## Vineet Loomba is teaching live on Unacademy Plus

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

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!!

APPLICATION OF DERIVATIVES CRITICAL POINTS AND STATIONARY POINTS 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)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) CRITICAL POINTS AND STATIONARY POINTS Critical Points are very vital points in differential calculus. It is important to understand its definition before we further go deep into this chapter Definition We say that x = c is a critical point of the function f (x) iff (c) exists and if either ofthe following are true. f'(c)=0 or f'(c) doesn't exist Stationary point: Definition: We can sayx c is a stationary point off(x) iff(c) exits and f" (c) 0. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) EXAMPLE Determine all the critical points for the function f(x) 6r 33 30r100 Differentiate w.r.t. x to get f'(x) 30 132r - 90x -6x (Sx222r - 15) -6r2 (5x -3) (r5) f' (r) is a polynomial and so will exist everywhere Therefore the only critical points will be those values ofx which make the derivative zero i.e. f'(x) = 0 6r2 (51-3) (x + 5)-0 x=-5, x=0, x=-are critical points off(x). MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration (A) 1 SOLUTION (B) The number of critical points off (x) = x2/3 (2-1) is : (B) 2 (C) 3 (D) 4 f, (x) is not defined when denominator= 0 Differentiate w.r.t.x to get, f' (x) = 10 x2/3-2 r-1/3-2 (5x-1) Now we can say that x = 0 and x =-are critical For critical points, f, (x) = 0 f' (x) is not defined. points as/(x) exists at both x = 0 and x = or Put f' (x)-0 to get > Critical points off(x) arex=0, x=-. Note : As f,|-| = 0, fl-l exists, x-is the stationary point off(x) MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration r+1 The number of critical points of f (x)= IS (A) 1 SOLUTION : (B) (B) 2 (C) 3 (D) 4 Differentiate w.r.tx to get As,f(x) exists for x =-7 5V2 these are critical points off(x) Case II f is not defined f' (x) is not defined where 0 f' (x)= 6) (i) -x2-14x+1 r-x- 6) For critical points, f' (x) 0 or f(x is not defined. Case '()0 x=3 and x=-2 At x = 3 and x-2,f(x) is NOT defined. Hence x = 3, x =-2 are NOT critical points. (ii) Combing (i) and (ii), x-7 5V2 are the only critical points off(x) =-7 5.5 MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKee)

FOR 100% SUCCESS IN JEE MAIN AND ADVANCED (IIT-JEE) Illustration The function: f(r)-xe (A) No critical point (C) No stationary point SOLUTION: (AC) has: (B) Only one critical point (D) Only one stationary point Differetiate w.rt.x to get: f' (x) = er" + xex" (2x)-ex" (1 + 2x2) This function will never be zero for any real value ofx. The exponential is also never zero for any real value of x. Therefore, this function will not have any critical points or stationary points. MATHEMATICS FOR IIT-JEE ER. VINEET LOOMBA (IIT RooRKEE)

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!!