Locate the extrema of... ?

f(x) = -x² + x + 3
f'(x) = -2x + 1
f"(x) = -2

When x = 1/2:
f(2) = -(1/2)² + (1/2) + 3 = 13/4
f'(2) = -2(1/2) + 1 = 0
f"(x) = -2 < 0
Hence, locate maximum at x = 1/2, f(1/2) = 13/4

f(0) = -(0)² + (0) + 3 = 3
f(3) = -(3)² + (3) + 3 = -3 < f(0)

On the interval [0, 3]:
Absolute maximum at x = 1/2, and f(1/2) = 13/4
Absolute minimum at x = 3, and f(3) = -3

We need to test critical points (obtained from the first derivative) and the endpoints of the interval [0,3]

For critical points, we set the derivative to zero to find the roots 

f(x) = -x² + x + 3
f'(x) = -2x + 1
0 = -2x + 1
x = 1/2

Insert the x value into the original function

f(1/2) = -(1/2)² + (1/2) + 3
f(1/2) = -0.25 + 0.5 + 3
f(1/2) = 3.25

Next test the endpoints of the interval. They are x-values and we want their respective y-values so we insert them into the original function

f(0) = -0² + 0 + 3
f(0) = 3

f(3) = -3² + 3 + 3
f(3) = -3

By comparing the points, we see that f(3) < f(0) < f(1/2)

So 
f(3) is the minima
f(1/2) is the maxima