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 - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.