Ex 6.3, 1 (iii) Class 12 - Find maximum and minimum value [Solutions]

Ex 6.3,1 (Method 1) Find the maximum and minimum values, if any, of the following functions given by (iii) 𝑓 (𝑥) = –(𝑥 – 1)^2+10 f(𝑥)=−(𝑥−1)^2+10 Finding f’ (x) Diff w.r.t 𝑥 f’(𝑥)=𝑑(−(𝑥−1)^2+10)/𝑑𝑥 f’(𝑥) = –2(𝑥−1)(𝑑(𝑥−1)/𝑑𝑥)+0 f’(𝑥) = –2(𝑥−1)(1−0) + 0 f’(𝑥)=−2(𝑥−1) Putting f’(𝒙)=𝟎 –2(𝑥−1)=0 (𝑥−1)=0 𝑥=1 Hence, 𝑥=1 is point of Maxima & No point of Minima Thus, f(𝑥) has maximum value at 𝑥=1 Putting x=1 in f(x) f(𝑥)=−(𝑥−1)^2+10 f(1)=−(1−1)^2+10 = 0 + 10 = 10 Maximum value of f(𝑥) is 10 There is no minimum value of f(𝒙) Ex 6.3,1 (Method 2) Find the maximum and minimum values, if any, of the following functions given by (iii) 𝑓 (𝑥) = – (𝑥 – 1)^2+10 f(𝑥)=−(𝑥−1)^2+10 Finding f’(x) Diff w.r.t 𝑥 f’(𝑥)=𝑑(−(𝑥−1)^2+10)/𝑑𝑥 f’(𝑥) = –2(𝑥−1)(𝑑(𝑥−1)/𝑑𝑥)+0 f’(𝑥) = –2(𝑥−1)(1−0) + 0 f’(𝑥)=−2(𝑥−1) Putting f’(𝒙)=𝟎 –2(𝑥−1)=0 (𝑥−1)=0 𝑥=1 Finding f’’(𝒙) f’(𝑥)=−2(𝑥−1) Again diff w.r.t 𝑥 f’’(𝑥)=𝑑(−2(𝑥 − 1))/𝑑𝑥 =−2 𝑑(𝑥 − 1)/𝑑𝑥 =−2(1−0) =−2 Since f’’(𝑥) < 0 for 𝑥=1 Hence, f(𝑥) has Maximum value at 𝑥=1 Finding maximum value of f(𝒙) f(𝑥)=−(𝑥−1)^2+10 Putting 𝑥=1 f(𝑥) =−(1−1)^2+10 = 0 + 10 = 10 Maximum value of f(𝒙) is 10 There is no minimum value of f(𝒙)

Ex 6.3, 1 (iii) - Chapter 6 Class 12 Application of Derivatives

Davneet Singh

Class 6

Class 7

Class 8

Class 9

Class 10

Class 11

Class 12

Courses

Interesting

Popular