Asked by Michael
Find the absolute extremes of the function π(π₯)=π₯β(π₯βπ₯^2)(provide answers as exact values).
Answers
Answered by
oobleck
y = xβ(x-x^2)
y' = x(3-4x)/(2β(x-x^2))
since y is defined only on the interval [0,1]
min is at (0,0) and (1,0)
max is at (3/4, 3β3/16)
y' = x(3-4x)/(2β(x-x^2))
since y is defined only on the interval [0,1]
min is at (0,0) and (1,0)
max is at (3/4, 3β3/16)