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March 31, 2015

March 31, 2015

Posted by **Mary** on Sunday, February 19, 2012 at 12:58pm.

g(t) = t sqrt6 − t , t < 13/3

I don't even know how to start out!

- Calculas -
**Reiny**, Sunday, February 19, 2012 at 2:18pmI 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)

any t-intercepts?

t(6-t)^(1/2) = 0

t = 0 or t = 6

look at the graph .....

http://www.wolframalpha.com/input/?i=t%286-t%29%5E%281%2F2%29

ignore the imaginary part of the graph (the red line)

the point (4,4√2) is a maximum point

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