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First Derivative Test (Part-2)
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This lesson explains the concept of first derivative test to find maxima and minima

## Vineet Loomba is teaching live on Unacademy Plus

Vineet Loomba
IITian | No. 1 Educator in IIT-JEE | 6 Million Minutes Watch Time | 8+ Years Experience | Youtube: Maths Wallah | vineetloomba.com

U
It shares border with Bhutan as well.... I missed it ... Sry :)
where is part 1
Vineet Loomba
2 months ago
See lesson 15 that is part 1
thanks sir
Vineet Loomba
3 months ago
Wlcm
Sir. Example m (close to 1) m sign jo hh left to right ( Negative se positive) ki traf ja rha hh to ye local minama ho gya... sir ese hm Right to left nhi dekh skte kyu ( Positive to negative ) ???
Vineet Loomba
a year ago
cram nhi krna ...concept ko smjho..jidhar se mrji dekho ..minima toh minima he rhega ...vna k dekho slope toh pta chlega
Raja gupta
a year ago
Sir. ye iska concept aapse lesson 13 ( Maxima minima ) m smjhaya hh kya...bcoz mene Abhi wo lesson nhi dekha hh bcoz mujhe usme difficulty feel ho rhe the...to m phle Graph A6e se smjhuga uske baad wo lesson dekhuga
Raja gupta
a year ago
Ooo!!! Yes sir concept to example se phle he bt mene skep kr diya tha bcoz mujse diffeculty ho rhe the smjne m... Thnq sir.
sir differentiation chapter of class 11and 12 please send
Sir is it possible that both the critical points of a function goes positive to negative only or negative to positive only?
Vineet Loomba
a year ago
its not possible in continuous functions and consecutive critical points
ok Thank you sir😃
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) FIRST DERIVATIVE TEST Working rule (First derivative test) : Calculate 0 and solve for x and say x - a, b, c etc. Put values of x slightly less than a in and values of x slightly greater than a. If changes sign from positive to negative, then maximum at x -a dx dy dx dy dx If changes sign from nagative to positive, then minimum at x - a. In case there is no change of sign, then neither a maximum nor a minimum. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

5. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) FIRST DERIVATIVE TEST point of local maxima point of non differentiability and point of local maxima f(c1)0 point of non differentiability and point of local minima of local minima C3 C Y' MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

6. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) ExAMPLE: Find all points of local maxima and local minima of the function f given by fx) 3x 3 f(x) 32 3 3 (r 1) (x1) (x) =0 at x = 1 and x =-1 Values of x to the right (say 1.1 etc.) to the left (say 0.9 etc.) to the right (say -0.9 etc.) Sign off'(x) 3(r 1) (r 1) >0 <0 <O >0 Close to 1 Close to -1 to the left (say -1.1 etc.) MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

7. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Ex. Find the local maxima or local minima, if any, of the function f(x) = sinx + cosx. 0 < x <-using the first derivative test So. We have, dy y f(x) sinx + cosx = cosx-sinx For a local maximum or a local minimum, we have dy 4 Now, we will see whether x =-is a point of local maximum or a point of local minimum or none of these In the left nbd of x =-, we have x <--> cosx > sinx cosx-sinx > 0 -> 0 In the right nbd of x- we Thus, So, f(x) attains a local maximum at x 2 4 have x > --> cosx < sinx cosx-sinx < 0 < 0 4 changes its sign from positive to negative as x increases through dx 4 + 4 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

8. FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Find the points at which the function f given by fx-(x-2f(x + 1 has i) local maxima Ex. (ii) local minima i) points of inflexion Sol.We have, 2 Now, f,(x)=0=> x=2,-1. Since (x 2(x +1) is always positive. So, sign of f(x) depends upon the sign of (x -2) (7x - 2). The changes in signs of f(x) as x increases through and 2 are shown in fig 2 Clearly, f(x) changes its sign from positive to negative as increases through 2/7 so, xis a point of local maximum. We observe that f(x) changes its sign from negative to positive as x increases through2 So, x 2 is a point of local minimum. There is no change in the sign of f(x) as increases through -1 So, x =-1 is a point of inflexion. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

9. 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)

10. 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)