Question

The derivative of f(x) = x^4/3 - x^5/5 attains its maximum value at x =

Answer 1

f' = (4/3)x^3 - x^4
f" = 0 for max or min of f'
f" = 4 x^2 - 4 x^3 = 4x^2 (1-x)
min or max at x = 0 or x = 1
next derivative says max or min
f''' = 8 x - 12 x^2
0 at 0 so saddle point
negative at x = 1 so MAXimum

Answer 2

AAAAAAAAAAAAAAAAAAAAA

Answer 3

f(x) = x^4/3 - x^5/5

simply the function
f(x) = 1/3(x^4) - 1/5(x^5)

take the derivative of f(x) using product rule
f' = 1/3(4x^3) - 1/5(5x^4)
f' = (4/3)x^3 - x^4

to find the maximum value of a DERIVATIVE you would have to take the second derivative
f'' = 4/3(3x^2) - 4x^3
f'' = 4x^2 - 4x^3

simplify again
f'' = 4x^2 (1 - x )

solve for x which is x = 1 and x = 0
plug in values of 1 and 0 into the first derivative
f'(1) = 4/3 - 1
f'(1) = 1/3
f'(0) = 0

the maximum value is x = 1
i think