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Ex 6.3, 1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (i) f (𝑥) = (2𝑥 – 1)^2 + 3 Square of number cant be negative It can be 0 or greater than 0 𝑓(𝑥)=(2𝑥−1)^2+3 Hence, Minimum value of (2𝑥−1)^2 = 0 Minimum value of (2𝑥−1^2 )+3 = 0 + 3 = 3 Also, there is no maximum value of 𝑥 ∴ There is no maximum value of f(x) Ex 6.3, 1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (i) f (𝑥) = (2𝑥 – 1)^2+3Finding f’(x) f(𝑥)=(2𝑥−1)^2+3 f’(𝑥)= 2(2𝑥−1) Putting f’(𝒙)=𝟎 2(2𝑥−1)=0 2𝑥 – 1 = 0 2𝑥 = 1 𝒙 = 𝟏/𝟐 Thus, x = 1/2 is the minima Finding minimum value f(𝑥)=(2𝑥−1)^2+3 Putting 𝑥 = 1/2 f(1/2)=(2 × 1/2−1)^2+3= (1−1)^2+3= 3 ∴ Minimum value = 3 There is no maximum value Ex 6.3, 1 (Method 3) Find the maximum and minimum values, if any, of the following functions given by (i) 𝑓 (𝑥)= (2𝑥 – 1)^2 + 3Double Derivative Test f(𝑥)=(2𝑥−1)^2+3 Finding f’(𝒙) f’(𝑥)=2(2𝑥−1)^(2−1) = 2(2𝑥−1) Putting f’(𝒙)=𝟎 2(2𝑥−1)=0 (2𝑥−1)=0 2𝑥 = 0 + 1 𝒙 = 𝟏/𝟐 Finding f’’(𝒙) f’(𝑥)=2(2𝑥−1) f’(𝑥) = 4𝑥 – 2 f’’(𝑥)= 4 f’’ (𝟏/𝟐) = 4 Since f’’ (𝟏/𝟐) > 0 , 𝑥 = 1/2 is point of local minima Putting 𝑥 = 1/2 , we can calculate minimum value f(𝑥) = (2𝑥−1)^2+3 f(1/2)= (2(1/2)−1)^2+3= (1−1)^2+3= 3 Hence, Minimum value = 3 There is no Maximum value

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.