Ex 6.2, 18 - Prove that f(x) = x3 - 3x2 + 3x - 100 is increasing

Ex 6.2,18 - Chapter 6 Class 12 Application of Derivatives - Part 2

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Ex 6.2, 18 Prove that the function given by f (š‘„) = š‘„3 – 3š‘„2 + 3š‘„ – 100 is increasing in R. We need to show f(š‘„) is strictly increasing on R i.e. we need to show f’(š’™) > 0 Finding f’(š’™) f’(š‘„)= 3x2 – 6x + 3 – 0 = 3(š‘„2āˆ’2š‘„+1) = 3((š‘„)2+(1)2āˆ’2(š‘„)(1)) = 3(š‘„āˆ’1)2 Since Square of any number is always positive (š‘„āˆ’1)2 > 0 3(š‘„āˆ’1)2>0 f’(š’™) > 0 Hence, f’(š‘„) > 0 for all values of š‘„ ∓ f(š‘„) is strictly increasing on R

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