Question
Find the critical numbers of y=x(4-x)^1/2.
The answer is x=8/3 but I got x=7/2.
The answer is x=8/3 but I got x=7/2.
Answers
Steve
y = x√(4-x)
y' = 1√(4-x) + x * 1/2 * 1/√(4-x) * -1
= √(4-x) - x/(2√(4-x))
= (2(4-x) - x)/(2√(4-x))
= (8-3x)/(2√(4-x))
assuming x≠4, we just have to find where the numerator is zero.
x = 8/3
y' = 1√(4-x) + x * 1/2 * 1/√(4-x) * -1
= √(4-x) - x/(2√(4-x))
= (2(4-x) - x)/(2√(4-x))
= (8-3x)/(2√(4-x))
assuming x≠4, we just have to find where the numerator is zero.
x = 8/3