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Second Derivative Test
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This lesson explains the second derivative test with examples

Vineet Loomba is teaching live on Unacademy Plus

Vineet Loomba
IITian | No. 1 Educator in IIT-JEE (Maths) | 4 Million Minutes Watch Time | 8+ Years Experience | Youtube: Maths Wallah | vineetloomba.com

U
Unacademy user
where are they posted ? 90% delhi chance hai ?
sir absolute extremum values ke liye agar open interval diya hua h toh hm end points ko bhi check krenge?
Sir if 2nd derivative is 0 then
Sir if on 2nd derivative
sir agar y=x^3 at x=0 ho toh ismein koun sa derivative test lagayenge
When 3rd derivative is a non zero then it is neither maxima or minima and when it is zero then we switch to 4th derivative test so is it says that we cant find the maxima or minima using the 3rd ,5th ,7th ...so on derivative test ?
Vineet Loomba
a year ago
its not like that ... watch it once again nth derivative portion
ok I ll try that . Can you share your email id ,so that I can send some questions which I am not able to solve.
  1. IIT-JEE VIDEO COURSE FOR SURE SHOT SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) IIT-JEE MATHS MADE EASY PREPARED BY: ER. VINEET LOOMBA IITian | IIT-JEE MENtOR Search vineet loomba unacademy" on GOOGLE


  2. REVISION COURSE FOR JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B.Tech. From IIT Roorkee IIT-JEE Mentor Since 2010 Doubts/Feedback in Comment Section * Comment other topics you want to revise. Follow me @ https://unacademy.com/user/vineetloomba to get updates or search me on Google * Share among your peers as SHARING is CARING!!


  3. APPLICATION OF DERIVATIVES MAXIMA-MINIMA IIT-IEE MATHS MADE EASY PREPARED BY: ER. VINEET LOOMBA IITiAN | IIT-JEE MENTOR FoR SURE SHoT SUcCESS IN JEE MAIN AND ADVANCED (IIT-JEE)


  4. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Second Method (second derivative test) Calculate -0 and solve for x. Suppose one root of dy <x d" dx2 = 0 is at x = a then find dx 2 If ve for x-a, then maximum at x-a dx If - +ve for x -a, then minimum at x a MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  5. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) NTH DeRIVATIVE TEST = 0 at x = a, then find dy', 0 at x = a, then neither maximum nor minimum at x = a. f 3 f0 at x-a, then find If > 0 i.e. positive at x a, then y is minimum at x a and 1 < 0 i.e. negative at x = a, then y is maximum at x = a and so on. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


  6. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) sinx+3 cosx is maximum when (1) x = 60 (3) x 30 Ex (2) x 45 (4) x - 0 dy Sol. (3) Let y= sinx + V3 cosx --sinxV3 COSX dy For maximum, or minimum =0 cosx=V3 sinx d2y -tanx- x 30 Since <0 for x 30 3 .. (3) is the correct answer. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


  7. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) 40 Ex The minimum value of the function is 3x4 +8x3-18x2 +60 (2) 3/2 (4) None of these (3) 40/53 Sol.(l) Let y -- 18x*+ 60) dy 1 dx 40 (12x3 24x2- 36x) d2y 1 dx 40 dy and- (36x2+ 48x-36) Now-=0 x3+ 2x2-3x = 0 or x(x-1) (x + 3) = 0 or x=0,1,-3 At x 0, 36 <0 y is maximum at x 0 the given function i.e.-is minimum at x-0 40 2 60 3 .. minimum value of the function- MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


  8. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Two parts of 10 such that the sum of the twice of first with the square of second is minimum, are- Ex (2) 5, 5 (3) 4, 6 Sol. (1) Let two parts bex and (10-x). If y=2x+(10-x)2 dy hen2 - 2 (10 - x) - 2x - 18 Now =0=> x-9 dx Also then - 2> 0. Hence when x 9 value of y is minimum. So required two parts of 10 are 9 and 1. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)


  9. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Example Find the absolute maximum and minimum values of a function fgiven by f(x) = 2x3-15x2 + 36x + 1 on the interval [1,5]. Solution We have f(x) = 2x3-15x2 + 36x + 1 f'(x) =6f-30x + 36 = 6(x-3) (x-2) or Note that f,(x) = 0 gives x = 2 and x = 3. We shall now evaluate the value of f at these points and at the end points of the interval [1, 5], i.e., at x = 1, x = 2, x = 3 and at x = 5. So /(1)=2(1)-15(12) +36 (1)+1=24 f (2) 2(23)- 15(22) +36(2) + 1 29 f(3) 2(33) 15(3) +36(3)+1 28 f (5) 2(53) 15(52)36 (5)1 56 Thus, we conclude that absolute maximum value of fon [1, 5] is 56, occurring at x=5, and absolute minimum value of/on [1, 5] is 24 which occurs at x= 1. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)


  10. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ABOUT ME B.Tech. From IIT Roorkee IIT-JEE Mentor Since 2010 Doubts/Feedback in Comment Section * Comment other topics you want to revise. Follow me @ https://unacademy.com/user/vineetloomba to get updates. * Share among your peers as SHARING is CARING!!