Find the absolute extreme values of the function on the interval. f(x)=5x2/3, -27 ≤x≤1 OA. absolute maximum is 5 at x 1; absolute minimum is 0 at x=0 B. absolute maximum is 9 at x=-27; absolute minimum is 0 at x=0 C. absolute maximum is 45 at x=-27; absolute minimum is 5 at x=1 D. absolute maximum is 45 at x=-27; absolute minimum is 0 at x=0 Find the value or values of c that satisfy the equation f(b)-f(a) b-a f(c) in the conclusion of the Mean Value Theorem for the function and interval. Round to the nearest thousandth. f(x) = tan x[-√3.√3] OA. 0, 0.523 OB. -0.523, 0, 0.523 OC. 0.523 OD. 10.523 Use the graph of the function f(x) to locate the local extrema and identify the intervals where the function is concave up and concave down. OA. Local minimum at x=2; local maximum at x=0; concave up on (1,00); concave down on (-00,1) OB. Local minimum at x=0; local maximum at x=2; concave down on (1,00); concave up on (-00,1) OC. Local minimum at x=2; local maximum at x=0; concave down on (1,00); concave up on (-00,1) OD. Local minimum at x 0; local maximum at x=2; concave up on (1,00); concave down on (-00,1) Find the largest open interval where the function is changing as requested. Increasing; f(x)=x2-2x+1 OA. (-00,1) OB. (0,00) OC. (0,0) OD. (1,00) Using the following properties of a twice-differentiable function y = f(x), select a possible graph of f. x y Derivatives x<-2 y>0,y" <0. -2 -2<x<0 0 0<x<2 12 y'=0, y' <0 y<0, y' <0 -4 y'<0, y"=0 y<0. y'>0 2 -20 y'=0, y'>0 x>2 y'>0, y'>0 OA OB. OC. A OD. Determine all critical points for the function. f(x)=x³-12x-5 O A. x=2 OB. x=-2, x = 0, and x = 2 OC. x=-2 OD. x= -2 and x = 2