You are here
Question 2 - Finding approximate value of function - Chapter 6 Class 12 Application of Derivatives
Last updated at April 16, 2024 by Teachoo
Finding approximate value of function
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
Transcript
Question 2 Find the approximate value of f (2.01), where f (x) = 4x2 + 5x + 2.Let x = 2 and â x = 0.01 Given f(x) = 4x2 + 5x + 2 fâ(x) = 8x + 5 Now, âđŚ = fâ(x) â x = (8x + 5) 0.01 Also, ây = f (x + âx) â f(x) f(x + â x) = f (x) + âđŚ f (2.01) = 4x2 + 5x + 2 + (8x + 5)(0.01) Putting value of x = 2 đ (2.01)=4 (2)^2+ 5(2)+2+(0.01)[8(2)+5] = (16+10+2)+(21) (0.01) = 28+0.21 =28.21 Hence, the approximate value of f (2.01) is 28.21
