Question 280076
f(x)=8x-3x^2 i must indicate if it has a maximum or minimum. how do i go about this
0 solutions


We need to find the "critical points" by taking the first derivative of f and setting that equal to zero:

f'(x) = 8 - 6x 
8 - 6x = 0
x = 4/3

So a max or min occurs at x = 4/3.

The second derivative test tells us if we have a max or min:

f''(x) = -6

Since f''(x) < 0 for all x including x = 4/3 we know x = 4/3 is where a maxiumm occurs. The actual value of the maximum is:

f(4/3) = 8*(4/3) - 3*(4/3)^2