Example 30 - Chapter 6 Class 12 Application of Derivatives (Term 1)
Last updated at May 6, 2021 by
Transcript
Example 30 (Method 1) Find all the points of local maxima and local minima of the function f given by f (π₯)=2π₯3 β6π₯2+6π₯+5.f (π₯)=2π₯3 β6π₯2+ 6π₯+5 Finding fβ (π) f β²(π₯)= π(2π₯3 β 6π₯2 + 6π₯ + 5)/ππ₯ f β²(π₯)=6π₯2 β12π₯+6+0 f β²(π₯)=6(π₯^2β2π₯+1) Putting f β²(π)= 0 6(π₯^2β2π₯+1)=0 π₯^2β2π₯+1=0 (π₯)^2+(1)^2β2(π₯)(1)=0 (π₯β1)^2=0 So, π=π is only critical point Hence π=π is point of inflexion Example 30 (Method 2) Find all the points of local maxima and local minima of the function f given by f (π₯)=2π₯3 β6π₯2+6π₯+5. f (π₯)=2π₯3 β6π₯2+ 6π₯+5 Finding fβ (π) f β²(π₯)= π(2π₯3 β 6π₯2+ 6π₯ + 5)/ππ₯ f β²(π₯)=6π₯2 β12π₯+6+0 f β²(π₯)=6(π₯^2β2π₯+1) Putting f β²(π)= 0 6(π₯^2β2π₯+1)=0 π₯^2β2π₯+1=0 π₯^2+1^2β2(π₯)(1)=0 (π₯β1)^2=0 So, π=π is only critical point Finding fββ(π) fββ(π₯)=6 π(π₯^2 β 2π₯ + 1)/ππ₯ fββ(π₯)=6(2π₯β2+0) fββ(π₯)=12(π₯β1) Putting π=π fββ(1)=12(1β1) = 12 Γ 0 = 0 Since fββ(1) = 0 Hence, π₯=1 is neither point of Maxima nor point of Minima β΄ π=π is Point of Inflexion.
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About the Author
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.
