do you mean what you wrote or
x sqrt (4-x) ????
That's because there is no maximum/minimum
g(x) = √(4-x) = (4-x)^(1/2)
g '(x) = (1/2)(4-x)^(-1/2) (-1)
This can never be zero
if the latter
y = x (4-x)^.5
dy/dx = x (.5)(4-x)^-.5)(-1) + (4-x)^.5
when is that zero
0 = -.5x /(4-x)^.5 + (4-x)^.5
0 = -.5 x + (4-x)
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
x * sqrt(4-x)
Here you go
x * sqrt(4-x)
and hit enter. You will see the max at 2.67
didn't see that x in front of √
so by the product rule
g'(x) = (1/2)(3x-8)(4-x)^(-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)
Why is there a .5 in your problem? And yes, it's x sqrt(4-x)
Is ^.5 supposed to be square root? Never heard it put that way before.
easy way to write square root - to the 1/2 power
especially when doing derivatives
d/dx x^(1/2) = (1/2) x^-(1/2)
d/dx x^.5 = .5 x^-.5
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