Ex 6.2, 15 - Let I be any interval disjoint from [–1, 1]. Prove

Ex 6.2,15 - Chapter 6 Class 12 Application of Derivatives - Part 2
Ex 6.2,15 - Chapter 6 Class 12 Application of Derivatives - Part 3

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Ex 6.2, 15 Let I be any interval disjoint from [–1, 1]. Prove that the function f given by š‘“(š‘„) = š‘„ + 1/š‘„ is strictly increasing on I.I is any interval disjoint from [–1, 1] Let I = (āˆ’āˆž, āˆ’šŸ)∪(šŸ, āˆž) Given f(š‘„) = š‘„ + 1/š‘„ We need to show f(š‘„) is strictly increasing on I i.e. we need to show f’(š’™) > 0 for š‘„ ∈ (āˆ’āˆž, āˆ’šŸ)∪(šŸ, āˆž) Finding f’(š’™) f(š‘„) = š‘„ + 1/š‘„ f’(š‘„) = 1 – 1/š‘„2 f’(š‘„) = (š‘„2 āˆ’ 1)/š‘„2 Putting f’(š’™) = 0 (š‘„2 āˆ’ 1)/š‘„2 = 0 š‘„2āˆ’1 = 0 (š‘„āˆ’1)(š‘„+1)=0 So, š’™=šŸ & š’™=āˆ’šŸ Plotting points on number line The point š‘„ = –1 , 1 into three disjoint intervals ∓ f(x) is strictly increasing on (āˆ’āˆž , āˆ’šŸ) & (šŸ , āˆž) Hence proved

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