Homework Soursefx 13x3 52x2 4x 6 determine the intervals where f is concav
f(x)= 1/3(x)^3 + 5/2(x)^2 +4x -6
determine the intervals where f is concave up and f is concave down
find the absolute maximum and the absolute minimum value of f on the interval [-6,2]
determine the intervals where f is concave up and f is concave down
find the absolute maximum and the absolute minimum value of f on the interval [-6,2]
Solution
f(-6) = -12
f(2) = 44/3 = 14.67
f\'(x) = x^2 + 5x + 4 = (x+1)(x+4) = 0 -> x = -4 , -1 are critical points
f(-4) = -10/3 = -3.33
f(-1) = -47/6 = -7.83
absolute maximum = 44/3
absolute minimum = -12
f\"(x) = 2x + 5 = 0 -> x = -5/2 = -2.5
x < -2.5 : f\"(x) < 0 -> f(x) is concave down on [-6,-2.5]
x > -2.5 : f\"(x) > 0 -> f(x) is concave up on [-2.5 , 2]
