Question

13. [-/3 Points) DETAILS LARCALC11 3.1.031. Find the absolute extrema of the function on the closed interval. g(x) - 2x2 X-6 (-3, 2] minimum (x, y) = (smaller x-value) (larger x-value) minimum (x, y) - maximum (x,y) = Need Help? Medit Talk to a Tutor 14. (-/2 points) DETAILS LARCALC11 3.1.508.XP. Find the absolute extrema of the function on the closed interval. 4 h(s) = [0, 1] 3 minimum (s, h) - maximum (s, h) = Need Help? Watch it Submit Answer 15. (-/2 Points) DETAILS LARCALC11 3.1.033. Find the absolute extrema of the function on the closed interval. y = 2 - It-21 (-9, 4) minimum (t, y) = maximum (t, y) = Need Help? Medit Wacht 16. (-/3 Points DETAILS LARCALC11 3.1.039.MI. Find the absolute extrema of the function on the closed interval. y = 7 cos x, [0, 2x] minimum (x, y) - maximum (x,y) - (smaller x-value) (x, y) - (larger x-value) Need Help? Wascht Masterit Thik to a Tube

Answer 1

Solution: #13. given that glul = 24 x-6 [-32 - first, we find derivative of glu) with respect to n. g'(x) = d dx ) UG 9x? (4-6) of (24) - 242 d (2.6) du (2.6? (2-6) (441 242 4x2 - 241-242 (1612 04-62 = 2* (1-12) (urole (2.62 Now, equate g'cul=0; to find critical Number & (if any). g'(x)=0 21(n-1) (nga 22-24 2(1.2)=0 x=0, 12 an Son we Now, 19 [-312 will neglect this. we will find value of g(x) at W2 -3,0 and 2. -9 -3.6 glolo 91-31 - 26-3) - 18 -a

9(9) = 2(2)? 2-6 -4 tenla minimum minimum (ny) = (wy) = (2-2) (ny) = (oo). maximum 14. h() = 4 i [0,1]. derisertive of ha) in giwn by 4. (8-3)? t(a)= 0 but į ato. for any s. Hence, we will find regla 4 value of h(x) at 4 12 h (o) - 4 0-3 } hl 1) = 4 4 1-3 - 2 minimum chih) = (1-2) maximum (Nahl 10,-4)