Find the absolute maximum and minimum values of f(x) on the interval [0, 4] if f(x) = (x^2) − 3x − 4

Answer 1

take the derivative, 2x-3 set it equal to zero and solve for x. 2x=3, x= 1.5, while y=-6,25

It's an upward opening parabola, so the place where the slope or 1st derivative = 0 is a minimum, local and global or "absolute minimum" there is no absolute maximum. It approaches infinity, positive infinity which is not a real number. But you postulated [0,4}. Just plug in those two values for x in f(x) to get -4 and 16-12-4 =0

end point min is -4 and endpoint max = 0, on that restricted range and domain

the local min is the global, absolute min = -6/25