Question

find all points of absolute minima and maxima on the given closed interval.

y = -x^2 + 4x - 3; [0,3]

Answer 1

we can draw the graph first
1) find x-intercept. (put y=0)
put y=0, -x^2 +4x -3=0
x^2 -4x+3=0
(x-1)(x-3)=0
x=1 or 3
2) find y-intercept (put x=0)
put x=0, y=-0+0-3
y=-3
3) Find the vertex of the curve (by completing square)
y=-x^2+ 4x -3=-(x^2 -4x)-3=-(x^2-4x+4)+4-3=-(x-2)^2 +1
so the vertex is (2,1)
4) find the opening of the curve.
as the coefficient of x^2 is -ve, the curve open downward.
Now, we can draw the curve.

For the range given,
[0,3]
for [], it means that the end points are also included.
Can you now find the max and min from the graph?
If yes, give me the answer. I can tell you if it is right or wrong.
If no, tell me what's your concern.