Ex 6.5,5 - Chapter 6 Class 12 Application of Derivatives (Term 1)

Transcript

Ex 6.5, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (i) f (𝑥) = 𝑥3, 𝑥∈[– 2, 2]Finding f’(𝒙) f’(𝑥)=𝑑(𝑥^3 )/𝑑𝑥 f’(𝑥)=3𝑥^2 Putting f’(𝒙)=𝟎 3𝑥^2=0 𝑥^2=0 𝑥=0 So, 𝑥=0 is critical point Since given interval is 𝑥 ∈ [−2 , 2] Hence calculating f(𝑥) at 𝑥=−2 , 0 , 2 Absolute Maximum value of f(x) is 8 at 𝒙 = 2 & Absolute Minimum value of f(x) is –8 at 𝒙 = –2 Ex 6.5, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (ii) f (𝑥) = sin⁡𝑥 + cos⁡𝑥 , 𝑥 ∈ [0, 𝜋 ] Finding f’(𝒙) f’(𝑥)=𝑑(𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥)/𝑑𝑥 f’(𝑥)=cos⁡〖𝑥 −sin⁡𝑥 〗 Putting f’(𝒙) cos⁡〖𝑥 −sin⁡𝑥 〗= 0 cos⁡〖𝑥=sin⁡𝑥 〗 1 = sin⁡𝑥/(cos⁡ 𝑥) 1 = tan 𝑥 tan 𝑥 = 1 We know that know tan θ = 1 at θ = 𝜋/4 ∴ 𝑥 = 𝜋/4 Since given interval 𝑥 ∈ [0 , 𝜋] Hence calculating f(𝑥) at 𝑥=0 , 𝜋/4 ,𝜋 Absolute Maximum value of f(𝑥) is √𝟐 at 𝒙 = 𝝅/𝟒 & Absolute Minimum value of f(𝑥) is –1 at 𝒙 = π Ex 6.5, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iii) f(𝑥) = 4𝑥 – 1/2 𝑥2 , 𝑥 ∈ [−2, 9/2] f(𝑥) = 4𝑥 – 1/2 𝑥2 Finding f’(𝒙) f’(𝑥)=𝑑(4𝑥 − 1/2 𝑥^2 )/𝑑𝑥 = 4 – 1/2 ×2𝑥 = 4 – 𝑥 Putting f’(𝒙)=𝟎 4 – 𝑥=0 𝑥=4 ∴ 𝑥=4 is only critical point Since given interval 𝑥 ∈ [−2 , 9/2] Hence , calculating f(𝑥) at 𝑥 = – 2 , 4 , 9/2 Absolute Maximum value of f(x) is 8 at 𝒙 = 4 & Absolute Minimum value of f(x) is –10 at 𝒙 = –2 Ex 6.5, 5 Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: (iv) f (𝑥) = (𝑥 −1)2 + 3, 𝑥 ∈ [−3,1] f (𝑥) = (𝑥 −1)2 + 3 Finding f’(𝒙) f’(x) = 𝑑((𝑥 − 1)^2+3)/𝑑𝑥 = 2(𝑥−1) Putting f’(𝒙)=𝟎 2(𝑥−1)=0 𝑥−1=0 𝑥=1 Since given interval 𝑥 ∈ [−3 , 1] Hence , calculating f(𝑥) at 𝑥 = – 3 , 1 Absolute Minimum value of f(x) is 3 at 𝒙 = 1 & Absolute Maximum value of f(x) is 19 at 𝒙 = – 3