Question

Consider the following. f(x) = 1 4 x4 + 1 2 x3 − 3x2 + 4 Find f '(x). f '(x) = x3+ 3x2 2​−6x Find f ''(x). f ''(x) = 3x2+3x−6 Find the x-values of the possible points of inflection. (Enter your answers as a comma-separated list.) x = Determine the intervals on which the function is concave up. (Enter your answer using interval notation.) Determine the intervals on which the function is concave down. (Enter your answer using interval notation.) Locate the points of inflection. (If an answer does not exist, enter DNE.)

Answer 1

solution: Given fonction fa)= 1404 + 12x3_327th let us find of'(x) .: f(1) = (14x+ 10 (79' 60' + ( - IW(4) x++ 12 (1) - 3(1) Xt o $60)- 56x+3637-621 to get inflection points +" (a) = 0 t" (0) 56(1)x+ 36 (2)x6=0 3) 1680++ +27620 5 28374 127-1=0 28m + 144-28-170 289(x+ )-2(x+5) 70 :(x+ 2) (29 1-2) 20 1. x+50 of 288-270 not ar 28x=2 x=-1 and it 14 9=-0.5 and 200.07 .. the inflection point are x=-0.5, 0.07 Ş-0.5, 0.07

(+) , ), (+) -0.5 0.07 161) = 26.10'+12(-1)-1 = 28-12-1 = 28-13 = 15 t the function is concave up (-2,-0.5] 28 (0) +12 (0) 1 2 - 1 the function is concane down Eoos, 0.07] f"(0.1) = 28.10.1)) + 12 (0.1)-1 - 0.48 the function • Concave up [0.07, as)