





Last updated at May 29, 2018 by Teachoo
Transcript
Ex 6.5,12 Find the maximum and minimum values of 𝑥 + sin 2𝑥 on [0, 2π ] Let f𝑥=𝑥 + sin 2𝑥 Step 1: Finding f’𝑥 𝑓’𝑥=𝑑𝑥 + sin 2𝑥𝑑𝑥 =1+2cos2𝑥 Step 2: Putting f’𝑥=0 1 + 2 cos 2𝑥=0 2 cos 2𝑥=−1 cos 2𝑥=−12 cos 2𝑥=cos2𝜋3 General solution for cos 2𝑥 is 2𝑥=2𝑛𝜋±2𝜋3 𝑥=2𝑛𝜋 ± 2𝜋32 𝑥= nπ ± 𝜋3 Putting 𝑛=0 𝑥=0𝜋±𝜋3 =±𝜋3 So, 𝑥=𝜋3,𝜋3 ⇒ 𝑥=𝜋3 only Putting 𝑛=1 𝑥=1𝜋±𝜋3 =𝜋±𝜋3 =3𝜋 + 𝜋3 , 3𝜋 − 𝜋3 =4𝜋3 , 2𝜋3 Putting 𝑛=2 𝑥=2𝜋±𝜋3 𝑥=2𝜋−𝜋3 & 2𝜋+𝜋3 𝑥=6𝜋 − 𝜋3 & 6𝜋 + 𝜋3 𝑥=5𝜋3 & 7𝜋3 So, 𝑥=5𝜋3 only Also, We are given interval 𝑥 ∈0 , 2𝜋 Hence , calculating f𝑥 at 𝑥=0 , 𝜋3 , 2𝜋3 , 4𝜋3 , 5𝜋3 , 2𝜋 Hence, f𝑥 is Maximum at 𝑥=2𝜋 Maximum value of f𝒙=𝟐𝝅 & f𝑥 is Minimum at 𝑥=0 Minimum value of f𝒙=𝟎
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