I will assume you meant
g(t) = t √(6-t) = t(6-t)^(1/2)
g'(t) = t (1/2)(6-t)^(-1/2) (-1) + (6-t)^(1/2) using the product rule
= 3)t-4)/(2√(6-t) )
=0 for any max/min values of the function
t/(2(6-t)^(1/2) = (6-t)^(1/2)
t = 2(6-t)
t = 12 - 2t
3t = 12
t = 4
so g(4) = 4√(6-4) = 4√2
to see if there are any points of inflection
differentiate 3(t-4)/(2√(6-t) ) , 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(t-8)/(4(6-t)^(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.
y-intercepts ? , let t =0
y = 0√6 = 0 , the origin (0,0)
t(6-t)^(1/2) = 0
t = 0 or t = 6
look at the graph .....
ignore the imaginary part of the graph (the red line)
the point (4,4√2) is a maximum point
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