Find the minimum and maximum value of the function f(x)=2sqrtx−3√^3 x on the interval [0,4]

Question

the square root with a ^3 has is a cubed square root

Answer 1

If this is the question:

f(X) = 2 sqrt of X - 3 cube root of X then the derivative is:

2 . 1/2 X ^(-1/2) - 3 . 1/3 X ^ (-2/3) =

1/ X^(1/2) - 1/ X^(2/3) =0 now solve equation and find the roots and the sign chart of the derivative. If the derivative changes the sign from negative to positive then that root is a local min but if from positive to negative then it is a local max (First Derivative Test).

X^(2/3) = X^( 1/2) make both sides to power 6:

X^4 = X^3 so

X^3 (X - 1) = 0 therefore X = 0 or X = 1. Since you need the absolute max and min (not local) in the closed interval between 0 to 4, you need to find the value of the function on 3 points: X = 0 , 1 and 4 then compare the Y values.

X = 0 Y =0

X = 1 Y = -1

X = 4 Y = 4 - 3 cube root of 4.

So the min is -1 and the max is 0.