Question

PLEASE ANSWER ALL.

(1 point) Library/Wiley/setAnton_Section_4.4/question 12.pg Find the absolute maximum and minimum values of f(x) = -(x2 – 2) over the interval (-4, 3). absolute maximum is -0.6299 absolute minimum is -7.368 and it occurs at x = and it occurs at x = 1/2 -4 Notes: If there is more than one z value, enter as a comma separated list.

Answer 1

be Salutist :Guliven Few = a (x2 -x)3, 0*0(-1,3] Diferentate equation Ce) with respeolelox we get fetei) = (1 633-] * Fe () 3+ (out) => F'(x)= -22 622-2355 (221) _() For coltical point pot f'(W5o , we get - (22-2543(2x-1) =0 33 1224)=0 > 9xco x=th is the colitical poort minima exists Now we knew that absolute moxina and at end point or at critical pointi : Beet a = -4 in equation weget fl-4) = -(+412_1-49133 = -(15+4143 = pagi - 12933

(4-4)= 420) = -7.3680 is obsoleade mig imam Now . : F(39 = -- [1332–373 :-(9–39273 F13) ==1653 -3.2014 Again Flak) = - (612-) 5-14-513 :7-133324 0.3968 TF1=0.3968 - Home ibido lucte o taxi anum. Absolute maximum is (0.3962 ) and it occuery at [2= 1/2 Absolute minimum is 17:363), and it occurs at 4.1