Check sibling questions

Slide77.JPG

Slide78.JPG

Slide79.JPG Slide80.JPG Slide81.JPG

Slide82.JPG Slide83.JPG Slide84.JPG

Β 


Transcript

Ex 6.5, 1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (i) f (π‘₯) = (2π‘₯ – 1)^2 + 3f(π‘₯)=(2π‘₯βˆ’1)^2+3 Hence, Minimum value of (2π‘₯βˆ’1)^2 = 0 Minimum value of (2π‘₯βˆ’1^2 )+3 = 0 + 3 = 3 Square of number cant be negative It can be 0 or greater than 0 Also, there is no maximum value of π‘₯ ∴ There is no maximum value of f(x) Ex 6.5, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (i) f (π‘₯) = (2π‘₯ – 1)^2+3Finding f’(x) f(π‘₯)=(2π‘₯βˆ’1)^2+3 f’(π‘₯)= 2(2π‘₯βˆ’1) Putting f’(𝒙)=𝟎 2(2π‘₯βˆ’1)=0 2π‘₯ – 1 = 0 2π‘₯ = 1 π‘₯ = 1/2 Thus, x = 1/2 is the minima Finding minimum value f(π‘₯)=(2π‘₯βˆ’1)^2+3 Putting π‘₯ = 1/2 f(1/2)=(2 Γ— 1/2βˆ’1)^2+3= (1βˆ’1)^2+3= 3 ∴ Minimum value = 3 There is no maximum value Ex 6.5, 1 (Method 3) Find the maximum and minimum values, if any, of the following functions given by (i) 𝑓 (π‘₯)= (2π‘₯ – 1)^2 + 3Double Derivative Test f(π‘₯)=(2π‘₯βˆ’1)^2+3 Finding f’(𝒙) f’(π‘₯)=2(2π‘₯βˆ’1)^(2βˆ’1) = 2(2π‘₯βˆ’1) Putting f’(𝒙)=𝟎 2(2π‘₯βˆ’1)=0 (2π‘₯βˆ’1)=0 2π‘₯ = 0 + 1 π‘₯ = 1/2 Finding f’’(𝒙) f’(π‘₯)=2(2π‘₯βˆ’1) f’(π‘₯) = 4π‘₯ – 2 f’’(π‘₯)= 4 So, f’’ (1/2) = 4 Since f’’ (1/2) > 0 , π‘₯ = 1/2 is point of local minima Putting π‘₯ = 1/2 , we can calculate minimum value f(π‘₯) = (2π‘₯βˆ’1)^2+3 f(1/2)= (2(1/2)βˆ’1)^2+3= (1βˆ’1)^2+3= 3 Hence, Minimum value = 3 There is no Maximum value

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.