Ex 6.2,11 - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 6.2, 11 Prove that the function f given by f (๐ฅ) = ๐ฅ^2 โ ๐ฅ + 1 is neither strictly increasing nor strictly decreasing on (โ 1, 1).Given f(๐ฅ) = ๐ฅ2 โ ๐ฅ + 1 Finding fโ(๐) fโ(๐ฅ) = 2๐ฅ โ 1 Putting fโ(๐) = 0 2๐ฅ โ 1 = 0 2๐ฅ = 1 ๐ฅ = 1/2 Since ๐ โ (โ๐ , ๐) So, our number line looks like Hence, f(x) is strictly decreasing for ๐ฅ โ (โ1 , 1/2) & f(x) is strictly increasing for ๐ฅ โ (1/2, 1) Hence, f(๐ฅ) is neither decreasing nor increasing on (โ๐ , ๐). Hence Proved
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Ex 6.2,11 You are here
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