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Last updated at Jan. 7, 2020 by Teachoo

Transcript

Example 23 Find the approximate value of f (3.02), where f (x) = 3x2 + 5x + 3.Let x = 3 and โ๐ฅ=0.02 Given f(x) = 3x2 + 5x + 3 fโ(x) = 6x + 5 Now, โณy = fโ(x) โ๐ฅ = (6x + 5) 0.02 Also, โ๐ฆ = f(x + โ๐ฅ) โ f(x) f (x + โ๐ฅ) = f(x) + โ๐ฆ f (3.02) = (3x2 + 5x + 3) + (6x + 5) 0.02 Putting x = 3 f (3.02) = [3(3)^2+5(3)+3]+[6(3)+5]0.02 = (27 + 15 + 3) + (23) 0.02 = 45 + 0.46 = 45.46 Hence, approximate value of f (3.02) is 45.46

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
- Absolute minima/maxima

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.