f(x)=x^3-5x^2+3x+1 find its critical points in the interval [0,1]?

y = F(x) = x³ - 5x² + 3x + 1
y' = 3x² - 10x + 3
y" = 6x - 10

When F'(x) = 0 :
3x² - 10x + 3 = 0
(3x - 1)(x - 3) = 0
x = 1/3 or x = 3

Only x = 1/3 lies on the interval [0, 1]

When x = 1/3 :
y = (1/3)³ - 5(1/3)² + 3(1/3) + 1 = 40/27
y' = 0
y" = 6(1/3) - 10 = -8 < 0

Hence, maximum at (1/3, 40/27)

f(x) = x³ - 5x² + 3x + 1


First, take the derivative:


f ' (x) = 3x² - 10x + 3


Set Derivative to Zero:


0 = 3x² - 10x + 3


This Factors to:


0 = (3x - 1)(x - 3)


Solve:


x = { 1/3 and 3 }


However our interval only wants 0 to 1, so we can exclude the three and only focus on one-third. Plug this x-value into the original function to find the coordinate where this CP takes place:


f(1/3) = (1/3)³ - 5(1/3)² + 3(1/3) + 1


f(1/3) = 1/27 - 5/9 + 1 + 1


f(1/3) = 1/27 - 15/27 + 54/27


f(1/3) = 40/27


[ As an decimal this is ≈ 1.481 ]


So your only relevant CP is located at (1/3 , 40/27 )