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Example 27 - Chapter 6 Class 12 Application of Derivatives
Last updated at August 2, 2026 by Teachoo
Chapter 6 Class 12 Application of Derivatives
Concept wise
- Finding rate of change
- To show increasing/decreasing in whole domain
- To show increasing/decreasing in intervals
- Find intervals of increasing/decreasing
- Finding slope of tangent/normal
- Finding point when tangent is parallel/ perpendicular
- Finding equation of tangent/normal when point and curve is given
- Finding equation of tangent/normal when slope and curve are given
- Finding approximate value of numbers
- Finding approximate value of function
- Finding approximate value- Statement questions
- Finding minimum and maximum values from graph
- Local maxima and minima
- Minima/ maxima (statement questions) - Number questions
- Minima/ maxima (statement questions) - Geometry questions
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Absolute minima/maxima
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Transcript
Example 27 Find the absolute maximum and minimum values of a function f given by š (š„) = 2š„3 ā 15š„2 + 36š„ +1 on the interval [1, 5]. f(š„) = 2š„3 ā 15š„2 + 36š„ + 1 Finding fā(š) fā(š„)=6š„^2ā30š„+36 Putting f ā(š)=š 6š„2 ā 30š„ + 36 = 0 6(š„^2ā5š„+6)=0 (š„^2ā5š„+6)=0 (š„^2ā2š„ā3š„+6)=0 š„(š„ā2)ā3(š„ā2)=0 (š„ā2)(š„ā3)=0 Hence. š = 2 or š = 3 Since we are given interval [š , š] Hence, calculating f(š„) at š„ = 1, 2, 3, 5 Hence, Absolute maximum value is 56 at š=š Absolute minimum value is 24 at š=š
