Posted by Ty on Monday, November 21, 2011 at 6:00pm.
Use a graphing utility to approximate any relative minimum or maximum values of the function g(x)=x sqrt4x Not getting this one, either. Thanks.

typo?  Damon, Monday, November 21, 2011 at 6:10pm
do you mean what you wrote or
x sqrt (4x) ????

Precalculus  Reiny, Monday, November 21, 2011 at 6:16pm
That's because there is no maximum/minimum
g(x) = √(4x) = (4x)^(1/2)
g '(x) = (1/2)(4x)^(1/2) (1)
= 1/(2√(4x)
This can never be zero

Precalculus  Damon, Monday, November 21, 2011 at 6:18pm
if the latter
y = x (4x)^.5
dy/dx = x (.5)(4x)^.5)(1) + (4x)^.5
when is that zero
0 = .5x /(4x)^.5 + (4x)^.5
0 = .5 x + (4x)
4 = 1.5 x
x = 2.67 for max or min
You should find a max or min around there with your grapher
You can find out with the second derivative but you should see it on your graph anyway
I will find a grapher online
for function
x * sqrt(4x)

Precalculus  Damon, Monday, November 21, 2011 at 6:19pm
Here you go
http://www.coolmath.com/graphit/
put in
x * sqrt(4x)
and hit enter. You will see the max at 2.67

oops  Precalculus  Reiny, Monday, November 21, 2011 at 6:23pm
didn't see that x in front of √
so by the product rule
I got
g'(x) = (1/2)(3x8)(4x)^(1/2)
which when we set that to zero
x = 8/3
your graphing calculator should show a max point near x = 8/3 , (near about x=3)

Precalculus  Damon, Monday, November 21, 2011 at 6:24pm
:)

Precalculus  Ty, Monday, November 21, 2011 at 6:25pm
Why is there a .5 in your problem? And yes, it's x sqrt(4x)

^.5???  Ty, Monday, November 21, 2011 at 6:35pm
Is ^.5 supposed to be square root? Never heard it put that way before.

Precalculus  Damon, Monday, November 21, 2011 at 6:40pm
easy way to write square root  to the 1/2 power

Precalculus  Damon, Monday, November 21, 2011 at 6:41pm
especially when doing derivatives
d/dx x^(1/2) = (1/2) x^(1/2)
or
d/dx x^.5 = .5 x^.5

Precalculus  Ty, Monday, November 21, 2011 at 8:33pm
Thanks!
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