Ex 6.5,12

Transcript

Ex 6.5,12 Find the maximum and minimum values of 𝑥 + sin 2𝑥 on [0, 2π ] Let f﷐𝑥﷯=𝑥 + sin 2𝑥 Step 1: Finding f’﷐𝑥﷯ 𝑓’﷐𝑥﷯=﷐𝑑﷐𝑥 + sin 2𝑥﷯﷮𝑑𝑥﷯ =1+2﷐cos﷮2𝑥﷯ Step 2: Putting f’﷐𝑥﷯=0 1 + 2 cos 2𝑥=0 2 cos 2𝑥=−1 cos 2𝑥=﷐−1﷮2﷯ cos 2𝑥=﷐cos﷮﷐2𝜋﷮3﷯﷯ General solution for cos 2𝑥 is 2𝑥=2𝑛𝜋±﷐2𝜋﷮3﷯ 𝑥=﷐2𝑛𝜋 ± ﷐2𝜋﷮3﷯﷮2﷯ 𝑥= nπ ± ﷐𝜋﷮3﷯ Putting 𝑛=0 𝑥=0﷐𝜋﷯±﷐𝜋﷮3﷯ =±﷐𝜋﷮3﷯ So, 𝑥=﷐𝜋﷮3﷯,﷐𝜋﷮3﷯ ⇒ 𝑥=﷐𝜋﷮3﷯ only Putting 𝑛=1 𝑥=﷐1﷯𝜋±﷐𝜋﷮3﷯ =𝜋±﷐𝜋﷮3﷯ =﷐3𝜋 + 𝜋﷮3﷯ , ﷐3𝜋 − 𝜋﷮3﷯ =﷐4𝜋﷮3﷯ , ﷐2𝜋﷮3﷯ Putting 𝑛=2 𝑥=2﷐𝜋﷯±﷐𝜋﷮3﷯ 𝑥=2𝜋−﷐𝜋﷮3﷯ & 2𝜋+﷐𝜋﷮3﷯ 𝑥=﷐6𝜋 − 𝜋﷮3﷯ & ﷐6𝜋 + 𝜋﷮3﷯ 𝑥=﷐5𝜋﷮3﷯ & ﷐7𝜋﷮3﷯ So, 𝑥=﷐5𝜋﷮3﷯ only Also, We are given interval 𝑥 ∈﷐0 , 2𝜋﷯ Hence , calculating f﷐𝑥﷯ at 𝑥=0 , ﷐𝜋﷮3﷯ , ﷐2𝜋﷮3﷯ , ﷐4𝜋﷮3﷯ , ﷐5𝜋﷮3﷯ , 2𝜋 Hence, f﷐𝑥﷯ is Maximum at 𝑥=2𝜋 Maximum value of f﷐𝒙﷯=𝟐𝝅 & f﷐𝑥﷯ is Minimum at 𝑥=0 Minimum value of f﷐𝒙﷯=𝟎