Answer:
The critical points of the function
f(x) = 2x(4 - x)³
are x = (1, 4)
Step-by-step explanation:
Given the function
f(x) = 2x(4 - x)³
To find the critical points, we differentiate f(x) to obtain f'(x), and then find the zeros of the resulting quadratic function.
Now, differentiate f(x).
f'(x) = 2(4 - x)³ + 2x(-1)×3(4 - x)²
= 2(4 - x)³ - 6x(4 - x)²
= 2(4 - x)²(4 - x- 3x)
= 2(4 - x)²(4 - 4x)
f'(x) = 8(4 - x)²(1 - x)
Now we set f'(x) = 0 and solve
8(4 - x)²(1 - x) = 0
=> 8(1 - x) = 0
=> 1 - x = 0
=> x = 1
Or
(4 - x)² = 0
x = 4 twice.
So x = (1, 4)
Therefore, the critical points are x = (1, 4)
