Question

Locate the absolute extrema of the function f(x) = 4x4-16x-4 on the closed interval (-4,4 absolute max: (-4, 124); no absolute min no absolute max or min no absolute max, absolute min: (-4, 124) absolute max: (14, 124); absolute min: (2, -20) absolute max: (2.-20), absolute min: (-4, 124) Question 3 1 pts The height of an object 1 seconds after it is dropped from a height of 350 meters is so - -4.91* +350. Find the average velocity of the object during the first 3 seconds. O 14.70 m sec 0-14.70 m/sec 0 -9.00 m sec 9.00 m/sec 0 -24.00 m sec

Answer 1

1+'(x) f(x) = 4x2-164-4 f(x)=84-16 but f(x)=0, 8x=16, [8=2] f"(x)=870 .,tve so x=2 is a point of minima 2 4 a f'(x) changes from negative to positive +14) = 4(4):-1610-4 = -4 f(2)=4(2) ² 16/20-4 - - a local mi local minima f(2)=-20 = -20 f(4) = 4(4)2-1664)-Y = 64+64-4 absolute max (-4,124) - 124 absolute min (2,-20) obtion D is correct - --- - - ------ - - -- - s(t) = -4.9 ++350 r(t) = -9.86 Average velocity = $13)-Flo) . 3-0 563) = -4.4 (3)?+350 = 305.9 islo2 = 350 Aug velocity = 3os. 9–350 =-4 99532-350 --444 =14.7 option B -14.7 mls