Ex 6.2, 2 - Show that f(x) = e2x is strictly increasing on R - Ex 6.2

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  1. Chapter 6 Class 12 Application of Derivatives
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Ex 6.2,2 (Method 1) Show that the function given by f (x) = e2x is strictly increasing on R. We need to show that f (š‘„) is strictly increasing on R f(š‘„) = e2š‘„ fā€™ (š‘„) = (e2š‘„)ā€™ fā€™(š‘„) = 2e2š‘„ Case 1 When š’™ = 0 fā€™ (0) = 2e2(0) fā€™ (0) = 2e0 fā€™ (0) = 2 . 1 fā€™ (0) = 2 > 0 āˆ“ fā€™ (š‘„) > 0 when š‘„ = 0 Case 2 When š’™ > 0 fā€™ (š‘„) = 2e2š‘„ fā€™ (š‘„) = 2ļ·1+ļ·2š‘„ļ·Æ+ ļ·ļ·2š‘„ļ·Æ2ļ·®2!ļ·Æ+ ļ·ļ·2š‘„ļ·Æ3ļ·®3!ļ·Æ+ ā€¦ļ·Æ Since š‘„ > 0 So, far every value of š‘„ greater than 0 fā€™ (š‘„) = 2 ļ·1+ļ·2š‘„ļ·Æ+ ļ·ļ·ļ·2š‘„ļ·Æļ·®2ļ·Æļ·®2!ļ·Æ+ ā€¦ļ·Æ > 0 Hence fā€™ (š‘„) > 0 when š‘„>0 ā‡’ Thus, f (x) is strictly increasing for x > 0 Case 3 When š’™ < 0 ā‡’ š‘„ = some negative value Let š‘„ =āˆ’ š‘¦ fā€™(š‘„) = 2ļ·eļ·®āˆ’2yļ·Æ fā€™(š‘„) = ļ·2ļ·®š‘’2š‘¦ļ·Æ fā€™(š‘„) = ļ·2ļ·®ļ·1+ļ·2š‘¦ļ·Æ+ ļ·ļ·2š‘¦ļ·Æ2ļ·®2ļ·Æļ·Æļ·Æ ā‡’ fā€™(š‘„) = ļ·2ļ·®š‘ š‘œš‘šš‘’ ļ·+ļ·Æ š‘žš‘¢š‘Žš‘›š‘”š‘–š‘”š‘¦ļ·Æ >0 āˆ“ f (š‘„) > 0 for š‘„ < 0 ā‡’ Thus fā€™ (š‘„) > 0 for all š‘„ āˆˆ R f is strictly increasing on R Ex 6.2,2 (Method 2) Show that the function given by f (x) = e2x is strictly increasing on R. Let š‘„1 & š‘„2 be any two real number such that š‘„1 < š‘„2 š‘„1 < š‘„2 ļ·š‘’ļ·®2š‘„1ļ·Æ<ļ·š‘’ļ·®2š‘„2ļ·Æ f (š‘„1) < f (š‘„2) Thus, if š‘„1 < š‘„2 then, f (š‘„1) < f (š‘„2) Hence f is strictly increasing on R

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.