Posted by Ty on .
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,
do you mean what you wrote or
x sqrt (4x) ???? 
Precalculus 
Reiny,
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,
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,
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,
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,
:)

Precalculus 
Ty,
Why is there a .5 in your problem? And yes, it's x sqrt(4x)

^.5??? 
Ty,
Is ^.5 supposed to be square root? Never heard it put that way before.

Precalculus 
Damon,
easy way to write square root  to the 1/2 power

Precalculus 
Damon,
especially when doing derivatives
d/dx x^(1/2) = (1/2) x^(1/2)
or
d/dx x^.5 = .5 x^.5 
Precalculus 
Ty,
Thanks!