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Ex 6.5, 5 - Find absolute maximum and min value of - Class 12 - Ex 6.5

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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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] Step 1: Finding f’﷐𝑥﷯ f’﷐𝑥﷯=﷐𝑑﷐﷐𝑥﷮3﷯﷯﷮𝑑𝑥﷯ f’﷐𝑥﷯=3﷐𝑥﷮2﷯ Step 2: Putting f’﷐𝑥﷯=0 3﷐𝑥﷮2﷯=0 ﷐𝑥﷮2﷯=0 𝑥=0 So, 𝑥=0 is critical point Step 3: Since given interval𝑥 ∈ ﷐−2 , 2﷯ Hence calculating f﷐𝑥﷯ at 𝑥=−2 , 0 , 2 Step 4: 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 ] Step 1: Finding f’﷐𝑥﷯ f’﷐𝑥﷯=﷐𝑑﷐𝑠𝑖𝑛𝑥 + 𝑐𝑜𝑠𝑥﷯﷮𝑑𝑥﷯ f’﷐𝑥﷯=﷐cos﷮𝑥 −﷐sin﷮𝑥﷯﷯ Step 2: 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﷯ Step 3: 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 Step 1: Finding f’﷐𝑥﷯ f’﷐𝑥﷯=﷐𝑑﷐4𝑥 − ﷐1﷮2﷯﷐𝑥﷮2﷯﷯﷮𝑑𝑥﷯ = 4 – ﷐1﷮2﷯ ×2𝑥 = 4 – 𝑥 Step 2: Putting f’﷐𝑥﷯=0 4 – 𝑥=0 𝑥=4 ∴ 𝑥=4 is only critical point Step 3: 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 Step 1: Finding f’﷐𝑥﷯ f’﷐𝑥﷯= ﷐𝑑﷐﷐﷐𝑥 − 1﷯﷮2﷯+3﷯﷮𝑑𝑥﷯ = 2﷐𝑥−1﷯ Step 2: Putting f’﷐𝑥﷯=0 2﷐𝑥−1﷯=0 𝑥−1=0 𝑥=1 Step 3: 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

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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