Find the relative extreme points of the function, if they exist. f(x)=1+3x-x^2?

f ' (x) = 3-2x = 0

x= 3/2

f(3/2) = 1+ 3 * 3/2 - 1.5^2 = 1+ 9/2 -9/4 = 5.5-2.25 = 3.25

rel max = (3/2,3.25)

local extrema are where the velocity or dy/dx (the derivative) is equal to zero...

dy/dx=-2x+3 and to quantify whether they are min or max we need to know the acceleration or d2y/dx2 (second deriviative) is at that point...

dy/dx=0 when 2x=3 or x=1.5

d2y/dx2=-2 so acceleration is negative meaning that when dy/dx=0 it is a maximum...

So the only extreme point is a maximum when x=1.5

f(1.5)=1+4.5-2.25=3.25

That point is thus (1.5, 3.25)