Question

1. Find the absolute maximum and absolute minimum of the function f(x) = x + 2 on the interval [16] 2. For the function f(x) = 3x48x3 +17, find a Intervals of increase, interrels of decrease, and local extrema. b. Intervals of concave upward, intends of concave dowward, and inflection points

Answer 1

Sals: fed = x + 3 Moro f'Q) = 1 - 3 on interval (1,6] f'Q) = 1- 34-2 and f'al = 0-3x-2x ple = 6.x² Kard takin f'a=0 for maxima or minima . 1 = 0 x² – 3 – 0 x2-3 = 0 {.: x (116) => 7? = 3 -) X = +13 = x= 33 or x= -13 € [1,8] Therefore a=55 because she (1.63 } Now f" (53) = 6(13)= > IT Therefore fa) has minima at x=5 Therefore absolute minimum value at n=83 is (53) = 83 + = S3 + 13 If 3) = 253) Auss.

x=0 , x= 4/ in interval (40) 1" (0) is positive it implies that the curve will concave upward. In interval that cure (0, 43) f(x) is negative will concave down wald, it implies it implies In that interval cure (-ooo) f(x) will concave is positive up would. Theufore, T (4/3 ) concave upward. Ang (0, 4) Concave downward. Ang (-o, o Concare up world. Ans.

since sign of f'a does not change as a increases through y o , then it is point of inflection Therefore, [x-o is point of inflection. And.