3(t8)/(4(6t)^(3/2 )
which when you set equal to zero will give an answer of
t = 8
BUT, t=8 is outside the domain of our function, so ... No Solution.
yintercepts ? , let t =0
y = 0√6 = 0 , the origin (0,0)
any tintercepts?
t(6t)^(1/2) = 0
t = 0 or t = 6
look at the graph .....
http://www.wolframalpha.com/input/?i=t%286t%29%5E%281%2F2%29
ignore the imaginary part of the graph (the red line)
the point (4,4√2) is a maximum point