Find the relative extrema of f (x) = 1+8x - 3x^2

F (x) = 1+8x-3x²

1) we have to calculate the first derived

f' (x) = 8-6x

2) we have to equalizate the first derived to "0", and find out the value of "x".

8-6x=0

-6x=-8

x=-8/-6=4/3

3) we have to calculate the second derived

f'' (X) = - 6<0 ⇒ at x=4/3 we have a maximum.

4) we find the value of "y"

f (4/3) = 1+8 (4/3) - 3 (4/3) ²=1 + (32/3) - (48/9) = 57/9

Therefore:

at (4/3, 57/9) have a maximum