Calculus
posted by Elizabeth on .
Find the critical numbers of y=x(4x)^1/2.
The answer is x=8/3 but I got x=7/2.

y = x√(4x)
y' = 1√(4x) + x * 1/2 * 1/√(4x) * 1
= √(4x)  x/(2√(4x))
= (2(4x)  x)/(2√(4x))
= (83x)/(2√(4x))
assuming x≠4, we just have to find where the numerator is zero.
x = 8/3