Could you please give me the answer? Question 3 of 10 - 4.6 Homewc X+assessments.macmillanlearning.com/sac/4949143#/4949143/2/-1PUpdate :3 of 10 QuestionsO Assignment Score:Give Up?Check Answer4%Resources1 Question40%Question 3 of 1O>21 of o AttemptsFind the transition points.2 Question0%O of o Attemptsy = 4x7 – 56x(Use symbolic notation and fractions where needed. Give your answers in the form of comma separated lists.)3 QuestionO of o Attempts0%4 Question0%O of o Attemptsthe critical point(s) at x =5 Question0%O of oo Attemptsthe point(s) of inflection at x =6 Question0%O of co Attempts7 QuestionO of oo Attempts0%Find the local maximum and local minimum for y = 4x' – 56xº.(Use symbolic notation and fractions where needed. Give your answers in the form of comma separated lists.)8 Question0%O of o Attempts9 Question0%local maximum at x =O of o Attempts10 Question0%O of o Attemptslocal minimum at x =Question Source: Rogawski 4e Calculus Early Transcendentals| Publisher: W.H. Freeman Question 2 of 10 - 4.6 Homewo X+assessments.macmillanlearning.com/sac/4949143#/4949143/1/-1PUpdate :2 of 10 QuestionsO Assignment Score:Give Up?Check Answer4%Resources1 Question40%Question 2 of 10>21 of o AttemptsFind the transition points.2 Question0%O of o Attemptsy = 5x – 75x3(Use symbolic notation and fractions where needed. Give your answers in the form of comma separated lists.)3 Question0%O of o Attempts4 Question0%O of o AttemptsThe critical point(s) at x =5 Question0%O of oo AttemptsThe point(s) of inflection at x =6 Question0%O of co Attempts7 QuestionO of oo Attempts0%Find the local maximum and local minimum for y =5x – 75x3.(Use symbolic notation and fractions where needed. Give your answers in the form of comma separated lists.)8 Question0%O of o Attempts9 Question0%Local maximum at x =O of o Attempts10 Question0%O of o AttemptsLocal minimum at x =Question Source: Rogawski 4e Calculus Early Transcendentals| Publisher: W.H. Freeman

(2) Given- y=5x5-75x3 To Find- The critical points and the points of inflection. Explanation- Rewrite the given expression, y=5x5-75x3 Differentiate the above expression w.r.t. x, we get, dydx=5·5x4-75·3x2dydx =25x4-225x2dydx=25x2x2-9 ......1 For critical points, equate dydx with zero, we get, 25x2x2-9=0 x=0 and x=±3 So, the critical points at x=0,3 and -3. Now, for the points of inflection, differentiate the equation (1) w.r.t. x, we get, d2ydx2=50xx2-9+25x22x =50x3-450x+50x3 =100x3-450x At x=3, d2ydx2=1350>0, function is concave up. At x=-3,d2ydx2=-1350<0 function is concave down.As from the above, the value of double derivative at x=3 is postive, so it is local minimum and at x=-3 it is negative, so it local maximum. Answer- So, the critical points at x=0,3 and -3. The points of inflections are 3 and -3. Local maximum at x=-3. Local minimum at x=3