Question

please help

Perform a first derivative test on the function f(x)=xw/36 - x2:1-6,6). a. Locate the critical points of the given function b. Use the first derivative test to locate the local maximuri and minimum values. c. Identify the absolute minimum and maximum values of the function on the given interval (when they exist): a. Locate the critical points of the given function. Select the correct choice below and, if necessary, fill in the answer box within your choice. O A. The critical point(s) are located at x = (Use a comma to separate answers as needed.) O B. There are no critical points. b. Locate the local maximum and minimum values. Select the correct choice below and, if necessary, fill in the answer box(es) within your choice O A. There are local maximum(s) at x= and local minimum(s) at x = (Use a comma to separate answers as needed.) OB. There are local minimum(s) at x = There are no local maximums. (Use a comma to separate answers as needed.) O C. There are local maximum(s) at x = . There are no local minimums. (Use a comma to separate answers as needed.) OD. There are no local maximums nor local minimums. c. Identify the absolute minimum and maximum values of the function. Select the correct choice below and, if necessary, fill in the answer Click to select and enter your answer(s). 2

Answer 1

fna ~ 36x² ; [-6, 6] For exitical points, f'(x) = 0 x_1 (0-20) +3t-n2 (1) 236-112 +36-X2 a 36-X 2 - 2n2 -n2 + 36-82 √36-X2 = -x2 + 36 -2 ✓36-X2 -2x²+36 √ 36-82 x2 f'(x)=0 36-2x² =0 36=2x2 √36x2 36 18 x=√4x2 = + 3√2. 81 The critical points located at X = +352-312. f'(x) tttttta- 350 +352 au fox) is decreasing X2-382 het x=-5 f'(-5) = -21-52+36 -50+36 -14 <0 √ 26-6-572 36-25 Hence on (-2, -352) for -35ã <x 2 30 het x=0 f110) -2(0)+ 36 36 1 > 6 W36-0 thence for is increasing on (-312, 32) For n7 +352 het x=+5 f(5)-215)2+36 -14 20 36-25 Hene tox) is (352, 2) decrearing on

Forom the sign chart We conclude that x = -352 Correspond to local minimum and x= +352 correspond to loce maximum. Ophon A There are local maximum at x =+352 and local minimum at X = -32. n f(x) -6 O - 32 -18 +352 +18 +6 From the above table that absolute maximum and absolute maximum can conclude olur at x = +352. Value +18 Absolute minimum Occur at X=-355 and absolute minimum value = -18 option (A) : Absolute maximum value is t18 and the absolute minimum value i - 18.