Help in College Level Calculus

a)

f(x) = 3x - 8

since this is a straight line, there are no critical points. Find f(x) at end points (-2, 1)

f(-2) = -6 - 8 = -14

f(1) = 3 - 8 = -5

so maximum is -5 at x = 1 ==> maximum = (-2, -14)

minimum is -14 at x = -2 ===> minimum = (1, -5)

b)

f(x) = x^3 - 4x^2 + 5x + 9 : [-2, 3]

f ' (x) = 3x^2 - 8x + 5

=> (3x - 5)(x - 1) = 0

critical points are 1 and 5/3

determine f(x) at critical points and end points.

f(x) at critical points

f(1) = 1 - 4 + 5 + 9 = 11

f(5/3) = (125/27) - 4(25/9) + 5(5/3) + 9 = 1/27(125 - 300 + 225 + 243) = 293/27 = 10.85

relative minimum = (5/3, 293/27)
relative maximum = (1, 11)

f(x) at end points

f(-2) = -8 - 16 - 10 + 9 = -25

f(3) = 27 - 36 + 15 + 9 = 15

absolute minimum = (-2, -25)
absolute maximum = (3, 15)